🧠:Okay, let's tackle this problem step by step. The question is asking for the sum of all natural numbers not exceeding 200 that satisfy two conditions: when divided by 11, they leave a remainder of 7, and when divided by 7, they leave a remainder of 5. Hmm, so we're looking for numbers that meet both these modular conditions. Let me break this down.First, let me restate the problem in mathematical terms. We need to find all numbers ( n ) such that:1. ( n equiv 7 mod 11 )2. ( n equiv 5 mod 7 )3. ( n leq 200 )And then sum all such numbers. Alright, so this is a system of congruences. To solve this, I think the Chinese Remainder Theorem (CRT) might be useful here. The CRT states that if the moduli are coprime, then there's a unique solution modulo the product of the moduli. Let me check if 11 and 7 are coprime. Since 11 and 7 are both prime numbers and different, their greatest common divisor is 1. So yes, they are coprime. That means there's a unique solution modulo 77 (since 11*7=77). Therefore, the solutions will be of the form ( n = 77k + x ), where ( x ) is the unique solution modulo 77, and ( k ) is a non-negative integer. Then, we need to find all such ( n ) up to 200 and sum them.So the first step is to find the smallest positive integer ( x ) that satisfies both congruences:1. ( x equiv 7 mod 11 )2. ( x equiv 5 mod 7 )Let me solve this system. Let's start with the first congruence: ( x = 11m + 7 ) for some integer ( m geq 0 ). Substitute this into the second congruence:( 11m + 7 equiv 5 mod 7 )Simplify this equation. Let's compute 11 mod 7 first. 11 divided by 7 is 1 with a remainder of 4, so 11 ≡ 4 mod 7. Similarly, 7 mod 7 is 0. Therefore:( 4m + 0 equiv 5 mod 7 )So, ( 4m equiv 5 mod 7 ). Now, we need to solve for ( m ). To do this, we can find the multiplicative inverse of 4 modulo 7. The inverse of 4 modulo 7 is a number ( a ) such that ( 4a equiv 1 mod 7 ).Testing values:4*2 = 8 ≡ 1 mod 7 (since 8-7=1). So, 2 is the inverse of 4 modulo 7. Therefore, multiplying both sides of the equation ( 4m ≡ 5 mod 7 ) by 2 gives:( m ≡ 5*2 mod 7 )Which is:( m ≡ 10 mod 7 )Simplify 10 mod 7: 10-7=3, so ( m ≡ 3 mod 7 ). Therefore, ( m = 7n + 3 ) for some integer ( n ≥ 0 ).Substitute back into the expression for ( x ):( x = 11m + 7 = 11*(7n + 3) + 7 = 77n + 33 + 7 = 77n + 40 )Therefore, the solutions to the system of congruences are numbers congruent to 40 modulo 77. So the general solution is ( x = 77n + 40 ), where ( n ) is a non-negative integer. Wait, but when we substituted m = 7n + 3 into x = 11m +7, we should get x = 11*(7n +3) +7 = 77n + 33 +7 = 77n +40, yes that's correct. So the smallest positive solution is 40. Let me verify this:Check 40 divided by 11: 11*3=33, 40-33=7, so remainder 7. Good. Then 40 divided by 7: 7*5=35, 40-35=5, remainder 5. Perfect. So 40 is indeed the smallest such number.Therefore, all numbers satisfying the two conditions are of the form 77n +40, where n is a non-negative integer. Now, we need to find all such numbers less than or equal to 200.So, we need to solve 77n +40 ≤ 200.Subtract 40 from both sides: 77n ≤ 160.Divide both sides by 77: n ≤ 160/77 ≈ 2.077. Since n must be an integer, the maximum possible n is 2. Wait, let me check 77*2 +40 = 154 +40 =194, which is less than 200. Then n=3 would be 77*3 +40=231 +40=271, which is over 200. So the possible values of n are 0,1,2.Therefore, the numbers are:For n=0: 77*0 +40=40n=1:77 +40=117n=2:154 +40=194So the numbers are 40,117,194. Now, let's check if these are indeed less than or equal to 200. 