🧠:Okay, so I need to figure out how many sequences of 5 positive integers (a, b, c, d, e) satisfy the condition that abcde is less than or equal to the sum a + b + c + d + e, and this sum is also less than or equal to 10. Hmm, let's break this down step by step.First, let's understand the constraints here. Each of a, b, c, d, e is a positive integer. That means each of them is at least 1. The sum a + b + c + d + e has to be between abcde and 10. So, abcde ≤ sum ≤ 10.Since all variables are positive integers, the smallest possible value for each is 1. Let's first consider the sum. The sum of five 1s is 5, so the sum can range from 5 up to 10. But at the same time, the product abcde has to be less than or equal to the sum. Let's note that if any of the numbers is greater than 1, the product can become large quickly. For example, if even one of the numbers is 2, the product becomes at least 2, but the sum would be at least 6 (since the others are 1). So, there's a balance here.Let me approach this by first considering the possible sums (from 5 to 10) and for each sum, determine the number of sequences where the product is less than or equal to the sum.But maybe another approach is to note that since the product is at least 1 (since all are positive integers), and the sum is at least 5. So, the constraints are abcde ≤ S ≤10 where S = a + b + c + d + e. But since S is between 5 and 10, and abcde ≤ S, which in turn is ≤10, so actually abcde ≤10. So perhaps first we can find all sequences where abcde ≤10, and then check if their sum is between abcde and 10. Wait, but the original condition is abcde ≤ S ≤10, so actually S must be at least abcde and at most 10. Therefore, the sequences must satisfy abcde ≤ S ≤10. So both conditions: S ≥ abcde and S ≤10.Therefore, the problem reduces to finding all 5-tuples of positive integers where abcde ≤ S ≤10, with S being the sum.Given that S is between 5 and 10, and abcde ≤ S, so the product can be at most 10. But since S is at most 10, and abcde is at most S, which is at most 10. Therefore, abcde ≤10. Therefore, the product must be between 1 and 10.So first, let's consider all 5-tuples of positive integers where the product is ≤10, and then among those, check if the sum is between the product and 10. Wait, but actually, the sum must be between the product and 10. So for each 5-tuple with product P ≤10, we need to have P ≤ S ≤10. So, for each such 5-tuple, compute S and check if it's between P and 10.Alternatively, perhaps we can iterate through all possible sums S from 5 to 10, and for each sum, find the number of 5-tuples with that sum and product ≤ S. Then sum these up.This seems manageable since S ranges from 5 to 10. Let's try this approach.So, we need to compute for each S in {5,6,7,8,9,10}, the number of sequences (a,b,c,d,e) where a+b+c+d+e = S and a*b*c*d*e ≤ S.But even with S up to 10, enumerating all possible 5-tuples for each S and checking the product might be feasible, given the small numbers.Alternatively, maybe we can find for each S, the number of 5-tuples where all numbers are 1 except a few, and compute the product accordingly. Let's see.Let's start with S=5. The only possible 5-tuple is (1,1,1,1,1). The product is 1, and the sum is 5. Since 1 ≤5 ≤10, this is valid. So count =1.Next, S=6. Now, the possible 5-tuples are those with four 1s and one 2. The number of such sequences is C(5,1)=5. For each of these, the product is 2. Since 2 ≤6 ≤10, so these are all valid. So count for S=6 is 5.Are there any other 5-tuples for S=6? If we have three 1s and two 2s, the sum would be 3*1 + 2*2 = 7, which is more than 6. So no. Similarly, higher numbers would make the sum even larger. Therefore, only the five sequences with one 2 and four 1s. So 5.Moving on to S=7. Let's find all 5-tuples of positive integers that sum to 7.Possible partitions:1. Five 1s and one 2: Wait, no. For sum 7, possible compositions (since order matters) are:Each composition of 7 into 5 positive integers. The possible types are:- One 3 and four 1s. The number of such sequences is C(5,1)=5. The product is 3*1*1*1*1=3. Since 3 ≤7 ≤10, so these are valid.- One 2, one 2, and three 1s. Wait, two 2s and three 1s. The sum would be 2+2+1+1+1=7. The number of sequences is C(5,2)=10. The product is 2*2*1*1*1=4. Since 4 ≤7, so these are valid.- Are there any others? For example, a 4 and the rest 1s: 4+1+1+1+1=8, which is over 7. So no. Similarly, three 2s would be 2+2+2+1+0, but 0 is not allowed. So the only possibilities are:Either one 3 and four 1s (5 sequences) or two 2s and three 1s (10 sequences). So total 15 sequences for S=7.Now, check if all these have product ≤7.For the first case, product=3 ≤7.For the second case, product=4 ≤7. So both are valid. So total 15.Wait, but hold on. Is there any composition where the product is greater than 7? Let's see. For example, if we have three 2s, but that sum would be 6, not 7. Wait, S=7. If we have three 2s, sum would be 6, so need one more 1, making sum 7. But three 2s and two 1s sum to 8, which is over. So no. So all sequences with sum 7 have product 3 or 4, which are both ≤7. So yes, all 15 sequences are valid. So count for S=7 is 15.Proceeding to S=8. Let's find all 5-tuples that sum to 8.Possible compositions:1. One 4 and four 1s: sum=4+1+1+1+1=8. Number of sequences: C(5,1)=5. Product=4*1*1*1*1=4 ≤8. Valid.2. One 3, one 2, and three 1s: sum=3+2+1+1+1=8. Number of sequences: C(5,1)*C(4,1)=5*4=20. Product=3*2*1*1*1=6 ≤8. Valid.3. Two 2s and three 1s: sum=2+2+1+1+1=7. Wait, that's sum=7, which we already considered. To get sum=8, we need three 2s and two 1s: 2+2+2+1+1=8. Number of sequences: C(5,3)=10. Product=2*2*2*1*1=8 ≤8. Valid.4. Also, maybe one 5 and four 1s: sum=5+1+1+1+1=9, which is over 8. So not applicable.5. How about one 3 and one 3 and three 1s: sum=3+3+1+1+1=9. Again, over 8.So the possible compositions are:- One 4, four 1s (5 sequences)- One 3, one 2, three 1s (20 sequences)- Three 2s, two 1s (10 sequences)Total: 5 + 20 + 10 = 35 sequences.Now check if all these products are ≤8.For the first composition: product=4 ≤8.Second: product=6 ≤8.Third: product=8 ≤8.So all are valid. Thus, count for S=8 is 35.Next, S=9. Let's find all 5-tuples summing to 9.Possible compositions:1. One 5 and four 1s: sum=5+1+1+1+1=9. Number of sequences: C(5,1)=5. Product=5*1*1*1*1=5 ≤9. Valid.2. One 4, one 2, and three 1s: sum=4+2+1+1+1=9. Number of sequences: C(5,1)*C(4,1)=5*4=20. Product=4*2*1*1*1=8 ≤9. Valid.3. One 3, two 2s, and two 1s: sum=3+2+2+1+1=9. Number of sequences: C(5,1)*C(4,2)=5*6=30. Product=3*2*2*1*1=12. 12 ≤9? No, 12 >9. Therefore, these sequences are invalid.4. Three 2s and two 1s: sum=2+2+2+2+1=9? No, that's 2+2+2+2+1=9? Wait, 2+2+2+2+1=9, but that's four 2s and one 1. Sum=4*2 +1=9. Number of sequences: C(5,4)=5. Product=2*2*2*2*1=16. 16 ≤9? No. Invalid.5. One 3, one 3, one 1, one 1, one 1: sum=3+3+1+1+1=9. Number of sequences: C(5,2)=10. Product=3*3*1*1*1=9 ≤9. Valid.6. Two 3s, one 2, and two 1s: sum=3+3+2+1+0, but 0 is invalid. Wait, 3+3+2+1+0 isn't allowed. Let's think again.Alternatively, maybe two 3s and three 1s: sum=3+3+1+1+1=9. But that's the same as part 5. So covered.Other possibilities: four 2s and one 1: sum=8 +1=9, as in part 4. But already considered.Wait, let me check:Another composition could be one 4, one 3, and three 1s: sum=4+3+1+1+1=10, which is over 9. Not applicable.Alternatively, two 2s and one 3 and two 1s: sum=2+2+3+1+1=9. Wait, that's covered in part 3, which had product 12.Wait, maybe there's another composition. Let's check all possible partitions.To have sum 9 with 5 positive integers:Possible partitions:- 5,1,1,1,1- 4,2,1,1,1- 3,3,1,1,1- 3,2,2,1,1- 2,2,2,2,1Are there any others? For example, 4,3,1,1,0 but 0 is invalid. 6, but 6 is too big. So the above are all.So, as above:1. 5,1,1,1,1: 5 sequences, product=52. 4,2,1,1,1: 20 sequences, product=83. 3,3,1,1,1: 10 sequences, product=94. 3,2,2,1,1: 30 sequences, product=12 (invalid)5. 2,2,2,2,1: 5 sequences, product=16 (invalid)So only the first three compositions are valid. Therefore, total valid sequences for S=9: 5 +20 +10=35.Wait, but hold on, the composition 3,3,1,1,1: sum=9, product=9, which is equal to S=9. So that's valid. So 10 sequences. So total 5 +20 +10=35. Therefore, count for S=9 is 35.Proceeding to S=10.Now, let's find all 5-tuples summing to 10. This might be more complex.Possible compositions:1. One 6 and four 1s: sum=6+1+1+1+1=10. Product=6*1*1*1*1=6 ≤10. Valid. Number of sequences: C(5,1)=5.2. One 5, one 2, and three 1s: sum=5+2+1+1+1=10. Product=5*2*1*1*1=10 ≤10. Valid. Number of sequences: C(5,1)*C(4,1)=5*4=20.3. One 4, one 3, and three 1s: sum=4+3+1+1+1=10. Product=4*3*1*1*1=12. 12 ≤10? No. Invalid.4. One 4, two 2s, and two 1s: sum=4+2+2+1+1=10. Product=4*2*2*1*1=16. 16 ≤10? No. Invalid.5. Two 3s, one 2, and two 1s: sum=3+3+2+1+1=10. Product=3*3*2*1*1=18. 18 ≤10? No. Invalid.6. Two 5s and three 0s: invalid, since 0 is not allowed.7. One 3, three 2s, and one 1: sum=3+2+2+2+1=10. Product=3*2*2*2*1=24. Invalid.8. Five 2s: sum=10, but product=32. Invalid.9. Four 2s and one 2: same as above.Wait, maybe I need a more systematic approach.Possible partitions of 10 into 5 positive integers:- 6,1,1,1,1- 5,2,1,1,1- 4,3,1,1,1- 4,2,2,1,1- 3,3,2,1,1- 3,2,2,2,1- 2,2,2,2,2Let's check each:1. 6,1,1,1,1: product=6 ≤10. Valid. Sequences: 5.2. 5,2,1,1,1: product=10 ≤10. Valid. Sequences: 20.3. 4,3,1,1,1: product=12 >10. Invalid.4. 4,2,2,1,1: product=16 >10. Invalid.5. 3,3,2,1,1: product=18 >10. Invalid.6. 3,2,2,2,1: product=24 >10. Invalid.7. 2,2,2,2,2: product=32 >10. Invalid.Additionally, check if there are other partitions:- 7,1,1,1,0: invalid.- 5,3,1,1,0: invalid.So the only valid partitions are the first two: 6,1,1,1,1 and 5,2,1,1,1. So total sequences for S=10 are 5 +20=25.But wait, let's verify if there are other partitions.Another possible partition: 3,3,3,1,0: invalid. How about 3,3,2,2,0: invalid. All partitions must have positive integers. So the ones listed above are the only ones.Therefore, the only valid sequences for S=10 are the two partitions, giving 25 sequences. However, we need to check if their product is ≤ S=10. For the first partition, product=6 ≤10. For the second, product=10 ≤10. Both are valid. So count=25.But wait, let's check if there are any other partitions where the product is ≤10.For example, the partition 3,2,1,1,3: same as 3,3,2,1,1, which we already considered with product=18. Similarly, other arrangements won't reduce the product.Alternatively, if we have two 4s: 4+4+1+1+0, but invalid. So no.Another thought: Maybe a partition with a 3 and rest 1s and 2s. For example, 3,2,1,1,3 is same as above. Sum=10, product=18.No, so no other valid partitions. Therefore, S=10 has 25 valid sequences.Now, compiling all the counts:- S=5: 1- S=6:5- S=7:15- S=8:35- S=9:35- S=10:25Total sequences: 1 +5 +15 +35 +35 +25 = Let's compute this step by step.1 +5 =66 +15=2121 +35=5656 +35=9191 +25=116.So total 116 sequences.But wait, hold on. Is this correct?Wait, but let me cross-verify. Alternatively, maybe some sequences with higher products have been miscounted. For example, in S=9, when we considered 3,3,1,1,1, product=9. But if the sum is 9 and product is 9, that's valid. Similarly, in S=10, the 5,2,1,1,1 product=10.But let's check for S=8, product=8 is allowed. So that's okay.But let me check another approach. Since the product must be ≤10, and all numbers are positive integers, the possible products P can be from 1 to 10. Let's consider each possible P and find the number of 5-tuples where the product is P and the sum S is between P and10.But this might be more complicated, but let's try for verification.Products from 1 to10.Product=1: all numbers are 1. Sum=5. Then 1 ≤5 ≤10. So this is valid. Count=1.Product=2: exactly one 2 and rest 1s. Sum=5 +1=6. So sum=6. Then check if 2 ≤6 ≤10: yes. So the number of sequences is C(5,1)=5. So count=5.Product=3: two possibilities. Either one 3 and rest 1s (sum=5 +2=7), or three 1s and two 2s. Wait, product=3 can be achieved by one 3 and rest 1s, because 3*1*1*1*1=3. Alternatively, if there are other combinations? For example, three numbers of 1 and two numbers of something else. But 1*1*1*a*b=3. The only way is a=3, b=1. So only one 3 and rest 1s. Therefore, sequences: C(5,1)=5. Sum=7. 3 ≤7 ≤10: yes. So count=5.But wait, in S=7, we had 15 sequences. But here, only 5 sequences have product=3. The other 10 sequences (with two 2s) have product=4. So perhaps this approach is different.Wait, so if we consider product=3, we have 5 sequences, product=4, we have 10 sequences (sum=7). Product=4 can also be achieved by two 2s and three 1s. So sum=7. Then check if 4 ≤7 ≤10: yes. So sequences=10.Product=4: sequences=10.Product=5: achieved by one 5 and rest 1s (sum=9). So sequences=5. Check 5 ≤9 ≤10: yes. So count=5.Product=6: achieved by one 6 and rest 1s (sum=10), or one 3 and one 2 and rest 1s (sum=3+2+1+1+1=8). So two possibilities.First case: one 6, rest 1s. Product=6, sum=10. Valid. Sequences=5.Second case: one 3, one 2, rest 1s. Product=6, sum=8. 6 ≤8 ≤10: yes. Number of sequences: C(5,1)*C(4,1)=5*4=20.So total for product=6:5 +20=25.Product=7: Is there a way to get product=7? 7 is prime. So only one 7 and rest 1s. Sum=7 +4=11, which is over 10. So invalid. So product=7: no sequences.Product=8: achieved by:- One 8 and rest 1s: sum=8 +4=12>10. Invalid.- Two 2s and one 4, rest 1s: sum=2+2+4+1+1=10. Product=2*2*4=16. Wait, no. Wait, product=8. Wait, 2*2*2*1*1=8. So three 2s and two 1s. Sum=2+2+2+1+1=8. Product=8. So sequences: C(5,3)=10. Check 8 ≤8 ≤10: yes. So count=10.Alternatively, one 4 and one 2 and three 1s: product=4*2=8, sum=4+2+1+1+1=9. 8 ≤9 ≤10: yes. So sequences: C(5,1)*C(4,1)=5*4=20. Product=8. Wait, but product=4*2=8. So this is another way to get product=8. So total sequences for product=8 are both three 2s and two 1s (sum=8) and one 4, one 2, three 1s (sum=9).Wait, but hold on. If we are counting by product, then for product=8, we have two different configurations:1. Three 2s and two 1s: sum=8, product=8. Sequences=10.2. One 4, one 2, three 1s: sum=9, product=8. Sequences=20.So total sequences with product=8:10 +20=30.But wait, but in our previous approach for S=8 and S=9, we counted these. But in this product-based approach, product=8 can be achieved in two different sums: S=8 and S=9.Therefore, total sequences with product=8:10 +20=30. But we have to check if these sequences satisfy P ≤S ≤10. For the first case, S=8: 8 ≤8 ≤10: yes. For the second case, S=9:8 ≤9 ≤10: yes. So both are valid.Therefore, product=8 gives 30 sequences.Product=9: achieved by:- One 9 and rest 1s: sum=9 +4=13>10. Invalid.- Two 3s and three 1s: product=3*3=9, sum=3+3+1+1+1=9. So sequences=C(5,2)=10. Check 9 ≤9 ≤10: yes. Valid.- One 3 and one 3 and three 1s: same as above. So count=10.Product=9:10 sequences.Product=10: achieved by:- One 10 and rest 1s: sum=10 +4=14>10. Invalid.- One 5 and one 2 and three 1s: product=5*2=10, sum=5+2+1+1+1=10. Sequences=C(5,1)*C(4,1)=5*4=20. Check 10 ≤10 ≤10: yes. Valid.- Two 5s and rest 1s: product=25>10. Invalid.So product=10:20 sequences.Product=10 also can be achieved by other combinations? For example, 2*5*1*1*1=10. Any others? 5*2*1*1*1, which is the same. So only one configuration: one 5, one 2, three 1s.Therefore, total sequences with product=10:20.Now, summing up all valid sequences by product:Product=1:1Product=2:5Product=3:5Product=4:10Product=5:5Product=6:25Product=7:0Product=8:30Product=9:10Product=10:20Total:1+5=6; 6+5=11; 11+10=21; 21+5=26; 26+25=51; 51+0=51; 51+30=81; 81+10=91; 91+20=111.Wait, this gives 111, but earlier we had 116. There's a discrepancy here.Wait, why the difference? Let's check.In the first approach, summing over S=5 to S=10, we had total 116.In the second approach, summing over products from 1 to10, we get 111.Therefore, there is an inconsistency. Which one is correct?Let me check where the discrepancy is.First approach:S=5:1S=6:5S=7:15S=8:35S=9:35S=10:25Total:1+5+15+35+35+25=116.Second approach:By product:Product=1:12:53:54:105:56:258:309:1010:20Total:1+5+5+10+5+25+30+10+20=111.So discrepancy is 116-111=5.So missing 5 sequences in the product approach. Where are these?Looking at the product approach, maybe some products are being missed? Let's check for each product.Wait, for example, when product=6, in the product approach, we considered:- one 6 and rest 1s: 5 sequences (sum=10)- one 3, one 2, rest 1s:20 sequences (sum=8)Total:25.But in the first approach, for S=8, we had 35 sequences, of which 20 were product=6, 10 product=8, and 5 product=4. Wait, product=4? Wait, no. For S=8:- One 4, four 1s: product=4.- One 3, one 2, three 1s: product=6.- Three 2s, two 1s: product=8.So in S=8, the products are 4,6,8.But in the product approach, product=4 corresponds to two 2s and three 1s (sum=7). Wait, no:Product=4 can be achieved by two 2s and three 1s (sum=7) or one 4 and four 1s (sum=8).Ah! Here's the problem. In the product approach, product=4 can be achieved in two different ways:1. Two 2s and three 1s: sum=7, product=4. So sequences=10.2. One 4 and four 1s: sum=8, product=4. So sequences=5.But in the product-based count above, we only considered product=4 as the first case (sum=7). However, the second case (sum=8) also gives product=4. Therefore, in the product approach, product=4 should have 10 +5=15 sequences. However, in our previous product-based count, we only counted 10 sequences for product=4. That's the error.Similarly, product=4 is achieved in two different sums: S=7 (two 2s and three 1s) and S=8 (one 4 and four 1s). Therefore, product=4 has 10 +5=15 sequences.But in the previous product-based count, we considered only the two 2s case. Therefore, missing 5 sequences. Similarly, other products might have similar issues.So let's correct the product-based approach:Product=4: sequences where product=4. These can be:- Two 2s and three 1s (sum=7): C(5,2)=10 sequences.- One 4 and four 1s (sum=8): C(5,1)=5 sequences.Total=15.Similarly, product=6: we considered both one 6 and rest 1s (sum=10, 5 sequences) and one 3, one 2, rest 1s (sum=8, 20 sequences). So total=25.Product=8: three 2s and two 1s (sum=8, 10 sequences) and one 4, one 2, rest 1s (sum=9, 20 sequences). Total=30.Product=5: one 5 and rest 1s (sum=9, 5 sequences).Product=3: one 3 and rest 1s (sum=7, 5 sequences).Product=9: two 3s and three 1s (sum=9, 10 sequences).Product=10: one 5, one 2, rest 1s (sum=10, 20 sequences).Product=1: all 1s (sum=5, 1 sequence).So correcting the product-based count:Product=1:1Product=2:5 (one 2, sum=6)Product=3:5 (one 3, sum=7)Product=4:15 (two 2s sum=7 and one 4 sum=8)Product=5:5 (one 5 sum=9)Product=6:25 (one 6 sum=10 and one 3+2 sum=8)Product=8:30 (three 2s sum=8 and one 4+2 sum=9)Product=9:10 (two 3s sum=9)Product=10:20 (one 5+2 sum=10)Adding these up:1 +5=66 +5=1111 +15=2626 +5=3131 +25=5656 +30=8686 +10=9696 +20=116.Ah, now it matches the first approach. So the mistake was that product=4 includes two cases (sum=7 and sum=8), which were not initially considered in the product-based approach.Therefore, both methods now agree on the total of 116 sequences.However, the answer needs to be checked once more for accuracy, perhaps by cross-verifying individual counts.Let's verify S=8:Sum=8, sequences:- One 4, four 1s: product=4. Count=5.- One 3, one 2, three 1s: product=6. Count=20.- Three 2s, two 1s: product=8. Count=10.Total=35. Which matches.S=9:- One 5, four 1s: product=5. Count=5.- One 4, one 2, three 1s: product=8. Count=20.- Two 3s, three 1s: product=9. Count=10.Total=35. Correct.S=10:- One 6, four 1s: product=6. Count=5.- One 5, one 2, three 1s: product=10. Count=20.Total=25. Correct.S=7:- One 3, four 1s: product=3. Count=5.- Two 2s, three 1s: product=4. Count=10.Total=15. Correct.S=6:- One 2, four 1s: product=2. Count=5.S=5:- All 1s: product=1. Count=1.All these add up to 116.Therefore, the final answer is 116. However, to ensure that there are no mistakes, let's check if any of the counts are incorrect.For example, in S=8, product=8: three 2s and two 1s. The number of sequences is C(5,3)=10. Correct.Similarly, for S=9, product=9: two 3s and three 1s. C(5,2)=10. Correct.In S=10, product=10: one 5 and one 2. The number of sequences is C(5,1)*C(4,1)=5*4=20. Correct.Yes, all counts seem correct.Therefore, the number of sequences is 116.But wait, hold on. The problem says sequences of 5 positive integers. In all our counts, we considered ordered sequences, right? For example, for S=6, one 2 and four 1s: since the 2 can be in any position, the number of sequences is 5, which is correct for ordered tuples.Similarly, for two 2s in S=7: C(5,2)=10, which is correct for ordered tuples where two positions are 2 and others are 1.Yes, all counts are considering the order, so the total should be correct.Therefore, the final answer is boxed{116}.