194 is, 117 is, 40 is. So three numbers in total.Now, sum them up: 40 +117 +194.Let me compute that step by step.40 +117 = 157157 +194 = 351So the sum should be 351. Wait, let me verify again:40 +117: 40+100=140, 140+17=157.157 +194: 157+200=357, 357-6=351. Yes, that's correct.But hold on, let me double-check each number's congruences to be sure.First, 40:40 ÷11 = 3 with remainder 7. Correct.40 ÷7 =5 with remainder 5. Correct.117:117 ÷11: 11*10=110, 117-110=7. Remainder 7. Good.117 ÷7: 7*16=112, 117-112=5. Remainder 5. Good.194:194 ÷11: 11*17=187, 194-187=7. Remainder 7.194 ÷7: 7*27=189, 194-189=5. Remainder 5.All correct. So the numbers are indeed 40,117,194, summing to 351.Therefore, the answer should be 351.Wait, but just to make sure that there are no other numbers below 200 that satisfy the conditions. The next number after 194 would be 194 +77=271, which is over 200, so indeed only 40,117,194. So the sum is 351.But let me think again if there's another way to approach this problem. Maybe by listing numbers that leave remainder 7 when divided by 11, and then check which of those leave remainder 5 when divided by 7. Let's try this alternative approach as a cross-verification.Numbers that leave remainder 7 when divided by 11: starting from 7, 18, 29, 40, 51, 62, 73, 84, 95, 106, 117, 128, 139, 150, 161, 172, 183, 194, 205,... but since we need numbers not exceeding 200, stop at 194.Now, from this list, we need numbers that also leave remainder 5 when divided by 7. Let's check each number:7: 7 ÷7 =1 remainder 0. Not 5.18:18 ÷7=2*7=14, 18-14=4. Remainder 4. Not 5.29:29 ÷7=4*7=28, remainder 1. Not 5.40:40 ÷7=5*7=35, remainder 5. Yes.51:51 ÷7=7*7=49, remainder 2. Not 5.62:62 ÷7=8*7=56, remainder 6. Not 5.73:73 ÷7=10*7=70, remainder 3. Not 5.84:84 ÷7=12*7=84, remainder 0. Not 5.95:95 ÷7=13*7=91, remainder 4. Not 5.106:106 ÷7=15*7=105, remainder 1. Not 5.117:117 ÷7=16*7=112, remainder 5. Yes.128:128 ÷7=18*7=126, remainder 2. Not 5.139:139 ÷7=19*7=133, remainder 6. Not 5.150:150 ÷7=21*7=147, remainder 3. Not 5.161:161 ÷7=23*7=161, remainder 0. Not 5.172:172 ÷7=24*7=168, remainder 4. Not 5.183:183 ÷7=26*7=182, remainder 1. Not 5.194:194 ÷7=27*7=189, remainder 5. Yes.205: Exceeds 200, so stop.So indeed, the numbers that satisfy both conditions are 40,117,194. Same as before. So sum is 351. Therefore, this cross-verification confirms the result. Therefore, the answer is 351.But let me also check if there's any mistake in the CRT approach. For example, when solving the congruence equations, did I correctly find x=40?Yes. Let me retrace:Starting with x ≡7 mod11. So x=11m +7. Then substituting into the second equation:11m +7 ≡5 mod7. Since 11≡4 mod7, so 4m +0 ≡5 mod7. Then 4m ≡5 mod7. The inverse of 4 mod7 is 2 because 4*2=8≡1 mod7. Therefore, m≡5*2=10≡3 mod7. Hence, m=7n +3. Then x=11*(7n +3)+7=77n +33 +7=77n +40. So x=77n +40. Correct.So the solutions are 40,117,194. Sum is 351.Another sanity check: the numbers are spaced 77 apart. 40+77=117, 117+77=194, 194+77=271. Correct. So up to 200, only three numbers. Sum is 351.Therefore, the answer should be 351. I think that's thorough. But just to be extra cautious, maybe compute 40+117+194 again.40 +117: 40 + 100=140, 140 +17=157.157 +194: 150 +190=340, 7 +4=11, so 340 +11=351. Yes. Correct.So I believe the answer is 351.Final AnswerThe sum is boxed{351}.