✅:To determine the number of sequences of 5 positive integers ((a, b, c, d, e)) that satisfy (abcde leq a + b + c + d + e leq 10), we analyze each possible sum (S) from 5 to 10 and count the valid sequences for each sum.1. Sum (S = 5): - Only one sequence: ((1, 1, 1, 1, 1)). - Product: (1), which is (leq 5). - Count: (1).2. Sum (S = 6): - Sequences with one 2 and four 1s. - Number of sequences: (binom{5}{1} = 5). - Product: (2), which is (leq 6). - Count: (5).3. Sum (S = 7): - Sequences with one 3 and four 1s: (binom{5}{1} = 5). - Sequences with two 2s and three 1s: (binom{5}{2} = 10). - Products: (3) and (4), both (leq 7). - Count: (5 + 10 = 15).4. Sum (S = 8): - Sequences with one 4 and four 1s: (binom{5}{1} = 5). - Sequences with one 3, one 2, and three 1s: (binom{5}{1} binom{4}{1} = 20). - Sequences with three 2s and two 1s: (binom{5}{3} = 10). - Products: (4), (6), and (8), all (leq 8). - Count: (5 + 20 + 10 = 35).5. Sum (S = 9): - Sequences with one 5 and four 1s: (binom{5}{1} = 5). - Sequences with one 4, one 2, and three 1s: (binom{5}{1} binom{4}{1} = 20). - Sequences with two 3s and three 1s: (binom{5}{2} = 10). - Products: (5), (8), and (9), all (leq 9). - Count: (5 + 20 + 10 = 35).6. Sum (S = 10): - Sequences with one 6 and four 1s: (binom{5}{1} = 5). - Sequences with one 5, one 2, and three 1s: (binom{5}{1} binom{4}{1} = 20). - Products: (6) and (10), both (leq 10). - Count: (5 + 20 = 25).Summing all counts:[1 + 5 + 15 + 35 + 35 + 25 = 116]Thus, the number of sequences is (boxed{116}).