✅:To find the sum of all natural numbers not exceeding 200 that leave a remainder of 7 when divided by 11 and a remainder of 5 when divided by 7, we need to solve the system of congruences:1. ( n equiv 7 pmod{11} )2. ( n equiv 5 pmod{7} )Using the Chinese Remainder Theorem (CRT), we first express ( n ) in terms of one modulus and substitute into the other. Starting with ( n = 11m + 7 ), we substitute into the second congruence:[11m + 7 equiv 5 pmod{7}]Since ( 11 equiv 4 pmod{7} ), this simplifies to:[4m equiv 5 pmod{7}]The multiplicative inverse of 4 modulo 7 is 2 (since ( 4 times 2 equiv 1 pmod{7} )). Multiplying both sides by 2:[m equiv 10 pmod{7} implies m equiv 3 pmod{7}]Thus, ( m = 7k + 3 ) for some integer ( k ). Substituting back, we get:[n = 11(7k + 3) + 7 = 77k + 40]The numbers ( n ) are of the form ( 77k + 40 ). We need to find all such numbers ( leq 200 ):[77k + 40 leq 200 implies 77k leq 160 implies k leq frac{160}{77} approx 2.077]Thus, ( k ) can be 0, 1, or 2, giving the numbers 40, 117, and 194. Summing these:[40 + 117 + 194 = 351]The sum of all natural numbers not exceeding 200 that satisfy the given conditions is (boxed{351}).

🔑:The problem requires finding the sum of all natural numbers that satisfy specified remainders when divided by certain parameters and are not greater than 200.1. Determine the numbers less than or equal to 200 that give a remainder of 7 when divided by 11: - We solve the congruence: ( x equiv 7 pmod{11} ). - The sequence of numbers is: ( x = 11k + 7 ) for integer ( k ) such that ( 11k + 7 leq 200 ).2. Calculate specific values of ( x ): - For ( k = 0, 1, 2, ldots ): [ x = 11(0) + 7 = 7, quad x = 11(1) + 7 = 18, quad x = 11(2) + 7 = 29, quad ldots, ] [ 196 + 7 = 203 quad Rightarrow text{stop at } 190 + 7 = 197 quad text{(within bounds)}. ]3. Determine the values of ( x ) less than or equal to 200 that give a remainder of 5 when divided by 7: - We solve the congruence: ( x equiv 5 pmod{7} ). - The sequence of numbers is: ( x = 7m + 5 ) for integer ( m ) such that ( 7m + 5 leq 200 ).4. Calculate specific values of ( x ): - For ( m = 0, 1, 2, ldots ): [ x = 7(0) + 5 = 5, quad x = 7(1) + 5 = 12, quad x = 7(2) + 5 = 19, quad ldots, ] [ 203 - 5 = 198 quad Rightarrow text{stop at } 196 - 5 = 191 quad text{(within bounds)}. ]5. Next, we seek a common solution to the system of congruences: - ( x equiv 7 pmod{11} ) - ( x equiv 5 pmod{7} )6. Use the Chinese Remainder Theorem (CRT) to find the smallest solution: - The solution to the system can be represented as ( x mod (11 times 7) = x mod 77 ), i.e., ( x = 77k + 40 ). - This follows because one checks ( x = 40, 117, 194 ): [ 40 equiv 7 pmod{11}, quad 40 equiv 5 pmod{7}, ] [ 117 = 77 + 40, ] and so on.7. Validate those values all lie within the given range: - Check ( 40, 117, 194 leq 200 ).8. Compute the final sum: [ text{Sum} = 40 + 117 + 194 = 351. ] Conclusion[boxed{351}]

🧠:Okay, let's try to figure out how to calculate the monthly rent per square meter needed to recoup the investment for this residential building. The construction cost is 250 yuan per square meter, the useful life is 50 years, and the annual interest rate is 5%. Hmm, I think this is a problem about capital recovery, maybe using the concept of the time value of money. First, I remember that when you have an initial investment and you want to recover it over a period with interest, you can use an annuity formula. The idea is to find the equal annual (or monthly) payments that, when discounted at the given interest rate, sum up to the initial cost. So, the key here is to calculate the equivalent annual cost and then convert that to a monthly rent.Let me start by recalling the formula for the capital recovery factor. The capital recovery factor (CRF) is