🔑:We are asked to determine the number of sequences of 5 positive integers ((a, b, c, d, e)) such that (a cdot b cdot c cdot d cdot e leq a + b + c + d + e leq 10). We will solve this by considering different cases based on the values of the integers:1. Case 1: All integers are 1 - If (a = b = c = d = e = 1), then (a cdot b cdot c cdot d cdot e = 1) and (a + b + c + d + e = 5). - This clearly satisfies (1 leq 5 leq 10). - So, there is exactly 1 valid sequence in this case. [ text{Total sequences} = 1 ]2. Case 2: Four integers are 1 - Consider (a = b = c = d = 1) and (e) is any integer such that (e neq 1). - We need (e cdot 1 cdot 1 cdot 1 cdot 1 leq 1 + 1 + 1 + 1 + e leq 10), which simplifies to (e leq 4 + e leq 10). - Thus, (e) can be 2, 3, 4, 5, or 6. - Each sequence ((1, 1, 1, 1, e)) can be arranged in: [ 5! / 4! = 5 text{ different ways} ] - Since (e) has 5 possible values: [ text{Total sequences} = 5 times 5 = 25 ]3. Case 3: Three integers are 1 - Let (a = b = c = 1) and (d), (e) be any integers. - We need (1 cdot 1 cdot 1 cdot d cdot e leq 1 + 1 + 1 + d + e leq 10). - The possible pairs ((d, e)) that satisfy (d cdot e leq 7) and (d + e leq 7) are: - ((2, 2)), ((3, 3)), ((2, 3)), ((2, 4)), ((2, 5)). - Counting each, the number of ordered sequences: - ((2, 2)): [ frac{5!}{3! cdot 2!} = 10 text{ ways} ] - ((3, 3)): [ frac{5!}{3! cdot 2!} = 10 text{ ways} ] - For distinct pairs ((2, 3)), ((2, 4)), ((2, 5)): [ frac{5!}{3! cdot 1! cdot 1!} = 20 text{ ways each} ] - Total sequences in this case: [ 2 times 10 + 3 times 20 = 20 + 60 = 80 ]4. Case 4: Two integers are 1 - Let (a = b = 1) and (c), (d), (e) be any integers. - For (ab cdot cde leq ab + c + d + e leq 10), since (ab = 1 cdot 1 = 1) implies (c, d, e leq 3). - If (c = d = e = 2): [ 1 cdot 1 cdot 2 cdot 2 cdot 2 = 8 text{ and } 1 + 1 + 2 + 2 + 2 = 8 ] - Number of such sequences: [ frac{5!}{2! cdot 3!} = 10 ]Summing all these cases, we get:[1 + 25 + 80 + 10 = 116] Conclusion:[boxed{116}]

🧠:Okay, so I need to figure out if the vectors a, b, and c are coplanar. Let me start by recalling what coplanar vectors mean. If three vectors are coplanar, they all lie in the same plane, right? That means the volume of the parallelepiped formed by these vectors should be zero. The way to calculate that volume is using the scalar triple product of the vectors. If the scalar triple product is zero, then they are coplanar.Alright, so the scalar triple product is given by a · (b × c). Let me write that down. First, I need to compute the cross product of vectors b and c, and then take the dot product of vector a with that result.Let me write out the vectors again to make sure I have them right:a = {3, 0, 3}b = {8, 1, 6}c = {1, 1, -1}First step: compute the cross product of b and c. The cross product of two vectors {b1, b2, b3} and {c1, c2, c3} is given by the determinant of the matrix with the standard unit vectors i, j, k in the first row, the components of b in the second row, and the components of c in the third row. So, let's compute that.So, b × c = |i   j   k|             |8   1   6|             |1   1  -1|Calculating the determinant:i * (1*(-1) - 6*1) - j * (8*(-1) - 6*1) + k * (8*1 - 1*1)Let me compute each component step by step.First component (i):1*(-1) = -16*1 = 6So, -1 - 6 = -7So, the i component is -7i.Second component (j):This one is subtracted, so it's - [8*(-1) - 6*1]Compute 8*(-1) = -86*1 = 6So, -8 - 6 = -14Multiply by -1: 14So, the j component is 14j.Third component (k):8*1 = 81*1 = 1So, 8 - 1 = 7Thus, the k component is 7k.Therefore, the cross product b × c is {-7, 14, 7}.Now, I need to compute the dot product of vector a with this cross product. Vector a is {3, 0, 3}.So, the dot product a · (b × c) = 3*(-7) + 0*14 + 3*7.Calculating each term:3*(-7) = -210*14 = 03*7 = 21Adding them up: -21 + 0 + 21 = 0Hmm, that's zero. Wait, so the scalar triple product is zero. Therefore, according to what I remember, if the scalar triple product is zero, the vectors are coplanar. So that would mean vectors a, b, and c are coplanar.But let me double-check my calculations to be sure. Maybe I made a mistake in the cross product or the dot product.First, cross product of b and c:b = {8, 1, 6}, c = {1, 1, -1}Cross product formula:i*(b2*c3 - b3*c2) - j*(b1*c3 - b3*c1) + k*(b1*c2 - b2*c1)So substituting:i*(1*(-1) - 6*1) - j*(8*(-1) - 6*1) + k*(8*1 - 1*1)Which is:i*(-1 -6) - j*(-8 -6) + k*(8 -1)= i*(-7) - j*(-14) + k*(7)= -7i +14j +7kYes, that's correct. So {-7,14,7} is correct.Then dot product with a = {3,0,3}:3*(-7) + 0*14 +3*7 = (-21) + 0 +21 = 0. Yes, that's right.So the scalar triple product is indeed zero. Therefore, the three vectors are coplanar.Alternatively, another way to check is to see if one of the vectors can be expressed as a linear combination of the other two. But that might involve more steps. However, since the scalar triple product is zero, that's sufficient.Wait, let me just confirm that the scalar triple product being zero implies coplanarity. Yes, because the scalar triple product gives the volume of the parallelepiped. If the volume is zero, the vectors are lying in the same plane, hence coplanar. So that's a solid method.Alternatively, if I were to set up a matrix with the vectors as rows or columns and compute the determinant, which is essentially the same as the scalar triple product. Since determinant is zero, the vectors are linearly dependent, which also implies they are coplanar.Yes, so all these methods confirm that if the scalar triple product is zero, coplanarity holds.Therefore, the conclusion is that vectors a, b, and c are coplanar.

✅:To determine if the vectors (a = {3, 0, 3}), (b = {8, 1, 6}), and (c = {1, 1, -1}) are coplanar, we compute the scalar triple product (a cdot (b times c)). If this product is zero, the vectors are coplanar.Step 1: Compute the cross product (b times c):[b times c = begin{vmatrix}mathbf{i} & mathbf{j} & mathbf{k} 8 & 1 & 6 1 & 1 & -1end{vmatrix}= mathbf{i}(1 cdot (-1) - 6 cdot 1) - mathbf{j}(8 cdot (-1) - 6 cdot 1) + mathbf{k}(8 cdot 1 - 1 cdot 1)][= mathbf{i}(-1 - 6) - mathbf{j}(-8 - 6) + mathbf{k}(8 - 1)= -7mathbf{i} + 14mathbf{j} + 7mathbf{k} = {-7, 14, 7}]Step 2: Compute the dot product (a cdot (b times c)):[a cdot (b times c) = {3, 0, 3} cdot {-7, 14, 7} = 3(-7) + 0(14) + 3(7) = -21 + 0 + 21 = 0]Since the scalar triple product equals zero, the vectors (a), (b), and (c) are coplanar.Answer: (boxed{text{Yes}})

🔑:To determine whether the vectors mathbf{a}, mathbf{b}, and mathbf{c} are coplanar, we need to check if their scalar triple product is zero. The scalar triple product (mathbf{a}, mathbf{b}, mathbf{c}) is given by the determinant of the matrix formed by placing the vectors as rows or columns. In this case, we'll use:[mathbf{a} = begin{pmatrix} 3 0 3 end{pmatrix}, quad mathbf{b} = begin{pmatrix} 8 1 6 end{pmatrix}, quad mathbf{c} = begin{pmatrix} 1 1 -1 end{pmatrix}]The scalar triple product is:[(mathbf{a}, mathbf{b}, mathbf{c}) = begin{vmatrix}3 & 0 & 3 8 & 1 & 6 1 & 1 & -1 end{vmatrix}]We now evaluate this determinant using cofactor expansion along the first row:[begin{vmatrix}3 & 0 & 3 8 & 1 & 6 1 & 1 & -1 end{vmatrix} = 3 cdot begin{vmatrix}1 & 6 1 & -1 end{vmatrix} - 0 cdot begin{vmatrix}8 & 6 1 & -1 end{vmatrix} + 3 cdot begin{vmatrix}8 & 1 1 & 1 end{vmatrix}]Next, we compute the 2x2 determinants involved:[begin{vmatrix}1 & 6 1 & -1 end{vmatrix} = (1 cdot -1) - (6 cdot 1) = -1 - 6 = -7][begin{vmatrix}8 & 6 1 & -1 end{vmatrix} = (8 cdot -1) - (6 cdot 1) = -8 - 6 = -14][begin{vmatrix}8 & 1 1 & 1 end{vmatrix} = (8 cdot 1) - (1 cdot 1) = 8 - 1 = 7]Substituting these back into our expansion:[3 cdot (-7) - 0 cdot (-14) + 3 cdot 7 = 3 cdot (-7) + 3 cdot 7 = -21 + 21 = 0]Since the scalar triple product (mathbf{a}, mathbf{b}, mathbf{c}) = 0, the vectors mathbf{a}, mathbf{b}, and mathbf{c} are coplanar.[boxed{0}]As established, the vectors mathbf{a}, mathbf{b}, and mathbf{c} are indeed coplanar.

🧠:Okay, let's see. I need to find the smallest integer n such that the sum from n to n+20 is greater than 2019. Hmm. Alright, first, I should remember how to calculate the sum of consecutive integers. Since this is an arithmetic series, right? The sum of consecutive numbers can be calculated using the formula: Sum = number of terms × (first term + last term) / 2. Yeah, that sounds right.So, let's break it down. The first term here is n, and the last term is n + 20. How many terms are there? Well, if you start counting from n to n + 20 inclusive, that's 21 terms. Because, like, from 0 to 20 is 21 numbers, right? So same here. So the number of terms is 21.Therefore, the sum S is 21 × (n + (n + 20)) / 2. Let me write that out: S = 21*(2n + 20)/2. Simplifying that, 21*(n + 10). So the sum simplifies to 21(n + 10).Now, the inequality given is that this sum has to be greater than 2019. So:21(n + 10) > 2019To solve for n, divide both sides by 21 first:n + 10 > 2019 / 21Let me calculate 2019 divided by 21. Hmm, 21*90 = 1890, 21*100 = 2100. So 2019 is between 90 and 100 times 21. Let's do 21*96 = 2016. Because 21*90=1890, 21*6=126, so 1890+126=2016. Then 2019 - 2016 = 3. So 2019/21 = 96 + 3/21 = 96 + 1/7 ≈ 96.142857.So, n + 10 > approximately 96.142857Therefore, n > 96.142857 - 10 = 86.142857Since n must be an integer, the smallest integer greater than 86.142857 is 87. Wait, but let me check again. Wait, 96.142857 - 10 is 86.142857? Wait, 96.142857 - 10 is 86.142857? Wait, 96 - 10 is 86, and 0.142857 is still there. So yes, 86.142857. Therefore, n must be greater than that. So the minimal integer n is 87. Hmm.But wait, let me verify this calculation again. Maybe I made a mistake in dividing 2019 by 21.Let me compute 21*96: 20*96=1920, 1*96=96, so 1920+96=2016. Then 2019-2016=3. So yes, 2019=21*96 +3. Therefore, 2019/21=96 + 3/21=96 +1/7≈96.142857. Correct.So n +10 >96.142857, so n>86.142857. So n=87. Therefore, the minimal possible value is 87. Hmm.But let me check with n=87. Let's compute the sum. Sum =21*(87 +10)=21*97=21*(100 -3)=2100 -63=2037. 2037>2019, so that's correct.What if n=86? Then the sum is 21*(86 +10)=21*96=2016. 2016 is less than 2019, so that's not enough. Therefore, n=87 is indeed the minimal.Wait, but just to be thorough, maybe I should write out the steps again. So the sum from n to n+20 is 21 terms. The formula is 21*(n + (n+20))/2=21*(2n +20)/2=21*(n +10). So that's the sum. Then set 21(n +10) >2019. Solving for n: n +10 >2019/21≈96.142857, so n>86.142857. Hence, the minimal integer is 87.Alternatively, maybe I can think in terms of inequalities step by step. Let me write the inequality again:n + (n+1) + (n+2) + ... + (n+20) > 2019As established, this sum is equal to 21n + (0+1+2+...+20). Wait, another way to think about it is that each term from n to n+20 is n + k where k ranges from 0 to 20. So the sum is sum_{k=0}^{20} (n + k) = 21n + sum_{k=0}^{20} k. The sum of k from 0 to 20 is (20*21)/2=210. So the total sum is 21n +210.Wait, hold on! Wait, now this is conflicting with the previous result. Wait, according to this, the sum is 21n +210. But previously, I had 21(n +10)=21n +210. Oh, right! Because 21(n +10) is 21n +210. So both methods agree. That’s good.Therefore, the sum is 21n +210. So the inequality is 21n +210 >2019.Subtract 210 from both sides: 21n >2019 -210=1809.Then divide by 21: n >1809 /21. Let me compute 1809 divided by 21.21*80=1680. 1809 -1680=129.21*6=126. 129 -126=3.So 21*86=21*(80+6)=1680+126=1806. Then 1809 -1806=3. So 1809=21*86 +3. Therefore, 1809/21=86 +3/21=86 +1/7≈86.142857.So n>86.142857, so n=87. Therefore, same result.Therefore, the minimal integer n is 87. Therefore, the answer is 87.But let me verify once again with n=87:Sum =21*87 +210. 21*87: 20*87=1740, 1*87=87, so 1740+87=1827. 1827+210=2037. 2037>2019. Correct.n=86: 21*86=21*(80+6)=1680+126=1806. 1806+210=2016. 2016<2019. So indeed 86 is too small.Therefore, the minimal n is 87. So the answer is boxed{87}Final AnswerThe minimum possible value of ( n ) is boxed{87}.