used to determine the annual payment required to recover the initial investment over a certain number of years with a given interest rate. The formula is:CRF = i(1 + i)^n / [(1 + i)^n - 1]Where:- i = annual interest rate- n = number of yearsOnce I find the CRF, multiplying it by the initial cost will give the annual rent per square meter. Then, to get the monthly rent, I divide that annual amount by 12.But wait, let me confirm if that's the right approach. The initial cost is 250 yuan per square meter. We need to spread this cost over 50 years with 5% interest. So yes, using the capital recovery factor makes sense here. The CRF gives the annual payment needed to pay off a loan (or recoup the investment) with interest over the period. Since the question asks for monthly rent, maybe I need to adjust the formula for monthly compounding? Hmm, the interest rate given is annual, 5%, so perhaps I should convert that to a monthly rate and use the number of monthly periods instead of annual.Wait, that's a good point. If we're calculating monthly rent, we might need to use a monthly interest rate and the total number of months. Let me think. The annual interest rate is 5%, so the monthly rate would be 5% divided by 12, which is approximately 0.4167% per month. The total number of periods would be 50 years times 12 months, so 600 months. Alternatively, if we use the annual CRF and then divide by 12 to get the monthly rent, that might be an approximation. But which method is more accurate? The exact method would require using the monthly rate and monthly payments. Let's check both approaches and see if there's a significant difference.First, let's try the annual CRF method:CRF_annual = (0.05*(1 + 0.05)^50) / [(1 + 0.05)^50 - 1]Calculating the denominator first: (1.05)^50. Let me approximate that. 1.05^50 is a large number. Using the rule of 72, 72/5 = 14.4 years to double. So in 50 years, it would double about 3.47 times. Starting with 1, doubling once is 2, twice is 4, three times is 8, 0.47 of a doubling would be approximately 8 * 1.47 = about 11.76. But that's a rough estimate. Alternatively, using a calculator: ln(1.05) ≈ 0.04879, so 50*0.04879 ≈ 2.4395. So e^2.4395 ≈ 11.467. So (1.05)^50 ≈ 11.467. So the denominator is 11.467 - 1 = 10.467. The numerator is 0.05 * 11.467 ≈ 0.5733. Therefore, CRF_annual ≈ 0.5733 / 10.467 ≈ 0.0548. So the annual payment per square meter is 250 yuan * 0.0548 ≈ 13.7 yuan per year. Then, monthly would be 13.7 / 12 ≈ 1.14 yuan per square meter per month.But wait, that seems low. Let me verify the calculations again. Maybe my estimation of (1.05)^50 was off. Let me check with a calculator. Actually, 1.05^50 is approximately 11.4674, as I had before. So 0.05*11.4674 = 0.57337. Divided by (11.4674 - 1) = 10.4674, gives 0.57337 / 10.4674 ≈ 0.05476. So the CRF is approximately 5.476%. Then 250 * 0.05476 ≈ 13.69 yuan per year, which is about 1.14 yuan per month. Hmm, but is this correct? Because 250 yuan over 50 years without interest would be 250 / 50 = 5 yuan per year, which is 0.42 yuan per month. But with 5% interest, the required payment is higher, which makes sense. 1.14 yuan per month seems plausible?Alternatively, let's try the monthly compounding approach. The monthly interest rate is 5% / 12 ≈ 0.0041667, and the number of periods is 50*12 = 600. The formula for the capital recovery factor in this case would be:CRF_monthly = (i_monthly*(1 + i_monthly)^n) / [(1 + i_monthly)^n - 1]Plugging in the numbers:i_monthly = 0.05 / 12 ≈ 0.0041667n = 600First, calculate (1 + 0.0041667)^600. Let's see, this is equivalent to (1 + 0.05/12)^(12*50) = e^(50*ln(1.05)) ≈ e^(50*0.048790164) ≈ e^(2.4395082) ≈ 11.4674, which is the same as (1.05)^50. So (1 + i_monthly)^n = 11.4674. Then, the denominator is 11.4674 - 1 = 10.4674. The numerator is i_monthly * 11.4674 ≈ 0.0041667 * 11.4674 ≈ 0.04778. Then, CRF_monthly ≈ 0.04778 / 10.4674 ≈ 0.004564. This is the monthly capital recovery factor. So the monthly payment would be 250 * 0.004564 ≈ 1.141 yuan per square meter per month.So both methods give approximately the same result, which is around 1.14 yuan per month. That seems consistent. Therefore, the required monthly rent is approximately 1.14 yuan per square meter.Wait a minute, but let me check with another approach to be sure. Another way to think about this is to calculate the present value of an ordinary annuity. The present value (PV) is the initial investment, which is 250 yuan. The payment (PMT) is what we need to find, and the present value of all those PMTs should equal 250. The formula for PV is:PV = PMT * [(1 - (1 + i)^-n)/i]But here, since we're dealing with monthly payments, we have to adjust i and n accordingly. So:250 = PMT * [(1 - (1 + 0.05/12)^-600)/(0.05/12)]First, calculate (1 + 0.05/12)^-600. That's the same as 1/(1.0041667)^600 ≈ 1/11.4674 ≈ 0.0872. So 1 - 0.0872 = 0.9128. Then, 0.9128 / (0.05/12) = 0.9128 / 0.0041667 ≈ 219.07. Therefore, 250 = PMT * 219.07, so PMT ≈ 250 / 219.07 ≈ 1.141 yuan per month. So again, we get the same answer.Therefore, regardless of whether I use the annual CRF and divide by 12 or use the monthly CRF directly, I end up with approximately 1.14 yuan per square meter per month. That seems to check out.But just to be thorough, let's consider if there are any other factors. The problem states "to recoup the entire investment," which implies that the present value of all the rental income over 50 years should equal the initial construction cost of 250 yuan per square meter. Since we have calculated the monthly rent that achieves this, it seems correct. Is there any possibility that we need to account for inflation or other factors? The problem doesn't mention inflation, so we can assume that the 5% interest rate is the real interest rate, and all values are in real terms. Therefore, no need to adjust for inflation.Another check: If the monthly rent is 1.14 yuan, then over 50 years (600 months), the total rent collected would be 1.14 * 600 = 684 yuan. But since we're discounting these payments at 5% annual interest, the present value of those 684 yuan is exactly 250 yuan. So even though the total nominal amount is higher, the present value equals the initial investment. This makes sense because of the time value of money.Alternatively, if there were no interest, the monthly rent would be 250 / 600 ≈ 0.4167 yuan. But with 5% interest, the required rent is higher, which aligns with our result of 1.14 yuan. So everything seems consistent.Therefore, I think the correct answer is approximately 1.14 yuan per square meter per month. Rounding to two decimal places, that would be 1.14 yuan. Depending on the required precision, maybe we need to compute it more accurately. Let me do a more precise calculation.Using the monthly approach:i = 0.05 / 12 ≈ 0.0041666667n = 600Compute (1 + i)^n = (1 + 0.0041666667)^600As before, this is (1.05)^50 ≈ e^(50*ln(1.05)) ≈ e^(50*0.048790164169432) ≈ e^2.4395082 ≈ 11.4673993Therefore, (1 + i)^n ≈ 11.4673993CRF_monthly = (0.0041666667 * 11.4673993) / (11.4673993 - 1) = (0.04778083) / (10.4673993) ≈ 0.00456415Monthly payment = 250 * 0.00456415 ≈ 1.1410375 yuanSo approximately 1.141 yuan per month. Rounded to two decimal places, 1.14 yuan. If we need more precision, maybe 1.141, but the problem doesn't specify, so 1.14 is acceptable.Alternatively, if the answer is expected in a typical currency format, it might be 1.14 yuan per square meter per month.So the final answer should be boxed{1.14} yuan per square meter per month.