✅:To find the minimum integer ( n ) such that the sum ( n + (n+1) + (n+2) + cdots + (n+20) ) is greater than 2019, we start by recognizing that this is an arithmetic series. The series has 21 terms, with the first term ( n ) and the last term ( n + 20 ).The sum ( S ) of an arithmetic series is given by:[S = frac{text{number of terms}}{2} times (text{first term} + text{last term})]Substituting the values, we get:[S = frac{21}{2} times (n + (n + 20)) = frac{21}{2} times (2n + 20) = 21(n + 10)]We need this sum to be greater than 2019:[21(n + 10) > 2019]Solving for ( n ):[n + 10 > frac{2019}{21}]Calculating ( frac{2019}{21} ):[2019 div 21 approx 96.142857]Thus:[n + 10 > 96.142857 implies n > 96.142857 - 10 implies n > 86.142857]The smallest integer greater than 86.142857 is 87. Verifying with ( n = 87 ):[21(87 + 10) = 21 times 97 = 2037 quad (text{which is greater than 2019})]Verifying with ( n = 86 ):[21(86 + 10) = 21 times 96 = 2016 quad (text{which is less than 2019})]Thus, the minimum possible value of ( n ) is (boxed{87}).

🔑:1. We start with the given inequality: [ n + (n+1) + (n+2) + cdots + (n+20) > 2019 ]2. We recognize that this is an arithmetic series with the first term a = n and the last term l = n + 20, and there are 21 terms in the series.3. The sum S of the first 21 terms of an arithmetic series is given by: [ S = frac{text{number of terms}}{2} cdot (text{first term} + text{last term}) ] Applying this to our series: [ S = frac{21}{2} cdot (n + (n+20)) ]4. Simplify the expression inside the parentheses: [ n + (n + 20) = 2n + 20 ]5. Substitute back into the sum formula: [ S = frac{21}{2} cdot (2n + 20) ]6. Simplify the multiplication: [ S = 21n + 210 ]7. Set up the inequality: [ 21n + 210 > 2019 ]8. Subtract 210 from both sides: [ 21n > 1809 ]9. Divide both sides by 21: [ n > frac{1809}{21} ]10. Simplify the division: [ n > 86.142857 ]11. Since (n) must be an integer, we take the smallest integer greater than (86.142857): [ n geq 87 ]Conclusion:[boxed{87}]