✅:To determine the monthly rent per square meter required to recoup the entire investment for a residential building with a construction cost of 250 yuan per square meter, a useful life of 50 years, and an annual interest rate of 5%, we use the capital recovery factor (CRF) approach. 1. Convert the annual interest rate to a monthly rate and calculate the number of monthly periods: - Monthly interest rate: ( i = frac{0.05}{12} approx 0.0041667 ) - Number of monthly periods: ( n = 50 times 12 = 600 )2. Calculate the capital recovery factor (CRF) for monthly payments: [ text{CRF}_{text{monthly}} = frac{i cdot (1 + i)^n}{(1 + i)^n - 1} ] - Compute ( (1 + i)^n approx 11.4674 ) - Numerator: ( 0.0041667 times 11.4674 approx 0.04778 ) - Denominator: ( 11.4674 - 1 = 10.4674 ) - CRF: ( frac{0.04778}{10.4674} approx 0.004564 )3. Calculate the monthly rent: [ text{Monthly Rent} = 250 times 0.004564 approx 1.141 ]4. Round to two decimal places for practical purposes: - The monthly rent is approximately 1.14 yuan per square meter.[boxed{1.14}]

🔑: Problem:A certain residential building has a construction investment of 250 RMB per square meter. Considering a service life of 50 years and an annual interest rate of 5%, calculate the monthly rent per square meter that will enable the entire investment to be recovered.Given:- Initial investment per square meter: ( = 250 ) RMB- Service life: ( n = 50 ) years- Annual interest rate: ( r = 0.05 )# Step 1: Calculate the annual annuity paymentThe formula to convert a lump sum into an annuity (regular payments) considering the time value of money (considering the interest rate) is given by:[x = P cdot frac{(1 + r)^n cdot r}{(1 + r)^n - 1}]Where:- ( P ) is the initial investment per square meter- ( r ) is the annual interest rate- ( n ) is the number of years- ( x ) is the annual rent per square meter that recoups the investmentSubstitute the given values into the formula:[x = 250 cdot frac{(1 + 0.05)^{50} cdot 0.05}{(1 + 0.05)^{50} - 1}]# Step 2: Numerical calculationFirst, compute ( (1 + 0.05)^{50} ):[(1 + 0.05)^{50} = 1.05^{50} approx 11.4674]Next, compute the numerator ( 11.4674 cdot 0.05 ):[11.4674 cdot 0.05 = 0.57337]Finally, compute the denominator:[11.4674 - 1 = 10.4674]So,[x = 250 cdot frac{0.57337}{10.4674} approx 250 cdot 0.05479 = 13.6975]Therefore, the annual rent per square meter is:[x approx 13.69 text{ RMB/year}]# Step 3: Convert the annual rent to monthly rentTo find the monthly rent, divide the annual rent by 12:[text{Monthly rent} = frac{13.69}{12} approx 1.14 text{ RMB/month}] Conclusion:The required monthly rent per square meter that enables the full recovery of the entire investment is:[boxed{1.14 , text{RMB/month}} ]