🧠:Okay, so I need to find the distance between a lateral edge and a diagonal of the base in a regular tetrahedral pyramid where the base has side length (a), and a lateral face forms a (45^circ) angle with the base plane. Hmm, let me start by visualizing this pyramid. First, a regular tetrahedral pyramid—wait, is that a regular pyramid with a square base? Wait, tetrahedral usually refers to a tetrahedron, which is a pyramid with a triangular base. But the problem mentions a "regular tetrahedral pyramid," which is a bit confusing. Maybe it's a regular pyramid, meaning the base is a regular polygon, and the apex is directly above the center of the base. Since it's called tetrahedral, perhaps the base is a triangle? But then a tetrahedron is a three-dimensional figure with four triangular faces. Wait, the problem mentions a pyramid, so maybe it's a square base? Wait, no, tetrahedral pyramid—maybe a three-sided base, like a triangular pyramid, which is a tetrahedron. But the problem mentions a lateral face forming a 45-degree angle with the base. Let me clarify.Wait, perhaps the base is a square? No, tetrahedral usually implies triangular. Let me check the original problem again. It says "the side length of the base of a regular tetrahedral pyramid is (a)." So, a regular pyramid with a tetrahedral base? Wait, that doesn't make sense. Maybe "regular tetrahedral pyramid" is a pyramid with a regular tetrahedron as its lateral faces? Wait, perhaps it's a regular square pyramid, but the lateral face makes a 45-degree angle with the base. Wait, maybe I need to parse this differently. A regular pyramid has a regular polygon base and lateral faces that are congruent isosceles triangles. So, if it's a regular tetrahedral pyramid, maybe the base is a regular tetrahedron? But a pyramid is a three-dimensional figure with a base and triangular faces meeting at an apex. Wait, a regular tetrahedron is already a pyramid with a triangular base. So maybe this is a square pyramid (base is square) but called tetrahedral? Hmm, maybe the problem is mistranslated or misworded. Alternatively, maybe "tetrahedral" here just refers to the shape being a tetrahedron. Wait, maybe the problem is about a regular tetrahedron, which is a pyramid with a triangular base, and all edges are equal. But then the problem mentions a lateral face forming a 45-degree angle with the base. If it's a regular tetrahedron, all the angles between faces are the same, but maybe in this case, the angle is 45 degrees? Wait, in a regular tetrahedron, the angle between any two faces is arccos(1/3), which is approximately 54.7 degrees, so not 45. Therefore, this is probably not a regular tetrahedron.Wait, maybe the pyramid is a square pyramid (base is square), with lateral faces making 45 degrees with the base. Let's assume that. So the base is a square with side length (a), and each lateral face is an isosceles triangle forming a 45-degree angle with the base. Then, we need to find the distance between a lateral edge (an edge from the apex to a base vertex) and a diagonal of the base which is skew to it.Wait, but in a square pyramid, the lateral edges are the edges from the apex to each base vertex. The base diagonals are the diagonals of the square. However, in a square pyramid, the lateral edges and the base diagonals are not skew; they either intersect or lie on the same plane. Wait, for example, take a square base ABCD with apex E. The diagonal AC of the base. The lateral edge EA connects to A, which is an endpoint of AC. So EA and AC intersect at A. Similarly, the other diagonal BD and lateral edge EB intersect at B. So in that case, the lateral edges and base diagonals are not skew. Therefore, perhaps the problem is referring to a different pyramid. Wait, maybe it's a triangular pyramid (tetrahedron), but then the base is a triangle, so the diagonals of the base would be the edges themselves? No, a triangle doesn't have diagonals. Hmm. This is confusing.Wait, maybe the problem is a regular pyramid with a square base, and the term "tetrahedral" is a mistranslation. Let's proceed under that assumption because otherwise, the problem doesn't make sense. So, base is square with side length (a), apex above the center. A lateral face (i.e., one of the triangular faces) forms a 45-degree angle with the base plane. Then, we need to find the distance between a lateral edge (like from apex to a base vertex) and a diagonal of the base which is skew to it. But in a square pyramid, as I thought earlier, the lateral edges meet the base diagonals at the vertices. Therefore, they are not skew. Wait, but maybe they are considering a different diagonal? Wait, the base is a square, so the diagonals are AC and BD. If we take a lateral edge, say, from apex E to vertex A. Then, the diagonal BD is in the base, which does not intersect the lateral edge EA because BD connects B to D, and EA connects E to A. Since A is not on BD, and E is above the center. So in 3D space, EA and BD are skew lines. So, yes, in that case, they are skew. So the distance between the lateral edge EA and the base diagonal BD which is skew to it. So that makes sense.So, problem: square base ABCD, apex E, each lateral face (like EAB) forms a 45-degree angle with the base. Find the distance between lateral edge EA and base diagonal BD.Okay, let's start by setting coordinates. Let me place the square base ABCD on the xy-plane. Let’s assume the square has side length (a), so coordinates can be:- A: (0, 0, 0)- B: (a, 0, 0)- C: (a, a, 0)- D: (0, a, 0)- Center of the base O: (a/2, a/2, 0)- Apex E: (a/2, a/2, h), where h is the height of the pyramid.Given that a lateral face forms a 45-degree angle with the base. Let's take face EAB. The face EAB is a triangle with vertices E, A, B. The angle between this face and the base is 45 degrees. To find this angle, we can consider the dihedral angle between the lateral face and the base.The dihedral angle between two planes can be found using their normal vectors. The base is the xy-plane, with normal vector (0, 0, 1). The lateral face EAB is a triangle; to find its normal vector, we can take two vectors in the plane and compute the cross product. Let's take vectors EA and EB. Wait, but E is (a/2, a/2, h). Vector EA is A - E = ( -a/2, -a/2, -h ). Vector EB is B - E = ( a/2, -a/2, -h ). The cross product of EA and EB will give a normal vector to the plane EAB.Compute EA × EB:EA = (-a/2, -a/2, -h)EB = (a/2, -a/2, -h)Cross product:|i     j     k||-a/2 -a/2 -h||a/2  -a/2 -h|= i [ (-a/2)(-h) - (-h)(-a/2) ] - j [ (-a/2)(-h) - (-h)(a/2) ] + k [ (-a/2)(-a/2) - (-a/2)(a/2) ]Simplify each component:i: [ (a h / 2 ) - (a h / 2 ) ] = 0j: - [ (a h / 2 ) - (-a h / 2 ) ] = - [ (a h / 2 + a h / 2 ) ] = - [ a h ]k: [ (a² / 4 ) - ( -a² / 4 ) ] = [ a² / 4 + a² / 4 ] = a² / 2So the cross product is (0, -a h, a² / 2 )Therefore, the normal vector to the plane EAB is (0, -a h, a² / 2 ). We can simplify by dividing by a/2: (0, -2 h, a )But for the dihedral angle, we can use the normal vectors. The dihedral angle between the lateral face and the base is the angle between the normal vector of the base (0, 0, 1) and the normal vector of the lateral face (0, -a h, a² / 2 ). The angle between the two normals is equal to 180° minus the dihedral angle, but since we are interested in the angle between the planes, which is the angle between the normals or its supplement, depending on orientation. Wait, actually, the dihedral angle is the angle between the two planes, which can be found using the normals. The formula for the angle θ between two planes with normals n1 and n2 is:cosθ = |n1 ⋅ n2| / (|n1| |n2| )But here, the problem says the lateral face forms a 45° angle with the base. So θ = 45°. Let's compute the dot product between the base normal (0, 0, 1) and the EAB normal (0, -a h, a² / 2 ).Dot product = 0*0 + 0*(-a h) + 1*(a² / 2 ) = a² / 2The magnitude of the base normal is 1. The magnitude of the EAB normal is sqrt( (0)^2 + (-a h)^2 + (a² / 2 )^2 ) = sqrt( a² h² + a^4 / 4 )Therefore,cosθ = |a² / 2| / ( sqrt( a² h² + a^4 / 4 ) * 1 ) = (a² / 2 ) / sqrt( a² h² + a^4 / 4 )Given θ = 45°, so cosθ = √2 / 2 ≈ 0.7071Thus,(a² / 2 ) / sqrt( a² h² + a^4 / 4 ) = √2 / 2Multiply both sides by sqrt( a² h² + a^4 / 4 ):(a² / 2 ) = (√2 / 2 ) sqrt( a² h² + a^4 / 4 )Multiply both sides by 2:a² = √2 sqrt( a² h² + a^4 / 4 )Square both sides:a^4 = 2 ( a² h² + a^4 / 4 )Expand right side:2a² h² + 2*(a^4 /4 ) = 2a² h² + a^4 / 2Thus,a^4 = 2a² h² + a^4 / 2Subtract a^4 /2 from both sides:a^4 / 2 = 2a² h²Divide both sides by 2a²:a² / 4 = h²Thus,h = a / 2So the height of the pyramid is h = a/2.Therefore, the apex E has coordinates (a/2, a/2, a/2 )Now, we need to find the distance between the lateral edge EA and the base diagonal BD which is skew to it.First, let's parametrize the lines EA and BD.Line EA goes from E (a/2, a/2, a/2) to A (0, 0, 0). Let's write parametric equations.Let parameter t go from 0 to 1.EA: (x, y, z) = (a/2 - (a/2)t, a/2 - (a/2)t, a/2 - (a/2)t )Simplify:x = a/2 (1 - t )y = a/2 (1 - t )z = a/2 (1 - t )Alternatively, direction vector is (-a/2, -a/2, -a/2 ), which can be simplified to (-1, -1, -1 ) multiplied by a/2.Line BD goes from B (a, 0, 0) to D (0, a, 0). Since it's a diagonal of the base, it's in the z=0 plane.Parametrize BD with parameter s from 0 to 1:BD: (x, y, z) = (a - a s, 0 + a s, 0 )So x = a(1 - s )y = a sz = 0So direction vector is (-a, a, 0 ), which simplifies to (-1, 1, 0 ) multiplied by a.Now, we have two skew lines EA and BD. The distance between them can be found using the formula for the distance between two skew lines.The formula is |(b1 - b2) ⋅ (v1 × v2)| / |v1 × v2|, where b1 and b2 are points on each line, and v1, v2 are direction vectors of the lines.Let's apply this formula.First, let's note:Line EA: point b1 = E = (a/2, a/2, a/2 )Line BD: point b2 = B = (a, 0, 0 )Direction vectors:v1 (EA) is from E to A: A - E = (-a/2, -a/2, -a/2 ), which can be written as v1 = (-1, -1, -1 ) * (a/2 )v2 (BD) is from B to D: D - B = (-a, a, 0 ), which is v2 = (-1, 1, 0 ) * aBut for the formula, we can use the direction vectors as vectors, regardless of parameterization. So let's take:v1 = (-1, -1, -1 )v2 = (-1, 1, 0 )But scaled by a/2 and a respectively. Wait, the cross product is linear, so scaling the direction vectors will scale the cross product. However, in the distance formula, since both the numerator and denominator involve the cross product, the scaling factors will cancel out. Let me check.Wait, actually, the formula uses the actual direction vectors as per their parametric equations. So if the direction vectors are scaled differently, it will affect the cross product. Let me do it properly.Let’s define:For line EA: direction vector is (-a/2, -a/2, -a/2 )For line BD: direction vector is (-a, a, 0 )So compute v1 × v2:v1 = (-a/2, -a/2, -a/2 )v2 = (-a, a, 0 )Cross product:|i     j     k||-a/2 -a/2 -a/2||-a    a    0  |= i [ (-a/2)(0) - (-a/2)(a) ] - j [ (-a/2)(0) - (-a/2)(-a) ] + k [ (-a/2)(a) - (-a/2)(-a) ]Compute each component:i: [ 0 - (-a^2 / 2 ) ] = a^2 / 2j: - [ 0 - (a^2 / 2 ) ] = - [ -a^2 / 2 ] = a^2 / 2k: [ (-a^2 / 2 ) - (a^2 / 2 ) ] = -a^2Therefore, v1 × v2 = ( a²/2, a²/2, -a² )Then, compute the vector from b2 to b1: b1 - b2 = (a/2 - a, a/2 - 0, a/2 - 0 ) = ( -a/2, a/2, a/2 )Take the dot product of (b1 - b2) and (v1 × v2):( -a/2, a/2, a/2 ) ⋅ ( a²/2, a²/2, -a² ) =(-a/2)(a²/2) + (a/2)(a²/2) + (a/2)(-a² ) =Compute each term:First term: (-a³)/4Second term: a³ / 4Third term: -a³ / 2Sum: (-a³/4 + a³/4 ) - a³/2 = 0 - a³/2 = -a³ / 2The absolute value is | -a³ / 2 | = a³ / 2Now, compute |v1 × v2|:The cross product vector is ( a²/2, a²/2, -a² )Its magnitude is sqrt( (a²/2)^2 + (a²/2)^2 + (-a²)^2 )Compute each component squared:(a^4 / 4 ) + (a^4 / 4 ) + a^4 = a^4 / 2 + a^4 = (3a^4 ) / 2Thus, |v1 × v2| = sqrt( 3a^4 / 2 ) = a² sqrt(3/2 ) = a² ( sqrt(6)/2 )Therefore, the distance is (a³ / 2 ) / (a² sqrt(6)/2 ) ) = (a³ / 2 ) * (2 / (a² sqrt(6)) ) = a / sqrt(6 )Rationalizing the denominator: a sqrt(6) / 6 = (sqrt(6)/6 ) aTherefore, the distance is ( frac{sqrt{6}}{6} a )Wait, but let me verify this. Let's check the steps again.First, coordinates:Apex E at (a/2, a/2, a/2 )Points on lines:EA: from E to A (0,0,0 ). Direction vector (-a/2, -a/2, -a/2 )BD: from B (a,0,0 ) to D (0,a,0 ). Direction vector (-a, a, 0 )Vector b1 - b2 = E - B = (a/2 - a, a/2 - 0, a/2 - 0 ) = (-a/2, a/2, a/2 )Cross product v1 × v2:As calculated earlier, ( a²/2, a²/2, -a² )Dot product of (b1 - b2) and (v1 × v2):(-a/2)(a²/2 ) + (a/2)(a²/2 ) + (a/2)(-a² )Which is:(-a³/4 ) + (a³/4 ) - (a³/2 ) = 0 - a³/2 = -a³/2Absolute value is a³/2Magnitude of cross product:sqrt( (a²/2 )^2 + (a²/2 )^2 + (-a² )^2 ) = sqrt( a^4 /4 + a^4 /4 + a^4 ) = sqrt( a^4 /2 + a^4 ) = sqrt( 3a^4 / 2 ) = a² sqrt(3/2 )Therefore, distance is (a³ / 2 ) / (a² sqrt(3/2 )) ) = (a / 2 ) / sqrt(3/2 ) = (a / 2 ) * sqrt(2/3 ) = a sqrt(2/3 ) / 2 = a sqrt(6)/6Yes, so the distance is ( frac{sqrt{6}}{6} a ). So that seems correct.But let me think again. Alternatively, maybe there's a different approach. For skew lines, another way is to find a vector that is perpendicular to both lines and then compute the projection.Alternatively, find the shortest distance which is along the common perpendicular.But the formula we used is standard and gives the correct answer.Alternatively, let me check with vectors.Suppose we have two skew lines:Line 1: r = b1 + t v1Line 2: r = b2 + s v2The distance is |(b2 - b1) ⋅ (v1 × v2)| / |v1 × v2|Wait, in our case, b1 is E, which is on line EA, and b2 is B, on BD. But in the formula, it's (b2 - b1) ⋅ (v1 × v2). Wait, but we used (b1 - b2) ⋅ (v1 × v2). Since dot product is commutative, the sign doesn't matter because we take absolute value. So our calculation was correct.Therefore, the distance is indeed ( frac{sqrt{6}}{6} a ), which can be written as ( frac{a}{sqrt{6}} ) but rationalized.Thus, the answer should be ( boxed{dfrac{asqrt{6}}{6}} ).But just to make sure, let's approach this another way. Let's parametrize the lines and find the minimum distance.For line EA: parameter t, point P(t) = (a/2 - (a/2)t, a/2 - (a/2)t, a/2 - (a/2)t )For line BD: parameter s, point Q(s) = (a - a s, a s, 0 )We need to find the minimum distance between P(t) and Q(s). The distance squared is:[ (a/2 - (a/2)t - (a - a s) )^2 + (a/2 - (a/2)t - a s )^2 + (a/2 - (a/2)t - 0 )^2 ]Simplify each component:First component (x-coordinate difference):a/2 - (a/2)t - a + a s = -a/2 - (a/2)t + a sSecond component (y-coordinate difference):a/2 - (a/2)t - a sThird component (z-coordinate difference):a/2 - (a/2)tSo distance squared D² is:[ (-a/2 - (a/2)t + a s )^2 + (a/2 - (a/2)t - a s )^2 + (a/2 - (a/2)t )^2 ]Let me factor out a/2 where possible:First term:(-a/2)(1 + t - 2 s ) ) → Wait, let's see:Original expression: -a/2 - (a/2)t + a s = - (a/2)(1 + t ) + a s = a( s - (1 + t)/2 )Second term:a/2(1 - t ) - a s = a/2(1 - t ) - a s = a( (1 - t )/2 - s )Third term:a/2(1 - t )Let me denote variables as u = t, v = s for simplicity.But perhaps it's easier to substitute variables. Let’s set u = t and v = s for parameters.But maybe we can let t and s be variables and take partial derivatives to find minimum.Alternatively, set up equations for the vector PQ being perpendicular to both direction vectors v1 and v2.Because the shortest distance occurs when the vector connecting P(t) and Q(s) is perpendicular to both lines.Therefore, vector PQ ⋅ v1 = 0 and PQ ⋅ v2 = 0.Let me compute PQ:PQ = Q(s) - P(t) = (a - a s - (a/2 - (a/2)t ), a s - (a/2 - (a/2)t ), 0 - (a/2 - (a/2)t ) )Compute each component:x: a - a s - a/2 + (a/2)t = a/2 - a s + (a/2)ty: a s - a/2 + (a/2)tz: -a/2 + (a/2)tSo PQ = ( a/2 - a s + (a/2)t, a s - a/2 + (a/2)t, -a/2 + (a/2)t )Direction vector of line EA is v1 = (-a/2, -a/2, -a/2 )Direction vector of line BD is v2 = (-a, a, 0 )So we need PQ ⋅ v1 = 0 and PQ ⋅ v2 = 0.First, PQ ⋅ v1:= ( a/2 - a s + (a/2)t )*(-a/2 ) + ( a s - a/2 + (a/2)t )*(-a/2 ) + ( -a/2 + (a/2)t )*(-a/2 )Let’s compute each term:First term:( a/2 - a s + (a/2)t )*(-a/2 ) = (-a²/4 + (a² s)/2 - (a² t)/4 )Second term:( a s - a/2 + (a/2)t )*(-a/2 ) = (-a² s /2 + a² /4 - (a² t)/4 )Third term:( -a/2 + (a/2)t )*(-a/2 ) = (a² /4 - (a² t)/4 )Sum all terms:First term: -a²/4 + a² s /2 - a² t /4Second term: -a² s /2 + a² /4 - a² t /4Third term: a² /4 - a² t /4Combine like terms:- a²/4 + a² s /2 - a² t /4 - a² s /2 + a² /4 - a² t /4 + a² /4 - a² t /4Simplify:- a²/4 + a²/4 + a²/4 (constants) = a²/4For the s terms: a² s /2 - a² s /2 = 0For the t terms: -a² t /4 -a² t /4 -a² t /4 = - (3a² t ) /4Thus, overall:a² /4 - (3a² t ) /4 = 0So,a² /4 - (3a² t ) /4 = 0 → Multiply both sides by 4/a²:1 - 3t = 0 → t = 1/3Now, compute PQ ⋅ v2 = 0.Vector PQ is the same as above with t = 1/3, but since we need to solve for s and t, let's do the dot product first.PQ ⋅ v2 = ( a/2 - a s + (a/2)t )*(-a ) + ( a s - a/2 + (a/2)t )*(a ) + ( -a/2 + (a/2)t )*0Compute each term:First term:( a/2 - a s + (a/2)t )*(-a ) = -a²/2 + a² s - (a² t )/2Second term:( a s - a/2 + (a/2)t )*(a ) = a² s - a²/2 + (a² t )/2Third term: 0Sum the terms:- a²/2 + a² s - (a² t )/2 + a² s - a²/2 + (a² t )/2Simplify:- a²/2 - a²/2 + a² s + a² s + (- (a² t )/2 + (a² t )/2 )= -a² + 2a² s + 0Set equal to 0:- a² + 2a² s = 0 → 2a² s = a² → s = 1/2So we have t = 1/3 and s = 1/2.Therefore, the points of closest approach are:On EA: P(1/3 ) = (a/2 - (a/2)(1/3 ), a/2 - (a/2)(1/3 ), a/2 - (a/2)(1/3 )) = (a/2 - a/6, a/2 - a/6, a/2 - a/6 ) = (a/3, a/3, a/3 )On BD: Q(1/2 ) = (a - a*(1/2 ), a*(1/2 ), 0 ) = (a/2, a/2, 0 )Now, compute the distance between these two points:P(1/3 ) = (a/3, a/3, a/3 )Q(1/2 ) = (a/2, a/2, 0 )Difference vector: (a/3 - a/2, a/3 - a/2, a/3 - 0 ) = (-a/6, -a/6, a/3 )Distance squared: (-a/6 )² + (-a/6 )² + (a/3 )² = (a²/36 + a²/36 + a²/9 ) = (2a²/36 + 4a²/36 ) = 6a²/36 = a²/6Therefore, distance is sqrt(a²/6 ) = a / sqrt(6 ) = a sqrt(6 ) / 6Same as before. Therefore, confirms that the distance is ( dfrac{asqrt{6}}{6} ).Therefore, the final answer is ( boxed{dfrac{asqrt{6}}{6}} ).Final AnswerThe distance between the lateral edge and the diagonal of the base is boxed{dfrac{asqrt{6}}{6}}.

✅:To find the distance between a lateral edge and a diagonal of the base which is skew to it in a regular tetrahedral pyramid with a square base of side length (a) and a lateral face forming a (45^circ) angle with the base, we proceed as follows:1. Determine the height of the pyramid: - The lateral face forms a (45^circ) angle with the base. Using the dihedral angle formula, we find the height (h) of the pyramid. - The normal vectors of the base (xy-plane) and a lateral face (EAB) are used to compute the angle between them. - The height (h) is found to be ( frac{a}{2} ).2. Set up coordinates: - The base vertices are (A(0, 0, 0)), (B(a, 0, 0)), (C(a, a, 0)), (D(0, a, 0)), and the apex (Eleft(frac{a}{2}, frac{a}{2}, frac{a}{2}right)).3. Parametrize the lines: - Lateral edge (EA) is parametrized from (Eleft(frac{a}{2}, frac{a}{2}, frac{a}{2}right)) to (A(0, 0, 0)) with direction vector ((-1, -1, -1)) scaled by (frac{a}{2}). - Base diagonal (BD) is parametrized from (B(a, 0, 0)) to (D(0, a, 0)) with direction vector ((-1, 1, 0)) scaled by (a).4. Calculate the distance between skew lines: - Using the formula for the distance between skew lines, we compute the cross product of the direction vectors and the dot product with the vector between points on each line. - The distance is found to be ( frac{sqrt{6}}{6} a ).The final answer is:[boxed{dfrac{asqrt{6}}{6}}]

🔑:Given:1. A right four-sided pyramid with a base side length (a).2. The side face of the pyramid forms an angle of (45^circ) with the base plane.3. We need to find the distance between a side edge and the skew diagonal of the base.Let's consider the pyramid (A B C D P) with vertex (P). Here, (A B = B C = C D = A D = a), and (M) is the center of the square base (A B C D). Let (K) be the midpoint of (A B). Since (P K perp A B) and (M K perp A B), the angle (P K M) is the linear angle between the plane containing the side face (A B P) and the plane of the base (A B C D). Given (angle P K M = 45^circ), we infer that (angle P K M = 45^circ).Because the pyramid is a regular one, its height passes through the center of the base, making (P M) the height of the pyramid. Consider the right isosceles triangle (P K M):[P M = M K = frac{a}{2}]Now, let (alpha) be the angle between the side edge and the base plane, i.e., (angle M A P). Through the right triangle property,[tan alpha = frac{P M}{A M} = frac{frac{a}{2}}{frac{asqrt{2}}{2}} = frac{1}{sqrt{2}}]This gives us: [alpha = arctan left( frac{1}{sqrt{2}} right)]Using trigonometric identities,[cos alpha = frac{1}{sqrt{1 + left( frac{1}{sqrt{2}} right)^2}} = sqrt{frac{2}{3}}, quad sin alpha = frac{1}{sqrt{3}}]Let's define (F) as the foot of the perpendicular dropped from (M) onto the diagonal (P C) of the base. The projection of the line segment (M F) on the plane of the base is orthogonal to (B D) (by the three perpendicular theorem). Thus, (M F perp B D), and (M F) is the common perpendicular to the skew lines (P C) and (B D). Considering the right triangle (C M F):[M F = C M sin alpha = frac{asqrt{2}}{2} cdot frac{1}{sqrt{3}} = frac{a sqrt{6}}{6}]Thus, the distance between the side edge and the skew diagonal of the base is:[boxed{frac{a sqrt{6}}{6}}]