🧠:Okay, let's tackle this problem. We need to find the sum of the floor of the base-2 logarithm of each integer from 1 to 2012. That is, compute the sum S = [log₂1] + [log₂2] + [log₂3] + ... + [log₂2012]. The floor function [x] gives the greatest integer less than or equal to x. First, I remember that the floor of a logarithm log₂n gives the exponent of the highest power of 2 that is less than or equal to n. For example, log₂8 = 3, but if n isn't a power of 2, say n=5, then log₂5 ≈ 2.32, and the floor of that is 2. So, [log₂n] = k when 2^k ≤ n < 2^(k+1). Therefore, for each integer k, all numbers n in the interval [2^k, 2^(k+1)) will have the same floor value k. The number of integers in this interval is 2^(k+1) - 2^k = 2^k. So, each k contributes k * 2^k to the sum. But wait, this is only up until a certain point where 2^(k+1) exceeds 2012. The last term might not have a full 2^k numbers.So, the strategy is to break the sum into intervals where [log₂n] is constant. For each integer k, find all n such that 2^k ≤ n < 2^(k+1). For each such n, [log₂n] = k. Then, sum over all k, multiplying each k by the number of terms in that interval.But we need to adjust for the last interval where 2^(k+1) might be greater than 2012. Let's formalize this.Let’s find the maximum k such that 2^k ≤ 2012. Then, the intervals will be from k=0 (since log₂1 = 0) up to this maximum k. However, the upper limit for each interval is 2^(k+1) - 1, but our upper limit is 2012. So, the last interval will not be a full interval if 2012 + 1 isn't a power of 2. Let's check.First, find the largest power of 2 less than or equal to 2012. Let's compute:2^10 = 10242^11 = 2048Ah, 2048 is greater than 2012, so the largest power of 2 less than or equal to 2012 is 2^10 = 1024. Wait, but 2^11 is 2048, which is larger than 2012, so the maximum k for which 2^k ≤ 2012 is k=10. However, the interval for k=10 is [2^10, 2^11) = [1024, 2048). But our upper limit is 2012, so the numbers from 1024 to 2012 will have [log₂n] = 10. So, the number of terms in this interval is 2012 - 1024 + 1 = 989. Wait, why +1? Because when you count from a to b inclusive, it's b - a + 1.Wait, so let me confirm:For each k, the number of integers n where [log₂n] = k is 2^(k) when the interval is [2^k, 2^(k+1) - 1], which has exactly 2^k numbers. But when we reach the maximum k, which in this case is 10, since 2^11 is 2048, which is larger than 2012, then the number of terms in the last interval is 2012 - 2^10 + 1 = 2012 - 1024 + 1 = 989. So, instead of 2^10 = 1024 terms for k=10, we have 989 terms.Therefore, the total sum S would be the sum over k from 0 to 10 of (number of terms with [log₂n] = k) multiplied by k. For k=0 to 9, the number of terms is 2^k, and for k=10, it's 989.Wait, let's check for k=0: [log₂1] = 0. The interval is [2^0, 2^1) = [1, 2), which includes only n=1. So, 1 term. 2^0 =1. Correct.For k=1: [2^1, 2^2) = [2,4), which includes n=2,3. That's 2 terms. 2^1=2. Correct.Similarly, for k=2: [4,8), numbers 4,5,6,7. 4 terms. 