🧠:Okay, so I have this equation to solve: (lg^{2} x - [lg x] - 2 = 0). I need to find the number of real solutions. Let me break this down step by step. First, let me make sure I understand the notation. The problem uses (lg x), which I believe is the logarithm base 10 of x, right? So (lg x = log_{10} x). The square here is on the logarithm, so (lg^{2} x) means ((log_{10} x)^2). The other part is ([lg x]), which is the greatest integer less than or equal to (lg x). That's the floor function, denoted with square brackets here. So, for example, if (lg x = 2.3), then ([lg x] = 2).The equation is ((log_{10} x)^2 - lfloor log_{10} x rfloor - 2 = 0). I need to find how many real x satisfy this equation. Since x must be positive (because logarithm is only defined for positive numbers), x > 0.Let me let (y = lg x), so the equation becomes (y^2 - lfloor y rfloor - 2 = 0). Now, I need to solve for y first, then translate back to x. The floor function complicates things, but maybe I can consider different intervals for y where (lfloor y rfloor) is constant. For any real y, we can write (y = n + t), where (n = lfloor y rfloor) is an integer, and (t in [0, 1)). So substituting into the equation: (y^2 - n - 2 = 0). So, (y^2 = n + 2). But since (y = n + t), we have ((n + t)^2 = n + 2). Expanding this: (n^2 + 2nt + t^2 = n + 2).So, rearranging: (n^2 + 2nt + t^2 - n - 2 = 0). Hmm. But since n is an integer and t is in [0,1), perhaps for each integer n, we can solve for t in [0,1), and check if the equation is satisfied. Then, the possible values of y would be n + t, and each such y corresponds to x = 10^y.Therefore, the strategy is: for each integer n, solve the equation (y^2 - n - 2 = 0) where y is in [n, n+1). Then, count the number of real solutions y, each of which gives a real solution x = 10^y.So let me consider different integer values of n. The equation is y^2 = n + 2. So, for each n, we can have y = sqrt(n + 2) or y = -sqrt(n + 2). But since y must be in [n, n + 1), we need to check whether sqrt(n + 2) or -sqrt(n + 2) lies within [n, n + 1). Wait, but sqrt(n + 2) is positive, and -sqrt(n + 2) is negative. So depending on n, maybe we can split into cases where n is positive or negative. Let's first consider possible n such that sqrt(n + 2) is in [n, n + 1).Case 1: Let's first consider the positive roots. So, y = sqrt(n + 2). For this to be in [n, n + 1), we need n <= sqrt(n + 2) < n + 1. Let's solve these inequalities for integer n.First inequality: n <= sqrt(n + 2). Squaring both sides (since both sides are non-negative if n is non-negative; but if n is negative, sqrt(n + 2) is real only if n + 2 >= 0, i.e., n >= -2. So possible n are integers >= -2. But sqrt(n + 2) is non-negative, so if n is negative, we have sqrt(n + 2) >= 0, but the left side n could be negative. So let's check for n >= -2.Let me split into subcases:Subcase 1: n >= 0. Then, sqrt(n + 2) is real and non-negative. So the inequality n <= sqrt(n + 2). Let's square both sides: n^2 <= n + 2. So, n^2 - n - 2 <= 0. Solving quadratic equation n^2 - n - 2 = 0: n = [1 ± sqrt(1 + 8)] / 2 = [1 ± 3]/2. So n = 2 or n = -1. So the inequality n^2 - n - 2 <= 0 holds for n between -1 and 2. But since we are in Subcase 1 where n >= 0, the integers n = 0, 1, 2. However, the original inequality n <= sqrt(n + 2). Let's check for n=0: 0 <= sqrt(2) ≈ 1.414, which is true. For n=1: 1 <= sqrt(3) ≈ 1.732, true. For n=2: 2 <= sqrt(4) = 2, which is true (since 2 <= 2). For n=3: 3 <= sqrt(5) ≈ 2.236, which is false. So n=0,1,2 are possible. So for n=0: Check if sqrt(0 + 2) = sqrt(2) ≈ 1.414 is in [0,1). But 1.414 is not in [0,1). So no solution here. For n=1: sqrt(1 + 2) = sqrt(3) ≈ 1.732. Check if it's in [1,2). Yes, it is. So y = sqrt(3) is in [1,2). So that's a valid solution. For n=2: sqrt(2 + 2) = sqrt(4) = 2. Check if 2 is in [2,3). Yes, since [2,3) includes 2 but not 3. So y=2 is a solution. So for positive roots, possible n=1 and n=2.Now, check the other inequality sqrt(n + 2) < n + 1. Let's verify for n=1: sqrt(3) ≈ 1.732 < 2? Yes. For n=2: sqrt(4) = 2 < 3? Yes. So both are valid.Now, check n=-2 and n=-1, since earlier we saw n >= -2. Let's check Subcase 2: n = -2. Then sqrt(n + 2) = sqrt(0) = 0. Check if 0 is in [-2, -1). 0 is not in that interval. For n=-1: sqrt(-1 + 2) = sqrt(1) = 1. Check if 1 is in [-1,0). No. So no solutions here.Therefore, for positive roots, possible n=1 and n=2, giving solutions y= sqrt(3) and y=2. So that's two solutions from positive roots.Case 2: Now consider negative roots. So y = -sqrt(n + 2). For y to be in [n, n + 1). But since y is negative here (because sqrt(n + 2) is non-negative, so -sqrt(n + 2) is non-positive). Let's see for which n this is possible.First, note that y = -sqrt(n + 2). Since sqrt(n + 2) is real, n + 2 >= 0, so n >= -2. So possible n are integers >= -2. Also, y must be in [n, n + 1). So we have n <= -sqrt(n + 2) < n + 1. Let's analyze these inequalities.First inequality: n <= -sqrt(n + 2). Since n is an integer >= -2, and y is negative or zero. Let's split into subcases:Subcase 1: n >= 0. Then, the right side -sqrt(n + 2) is negative, but n >=0, so n <= negative number? That can't happen. So no solutions for n >=0.Subcase 2: n = -2, -1. Let's check n=-2 first.For n=-2: y = -sqrt(-2 + 2) = -sqrt(0) = 0. Check if 0 is in [-2, -1). No. So no solution here.For n=-1: y = -sqrt(-1 + 2) = -sqrt(1) = -1. Check if -1 is in [-1, 0). Yes, since -1 is included in [-1, 0). So y=-1 is a solution here. So n=-1 gives y=-1.Check the inequalities for n=-1:First inequality: n <= -sqrt(n + 2) => -1 <= -sqrt(-1 + 2) => -1 <= -1. Since sqrt(1)=1, so -1 <= -1, which is true. Second inequality: -sqrt(n + 2) < n + 1 => -sqrt(1) < -1 + 1 => -1 < 0. Which is true. So this works.Any other n? Let's check n= -1 is the only possible here. What about n=-3? Wait, n >= -2, so n=-2 and -1. So only n=-1 gives a solution here.Therefore, from negative roots, we have y=-1, which corresponds to n=-1. So that's one solution from negative roots.So in total, from the equation y^2 - [y] - 2 = 0, we have solutions at y = sqrt(3) ≈1.732 (n=1), y=2 (n=2), and y=-1 (n=-1). Each of these y values corresponds to x via x = 10^y. So let's check each:For y = sqrt(3): x = 10^{sqrt(3)} ≈10^1.732 ≈ 53.957.For y = 2: x = 10^2 = 100.For y = -1: x = 10^{-1} = 0.1.So that's three solutions. But wait, is that all? Wait, perhaps there are more solutions. Let me check again. Because maybe in some intervals, there could be two solutions? Wait, the original equation is y^2 - n - 2 = 0, where n = floor(y). For each n, this is a quadratic in y. Maybe there are two solutions in some intervals? Wait, let's think again.Wait, in my previous approach, I considered y = sqrt(n + 2) and y = -sqrt(n + 2). But depending on the interval [n, n+1), maybe both roots could lie in that interval for some n?Wait, for example, suppose n is such that both sqrt(n + 2) and -sqrt(n + 2) lie in [n, n + 1). But given that sqrt(n + 2) is positive and -sqrt(n + 2) is negative, it's possible only if n is negative. Let me check.Wait, let's take n=-1. Then sqrt(n + 2) = sqrt(1) = 1, and -sqrt(1) = -1. So y = -1 is in [n, n +1) = [-1, 0). But y=1 is not in [-1,0). So only the negative root is valid here.Alternatively, for some n, maybe both roots lie in the interval? For example, if n is negative, but let's see. Suppose n=-3. But n >= -2. So n=-2: sqrt(0) = 0, and -0 = 0. But 0 is not in [-2, -1). So no. For n=-1, we've checked. For n=0: sqrt(2) ≈1.414 and -sqrt(2) ≈-1.414. The interval [0,1). sqrt(2) is not in [0,1), and -sqrt(2) is not in [0,1). So no solutions here. For n=1: sqrt(3)≈1.732 and -sqrt(3)≈-1.732. The interval [1,2). sqrt(3) is in [1,2), but -sqrt(3) is not. So only positive root. For n=2: sqrt(4)=2 and -2. The interval [2,3). sqrt(4)=2 is there, but -2 is not. So only positive root.So in each interval, only one solution can come from either the positive or negative root. So in total, from my previous analysis, we have three solutions: y=-1, sqrt(3), and 2, leading to x=0.1, ~53.957, and 100. Therefore, three real solutions. Wait, but the problem is asking for the number of real solutions. But maybe I missed something.Wait, let's verify each of these solutions:1. For y=-1 (x=0.1): Let's check the original equation. (lg^{2} x - [lg x] -2 = (-1)^2 - [-1] -2 = 1 - (-1) -2 = 1 +1 -2 =0. Yes, works.2. For y=sqrt(3) (x=10^{sqrt(3)}): [lg x] = [sqrt(3)] ≈1.732, so floor is 1. Then equation: (sqrt(3))^2 -1 -2 = 3 -1 -2=0. Works.3. For y=2 (x=100): [lg x] = [2] =2. Equation: 2^2 -2 -2=4 -2 -2=0. Works.So these three solutions are valid. But wait, maybe there are more? Let me check other possible n.Wait, let me consider n=3. For n=3, the equation would be y^2 = 3 + 2 =5, so y= sqrt(5)≈2.236 or -sqrt(5). Check if sqrt(5) is in [3,4). No, since sqrt(5)≈2.236 <3. Similarly, -sqrt(5)≈-2.236, check if in [3,4)? No. So no solution for n=3.n=4: y= sqrt(6)≈2.449, which is not in [4,5). Similarly, no.Negative n beyond n=-1: n=-2. We saw earlier that y=0, which is not in [-2, -1). n=-3 not allowed since n >=-2.Wait, maybe n=0. Let's check n=0 again. The equation is y^2 =0 +2=2. So y= sqrt(2)≈1.414 or -sqrt(2)≈-1.414. Check if sqrt(2) is in [0,1). No. Check if -sqrt(2) is in [0,1). No. So no solutions here.Hmm. So only three solutions. But wait, let me think again. Maybe I missed that for a given n, the quadratic equation could have two solutions in the interval [n, n+1). For example, if n is such that both roots of y^2 = n + 2 lie in [n, n +1). But given that y = sqrt(n + 2) and y = -sqrt(n + 2), unless n is negative and the interval includes both positive and negative numbers. Wait, intervals [n, n +1) for n negative can include both negative and positive numbers if n +1 >0. For example, if n=-1, the interval is [-1, 0). If n=0, it's [0,1). But for n=-1, the negative root y=-1 is in [-1,0). The positive root y=1 is not. For n=-2, interval is [-2, -1), which is entirely negative. So maybe in some intervals both roots could be present? Wait, but sqrt(n + 2) must be positive, so y = sqrt(n + 2) would only be in an interval [n, n +1) if n <= sqrt(n +2) <n +1. Similarly, the negative root y=-sqrt(n +2) would be in [n, n +1) if n <= -sqrt(n +2) <n +1.Wait, let's take n=-1.5, but n must be integer, so n=-1. For n=-1, y=-sqrt(1) = -1, which is in [-1, 0). If n=-0.5, but n must be integer. So, only integer n's. So the only case where the negative root is in the interval is n=-1. For other n, the negative root is either outside the interval or the interval is positive. Similarly, positive roots can only be in intervals where n is positive or zero, but as we saw, only n=1 and 2 yield positive roots in their intervals.Wait, but let's think differently. Maybe for some n, both positive and negative roots lie in the interval [n, n+1). But since the interval [n, n+1) is a half-open interval of length 1, and the positive and negative roots are separated by 2*sqrt(n +2), which is at least 2*sqrt(0 +2)=2*sqrt(2)≈2.828 for n=0. So unless n is such that the interval is wide enough to contain both roots, but since each interval is only length 1, and the distance between roots is 2*sqrt(n +2), which is increasing as n increases. So probably impossible. Hence, each interval can contain at most one root.Therefore, only three solutions: y=-1, sqrt(3), and 2. Therefore, three real solutions. But wait, wait. Wait, when I let y = n + t, where n is integer and t ∈ [0,1), then for each n, we have the equation (n + t)^2 - n - 2 =0. So expanding, n² + 2nt + t² -n -2=0. So this is quadratic in t: t² + 2n t + (n² -n -2) =0. So solving for t:t = [-2n ± sqrt(4n² -4*1*(n² -n -2))]/2Simplify discriminant:sqrt(4n² -4(n² -n -2)) = sqrt(4n² -4n² +4n +8) = sqrt(4n +8) = 2*sqrt(n +2)Thus, t = [-2n ± 2*sqrt(n +2)]/2 = -n ± sqrt(n +2)So t = -n + sqrt(n +2) or t = -n - sqrt(n +2)But since t ∈ [0,1), we require 0 <= t <1.So, for each n, check if t is in [0,1):Case 1: t = -n + sqrt(n +2)Case 2: t = -n - sqrt(n +2)So for each integer n, check both t's.But since sqrt(n +2) is real, n +2 >=0 => n >= -2.So possible n >= -2.Let's analyze both cases for each n:Start with n=-2:n=-2:t1 = -(-2) + sqrt(-2 +2) = 2 + 0 =2. Not in [0,1).t2 = -(-2) - sqrt(0) =2 -0=2. Not in [0,1). So no solutions.n=-1:t1 = -(-1) + sqrt(-1 +2)=1 +1=2. Not in [0,1).t2= -(-1) - sqrt(1)=1 -1=0. t=0 is in [0,1). So t=0. Thus, y = n + t = -1 +0= -1. So this is the solution we found earlier.n=0:t1= -0 + sqrt(0 +2)=sqrt(2)≈1.414. Not in [0,1).t2= -0 - sqrt(2)≈-1.414. Not in [0,1). So no solutions.n=1:t1= -1 + sqrt(1 +2)= -1 + sqrt(3)≈-1 +1.732≈0.732. Which is in [0,1). So valid.t2= -1 - sqrt(3)≈-1 -1.732≈-2.732. Not in [0,1). So only t=0.732. So y =1 +0.732≈1.732= sqrt(3).n=2:t1= -2 + sqrt(2 +2)= -2 +2=0. Which is in [0,1). So t=0. So y=2 +0=2.t2= -2 -2= -4. Not in [0,1).n=3:t1= -3 + sqrt(5)≈-3 +2.236≈-0.764. Not in [0,1).t2= -3 - sqrt(5)≈-5.236. Not in [0,1). No solutions.n=4:t1= -4 + sqrt(6)≈-4 +2.449≈-1.551. Not in [0,1). Similarly, t2 is worse.Similarly, for n >=3, t1 and t2 will be negative or not in [0,1). So no solutions.Therefore, the solutions are:For n=-1: y=-1.For n=1: y≈1.732.For n=2: y=2.Thus, three solutions. Therefore, three real solutions for x: 10^{-1}=0.1, 10^{sqrt(3)}, and 10^{2}=100. But wait, the problem is asking for the number of real solutions. So the answer is 3. But let me confirm once again. Is there any other n that I missed? For example, n=3, but as above, no. n=-1,0,1,2,3 etc. Checked all from n=-2 upwards. So only three solutions. Therefore, the answer should be 3. However, wait, let me check again for n=1:When n=1, t= -1 + sqrt(3) ≈0.732, which is in [0,1), so y=1 +0.732≈1.732. Which is correct.For n=2, t=0, which is allowed, since t ∈ [0,1). So y=2+0=2.For n=-1, t=0, so y=-1 +0=-1.No other solutions. Therefore, three real solutions. So the answer is 3. But wait, wait. Wait, is there a possible case where for the same n, both t1 and t2 are in [0,1)? For example, can there be two solutions in the same interval?Let me check for n=1: t1≈0.732 and t2≈-2.732. Only t1 is valid. For n=0: t1≈1.414 and t2≈-1.414: neither in [0,1). For n=2: t1=0 and t2=-4: only t1=0. For n=3: t1≈-0.764, etc. So no, for no n do both t1 and t2 fall into [0,1). Therefore, only three solutions. So the answer should be 3.But wait, let me check the original equation again. Wait, in the original substitution, y = lg x. So x =10^y. Each real solution y corresponds to exactly one real solution x. So if there are three real y's, then three real x's. Therefore, the number of real solutions is 3. However, I have a feeling that maybe there's a mistake here. Let me check some references.Wait, let's plug x=100 into the original equation: (lg^{2} 100 - [lg 100] -2 = (2)^2 - [2] -2 =4 -2 -2=0. Correct.x=0.1: (lg^{2} 0.1 - [lg 0.1] -2 = (-1)^2 - [-1] -2=1 - (-1) -2=0. Correct.x=10^{sqrt(3)}: (lg x = sqrt(3), so [lg x] =1. Then equation: (sqrt(3))^2 -1 -2=3 -1 -2=0. Correct.So all three solutions work. Therefore, three real solutions. But I recall that sometimes floor function equations can have more solutions, so maybe I missed something. Let me check for n=1. When n=1, the interval is [1,2). The solution y≈1.732 is in [1,2). But sqrt(3) is ≈1.732, which is in [1,2). So correct. For n=2, the solution y=2 is in [2,3). Wait, but 2 is the left endpoint. But floor(y)=2 when y is in [2,3). So correct.Alternatively, let's graph the functions to see. The equation is y² - floor(y) -2=0. Let me sketch y² -2 and floor(y). The intersections would be the solutions. For y in different intervals:For y in [-1,0): floor(y)=-1. So equation becomes y² - (-1) -2= y² -1=0 => y=±1. But y must be in [-1,0). So y=-1 is in [-1,0). So solution at y=-1.For y in [0,1): floor(y)=0. Equation: y² -0 -2= y² -2=0 => y=±sqrt(2). But sqrt(2)≈1.414 not in [0,1). Negative root -sqrt(2)≈-1.414 not in [0,1). So no solution.For y in [1,2): floor(y)=1. Equation: y² -1 -2= y² -3=0 => y=±sqrt(3). sqrt(3)≈1.732 in [1,2). Negative root not. So solution y=sqrt(3).For y in [2,3): floor(y)=2. Equation: y² -2 -2= y² -4=0 => y=±2. y=2 is in [2,3). So solution y=2.For y >=3: floor(y)>=3. Equation: y² -3 -2= y² -5=0 => y=±sqrt(5)≈±2.236. But y >=3, so no solution.For y < -1: floor(y)<=-2. Equation: y² - (-2) -2= y²=0 => y=0. But y < -1, so no solution.Thus, the solutions are at y=-1, sqrt(3), and 2. So three solutions. Therefore, the number of real solutions is 3.But wait, according to this analysis, in each interval [n, n+1), we check if the equation y² -n -2=0 has solutions. For n=-1: solution at y=-1. For n=1: solution at y=sqrt(3). For n=2: solution at y=2. So three solutions. So answer is 3. However, in some references, similar problems might have more solutions, so I need to be careful.Wait, perhaps I missed an interval. Let's check n=3, but no. For n=3, the equation is y²=5, which would be y≈2.236, but that's in [3,4)? No. So no solution. Similarly, n=-2: equation y²=0, but y=0 not in [-2,-1). So no. Therefore, three solutions. Therefore, the answer should be 3. But the problem says "number of real solutions", so I need to confirm once more.Alternatively, suppose the equation is (lg^2x - lfloor lg x rfloor -2 =0). Let me consider the graph of the function f(x) = (lg x)^2 - floor(lg x) -2. The number of times this crosses zero is the number of solutions.Let me analyze the behavior in different intervals of x determined by powers of 10.For x in [10^n, 10^{n+1}), where n is integer, then lg x is in [n, n+1), so floor(lg x)=n. Thus, in this interval, the equation becomes (lg x)^2 -n -2=0. So lg x = sqrt(n +2) or lg x=-sqrt(n +2). But lg x is in [n, n+1). So we need sqrt(n +2) ∈ [n, n+1) or -sqrt(n +2) ∈ [n, n+1).So same as before. So for each interval [10^n,10^{n+1}), check if sqrt(n +2) ∈ [n, n+1) or -sqrt(n +2) ∈ [n, n+1).For each integer n:If sqrt(n +2) ∈ [n, n+1):- Which requires n <= sqrt(n +2) <n +1.Similarly, if -sqrt(n +2) ∈ [n, n+1):- Which requires n <= -sqrt(n +2) <n +1.So let's check for which n these are possible.First case: sqrt(n +2) ∈ [n, n+1). We did this before, for n=1 and 2.Second case: -sqrt(n +2) ∈ [n, n+1). Let's solve this.This requires n <= -sqrt(n +2) <n +1.Note that sqrt(n +2) is non-negative, so -sqrt(n +2) is non-positive. So the interval [n, n+1) must contain a non-positive number. So n must be <=0.Moreover, n must satisfy n <= -sqrt(n +2) <n +1.Let's solve for integer n <=0.Start with n=0:Check if 0 <= -sqrt(0 +2) <1. But -sqrt(2)≈-1.414. So -1.414 is not in [0,1). No.n=-1:Check if -1 <= -sqrt(-1 +2)= -1 <= -1 <0. So -1 <= -1 is true, and -1 <0 is true. So yes. So -sqrt(n +2)= -sqrt(1)= -1 ∈ [-1,0). So yes. Which gives solution y=-1.n=-2:Check if -2 <= -sqrt(-2 +2)= -sqrt(0)=0 < -1. 0 is not in [-2,-1). So no.So only n=-1 gives a solution in this case.Therefore, in each interval [10^n,10^{n+1}), we have:- For n=-1: solution at y=-1, x=0.1.- For n=1: solution at y=sqrt(3), x=10^{sqrt(3)}.- For n=2: solution at y=2, x=100.Therefore, three solutions. Hence, the number of real solutions is 3.But wait, another way to think about it is to analyze the equation in different intervals. For example:When x ∈ [10^n,10^{n+1}), floor(lg x)=n. So the equation becomes (lg x)^2 -n -2=0.Solve for lg x: (lg x)^2 =n +2. So lg x=±sqrt(n +2). But since lg x ∈ [n, n+1), we need:Either sqrt(n +2) ∈ [n, n+1) or -sqrt(n +2) ∈ [n, n+1).Therefore, the solutions are when:1. sqrt(n +2) ∈ [n, n+1). Which happens for n=1,2.2. -sqrt(n +2) ∈ [n, n+1). Which happens for n=-1.Hence, three solutions. Therefore, the answer is 3. However, I just recalled that when n=2, sqrt(n +2)=2, which is the left endpoint of [2,3). Since the interval is [2,3), which includes 2, it is valid. Similarly, when n=-1, -sqrt(n +2)= -1, which is the left endpoint of [-1,0), which is included. So these are valid. Therefore, three solutions. But now I'm a bit confused because sometimes people might think floor(lg x) is the integer part, but in this case, the analysis seems consistent. So I think the answer is three. However, I remember once solving a similar problem and getting 4 solutions. Maybe I'm missing something here.Wait, let's check another angle. For example, consider x=10^{sqrt(3)}. This is approximately 10^{1.732} ≈53.957. So this x is in [10^1,10^2)= [10,100). Therefore, in this interval, floor(lg x)=1. So the equation becomes (lg x)^2 -1 -2=0 => (lg x)^2=3 => lg x=±sqrt(3). But since x is in [10,100), lg x is in [1,2), so only positive sqrt(3)≈1.732 is valid. So that's correct.x=100 is in [100,1000), floor(lg x)=2. Equation becomes (lg x)^2 -2 -2=0 => (lg x)^2=4 => lg x=±2. But lg x is in [2,3). So only +2 is valid. Correct.x=0.1 is in [0.1,1), floor(lg x)= -1. Equation: (lg x)^2 -(-1) -2= (lg x)^2 +1 -2= (lg x)^2 -1=0 => lg x=±1. But lg x is in [-1,0). So only -1 is valid. Correct.So three solutions. Therefore, three real solutions. Therefore, the answer is 3. So I think this is the correct answer. Therefore, the number of real solutions is three.Final Answerboxed{3}

✅:To solve the equation (lg^{2} x - [lg x] - 2 = 0), we start by letting (y = lg x). This transforms the equation into (y^2 - lfloor y rfloor - 2 = 0), where (lfloor y rfloor) denotes the greatest integer less than or equal to (y).We analyze the equation by considering different intervals for (y) where (lfloor y rfloor) is constant. For each integer (n), we solve the equation (y^2 - n - 2 = 0) within the interval ([n, n+1)).1. Case (n = -1): - The equation becomes (y^2 - (-1) - 2 = y^2 - 1 = 0), giving (y = pm 1). - Only (y = -1) lies in the interval ([-1, 0)). This corresponds to (x = 10^{-1} = 0.1).2. Case (n = 1): - The equation becomes (y^2 - 1 - 2 = y^2 - 3 = 0), giving (y = pm sqrt{3}). - Only (y = sqrt{3} approx 1.732) lies in the interval ([1, 2)). This corresponds to (x = 10^{sqrt{3}}).3. Case (n = 2): - The equation becomes (y^2 - 2 - 2 = y^2 - 4 = 0), giving (y = pm 2). - Only (y = 2) lies in the interval ([2, 3)). This corresponds to (x = 10^2 = 100).After verifying all possible intervals for (n), we find three valid solutions for (y), which translate to three real solutions for (x). Therefore, the number of real solutions to the equation is:[boxed{3}]

🔑:1. Define ( [x] ) as the greatest integer less than or equal to a real number ( x ). Let's start by writing the given equation: [ (lg x)^2 - [lg x] - 2 = 0 ]2. By using the inequality ( [lg x] leq lg x ), we can state: [ (lg x)^2 - lg x - 2 leq 0 ]3. Solve the inequality: [ (lg x)^2 - lg x - 2 leq 0 ] This is a quadratic inequality in ( lg x ). Solve the corresponding quadratic equation: [ (lg x)^2 - lg x - 2 = 0 ] Using the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) where ( a = 1 ), ( b = -1 ), and ( c = -2 ): [ lg x = frac{-(-1) pm sqrt{(-1)^2 - 4 cdot 1 cdot (-2)}}{2 cdot 1} ] [ lg x = frac{1 pm sqrt{1 + 8}}{2} ] [ lg x = frac{1 pm 3}{2} ] Hence: [ lg x = 2 quad text{or} quad lg x = -1 ]4. Determine the interval for ( lg x ): [ -1 leq lg x leq 2 ] Consider different intervals: - When ( -1 leq lg x < 0 ): [ [lg x] = -1 ] Substitute ( [lg x] = -1 ) back into the original equation: [ (lg x)^2 + 1 - 2 = 0 ] [ (lg x)^2 - 1 = 0 ] [ lg x = pm 1 ] [ lg x = 1 quad (text{not valid as } lg x in [-1, 0)) ] [ lg x = -1 quad (text{valid}) ] Hence: [ x = 10^{-1} = frac{1}{10} ] - When ( 0 leq lg x < 1 ): [ [lg x] = 0 ] Substitute ( [lg x] = 0 ) back into the original equation: [ (lg x)^2 - 2 = 0 ] [ (lg x)^2 = 2 ] [ lg x = pm sqrt{2} ] [ lg x = sqrt{2} quad text{(not valid as } lg x in [0, 1)) ] ] [ lg x = -sqrt{2} quad text{(not valid as this value does not fall into } [0, 1)) ] - When ( 1 leq lg x < 2 ): [ [lg x] = 1 ] Substitute ( [lg x] = 1 ) back into the original equation: ] [ (lg x)^2 - 1 - 2 = 0 ] ] ] [ (lg x)^2 - 3 = 0 ] ] [ (lg x) = pm sqrt{3} ] [ lg x = sqrt{3} quad text{(Valid as in [1,2))} ] ] Hence: ) [ x = 10^{sqrt{3}} ] ( (tex )5. When ( lg x = 2 ): ([ x = 10^{2} ) or ( 100 )6. The roots of the original equation are ( x_1 = frac{1}{10}, x_2 = 10^{sqrt{3}}, x_3 = 100 ) Conclusion:[boxed{3}]

🧠:Okay, so I need to find a four-digit number n that's equal to 20A2, where A is some digit between 0 and 9. Then, this number n should be equal to 12 times a positive integer m. So essentially, 12 * m = 20A2. My goal is to figure out what A is and therefore determine the exact value of n.First, let me understand the structure of the number n. It's a four-digit number where the thousands place is 2, the hundreds place is 0, the tens place is A, and the units place is 2. So n can be written as 2000 + 0*100 + A*10 + 2 = 2000 + 10A + 2 = 2002 + 10A. So n = 2002 + 10A. Since n is equal to 12 times m, we can write 12m = 2002 + 10A. Therefore, m = (2002 + 10A)/12. Now, since m has to be a positive integer, (2002 + 10A) must be divisible by 12. So, the key here is to find a digit A (0 through 9) such that 2002 + 10A is a multiple of 12.To check for divisibility by 12, a number must be divisible by both 3 and 4. So, perhaps I can check the divisibility rules for 3 and 4 separately.First, let's check divisibility by 4. A number is divisible by 4 if its last two digits form a number that's divisible by 4. In this case, the last two digits of n are A2. So, the number A2 must be divisible by 4. Since A is a digit from 0 to 9, the possible values for A2 are 02, 12, 22, 32, 42, 52, 62, 72, 82, 92. Let's check which of these are divisible by 4:- 02: 2 ÷ 4 = 0.5 → Not divisible- 12: 12 ÷ 4 = 3 → Divisible- 22: 22 ÷ 4 = 5.5 → Not divisible- 32: 32 ÷ 4 = 8 → Divisible- 42: 42 ÷ 4 = 10.5 → Not divisible- 52: 52 ÷ 4 = 13 → Divisible- 62: 62 ÷ 4 = 15.5 → Not divisible- 72: 72 ÷ 4 = 18 → Divisible- 82: 82 ÷ 4 = 20.5 → Not divisible- 92: 92 ÷ 4 = 23 → DivisibleSo the possible values of A that make A2 divisible by 4 are A = 1, 3, 5, 7, 9. So A can be 1, 3, 5, 7, or 9.Next, we need to check divisibility by 3. A number is divisible by 3 if the sum of its digits is a multiple of 3. The digits of n are 2, 0, A, and 2. So the sum is 2 + 0 + A + 2 = 4 + A. Therefore, 4 + A must be divisible by 3.Since A is 1, 3, 5, 7, or 9, let's check each:- A = 1: 4 + 1 = 5 → 5 ÷ 3 = 1.666... Not divisible- A = 3: 4 + 3 = 7 → 7 ÷ 3 ≈ 2.333... Not divisible- A = 5: 4 + 5 = 9 → 9 ÷ 3 = 3 → Divisible- A = 7: 4 + 7 = 11 → 11 ÷ 3 ≈ 3.666... Not divisible- A = 9: 4 + 9 = 13 → 13 ÷ 3 ≈ 4.333... Not divisibleSo the only value of A that satisfies both divisibility by 4 and 3 is A = 5. Therefore, the number n must be 2052.Wait, but let me double-check. If A is 5, then n is 2052. Let's confirm if 2052 divided by 12 gives an integer. 2052 ÷ 12. Let's compute:12 * 170 = 2040. 2052 - 2040 = 12. So 170 + 1 = 171. 12 * 171 = 2052. Correct. So m is 171, which is an integer. So that works.But let me check if there are any other possible As. Wait, according to the previous checks, only A=5 works. But perhaps I should verify if 2052 is indeed the only possible number here.Alternatively, another approach: Let's compute 2002 + 10A divisible by 12. So 2002 + 10A ≡ 0 mod 12.Let me compute 2002 mod 12 first. 12*166 = 1992. 2002 - 1992 = 10. So 2002 ≡ 10 mod 12. Then, 10A mod 12. So 10A + 10 ≡ 0 mod 12 ⇒ 10(A + 1) ≡ 0 mod 12. Therefore, 10(A + 1) ≡ 0 mod 12. Since 10 and 12 have a GCD of 2, so we can divide both sides by 2: 5(A + 1) ≡ 0 mod 6. Therefore, 5(A + 1) ≡ 0 mod 6.Since 5 and 6 are coprime, this implies that (A + 1) ≡ 0 mod 6. Therefore, A + 1 ≡ 0 mod 6 ⇒ A ≡ 5 mod 6. Since A is a digit (0-9), the possible values are A=5 and A=11, but A=11 is not a digit, so only A=5. Therefore, A must be 5, which gives n=2052. So that confirms the previous conclusion.Therefore, the four-digit number n is 2052.Just to make sure, let's verify again:2052 divided by 12: 12 * 170 = 2040. 2052 - 2040 = 12. So 170 + 1 = 171. 12 * 171 = 2052. Correct.Another check: sum of digits of 2052 is 2 + 0 + 5 + 2 = 9, which is divisible by 3. Last two digits 52, 52 ÷ 4 = 13, so divisible by 4. Hence, 2052 is divisible by 12. So yes, it checks out.Therefore, the answer is 2052.

✅:To determine the four-digit number ( n = 20A2 ) such that ( n = 12m ) for a positive integer ( m ), we follow these steps:1. Express ( n ) algebraically: ( n = 2002 + 10A ).2. Divisibility by 12: ( n ) must be divisible by both 3 and 4. - Divisibility by 4: The last two digits ( A2 ) must form a number divisible by 4. Valid ( A ) values: 1, 3, 5, 7, 9 (since 12, 32, 52, 72, 92 are divisible by 4). - Divisibility by 3: The sum of the digits ( 2 + 0 + A + 2 = 4 + A ) must be divisible by 3. Testing valid ( A ): - ( A = 1 ): ( 4 + 1 = 5 ) (not divisible by 3) - ( A = 3 ): ( 4 + 3 = 7 ) (not divisible by 3) - ( A = 5 ): ( 4 + 5 = 9 ) (divisible by 3) - ( A = 7 ): ( 4 + 7 = 11 ) (not divisible by 3) - ( A = 9 ): ( 4 + 9 = 13 ) (not divisible by 3) Only ( A = 5 ) satisfies both conditions.3. Verification: Substituting ( A = 5 ), ( n = 2052 ). - ( 2052 div 12 = 171 ), an integer. - Sum of digits: ( 2 + 0 + 5 + 2 = 9 ) (divisible by 3). - Last two digits: 52 (divisible by 4).Final Answer: The four-digit number ( n ) is (boxed{2052}).

🔑:Step 1: Analyze the given information and required form of the four-digit number nGiven:- The number n = 20A2, where A is a digit.- The number n multiplied by 12 gives the product m, a positive integer. We can write:[ n = 12m ]Step 2: Conditions for divisibility- Divisibility by 3: For n to be divisible by 3, the sum of its digits must be a multiple of 3. [ 2 + 0 + A + 2 = A + 4 ] We need ( A + 4 equiv 0 pmod{3} ): [ A + 4 equiv 0 pmod{3} implies A equiv -4 equiv 2 pmod{3} ] This gives possible values: [ A in {2, 5, 8} ]- Divisibility by 4: For n to be divisible by 4, the number formed by the last two digits must be divisible by 4. [ 10A + 2 ] Let's check possible values of A: [ begin{aligned} &A = 2 : &10 cdot 2 + 2 = 22 quad &text{(not divisible by 4)} &A = 5 : &10 cdot 5 + 2 = 52 quad &text{(divisible by 4)} &A = 8 : &10 cdot 8 + 2 = 82 quad &text{(not divisible by 4)} end{aligned} ] Hence, the only value satisfying both conditions is: [ A = 5 ]Conclusion:Putting A = 5 into the number:[n = 2052]To verify, we see that ( 2052 div 12 = 171 ), which is an integer, confirming our solution is correct.Thus, the four-digit number n is:[boxed{2052}] 2:Step 1: Given the structure of the number and form to analyzeWe know n should be structured as:[[n = 60k + 12] ]Step 2: Calculating with necessary checksLet's find k where ( k cdot 60 + 12 ) forms a number starting with 2000 (because n is four digits between 2000 - 2999)- Set up for k: [ 2000 leq 60k + 12 < 3000 ] Solving the inequalities: [ 2000 leq 60k + 12 implies 1988 leq 60k implies k geq frac{1988}{60} implies k geq 33.13 implies k geq 34 ] [ 60k + 12 < 3000 implies 60k < 2988 implies k < frac{2988}{60} implies k < 49.8 implies k leq 49 ] Thus, [- 34 should be the feasible to test for the most probable values. ]- Calculate ( 60 times 34 + 12 = 2040 + 12 = 2052 )Conclusion:Introducing:[n = 2052]Both provided solutions verify independently and consistently the value of the four-digit number n, which is:boxed{2052}