2^2=4. Correct.Continuing up to k=10: [1024, 2048), but since we stop at 2012, the count is 2012 - 1024 + 1 = 989.So, the sum S is Σ (k * 2^k) from k=0 to 9, plus 10 * 989.Therefore, we need to compute:Sum = Σ (k * 2^k) from k=0 to 9 + 10 * 989.But first, let's compute Σ (k * 2^k) from k=0 to 9.I recall that the sum Σ k * r^k from k=0 to n has a closed-form formula. The formula is r*(1 - (n+1)*r^n + n*r^(n+1)) / (1 - r)^2, for r ≠ 1. Since here r=2, so substituting:Sum = 2*(1 - (n+1)*2^n + n*2^(n+1)) / (1 - 2)^2Simplify denominator: (1 - 2)^2 = 1So, Sum = 2*(1 - (n+1)*2^n + n*2^(n+1))But let's check with n=9:Sum = 2*(1 - 10*2^9 + 9*2^{10}) Compute each term:10*2^9 = 10*512 = 51209*2^{10} = 9*1024 = 9216So, inside the brackets: 1 - 5120 + 9216 = 1 + (9216 - 5120) = 1 + 4096 = 4097Multiply by 2: 4097*2 = 8194Wait, let me compute step by step:Original formula: Σ (k=0 to n) k*2^k = 2*(1 - (n+1)*2^n + n*2^{n+1}) )Wait, let's plug in n=9:Sum = 2*(1 - 10*512 + 9*1024) = 2*(1 - 5120 + 9216) = 2*(1 + 4096) = 2*4097 = 8194.Wait, but let's verify this sum manually for small n to check if the formula is correct.For example, n=1:Σ k*2^k from 0 to 1 = 0*1 + 1*2 = 2.Using formula: 2*(1 - 2*2^1 + 1*2^2) = 2*(1 - 4 + 4) = 2*1 = 2. Correct.n=2:Σ 0*1 +1*2 +2*4 = 0 +2 +8=10.Formula: 2*(1 -3*4 + 2*8)=2*(1 -12 +16)=2*(5)=10. Correct.So formula works. Therefore, for n=9, Sum=8194.But let me also check with a step-by-step calculation for a few terms to make sure.Compute Σ k*2^k from k=0 to 9:k=0: 0*1=0k=1:1*2=2k=2:2*4=8k=3:3*8=24k=4:4*16=64k=5:5*32=160k=6:6*64=384k=7:7*128=896k=8:8*256=2048k=9:9*512=4608Now sum these up:Start adding:0 +2=22 +8=1010 +24=3434 +64=9898 +160=258258 +384=642642 +896=15381538 +2048=35863586 +4608=8194Yes, matches the formula result. So Σ (k=0 to 9) k*2^k=8194.Then, the sum for k=10 is 10*989=9890.Therefore, total sum S=8194 +9890=18084.Wait, 8194 +9890: 8000 + 9000=17000; 194 + 890=1084; total=17000+1084=18084.But wait, let me double-check the arithmetic:8194 + 9890:8000 + 9000 = 17000194 + 890 = 108417000 + 1084 = 18084. Correct.Therefore, the answer is 18084.But hold on, let me verify another way.Alternatively, let's check the number of terms in each interval.For k=0: n=1. 1 term. So, 0*1=0.k=1: n=2,3. 2 terms. 1*2=2.k=2: n=4-7. 4 terms. 2*4=8.k=3: n=8-15. 8 terms. 3*8=24.k=4: 16-31. 16 terms. 4*16=64.k=5: 32-63. 32 terms. 5*32=160.k=6: 64-127. 64 terms. 6*64=384.k=7: 128-255. 128 terms.7*128=896.k=8: 256-511. 256 terms.8*256=2048.k=9: 512-1023. 512 terms.9*512=4608.k=10: 1024-2012. Number of terms: 2012 -1024 +1=989. 10*989=9890.Adding all these contributions:0 +2 +8 +24 +64 +160 +384 +896 +2048 +4608 +9890.Let's add step by step:Start with 0.0 +2=22 +8=1010 +24=3434 +64=9898 +160=258258 +384=642642 +896=15381538 +2048=35863586 +4608=81948194 +9890=18084.Same result. So that seems consistent.But wait, let me check if the count for k=10 is correct. The interval for k=10 is [2^10, 2^11) = [1024, 2048). But our upper limit is 2012, so the numbers are from 1024 to 2012 inclusive. The number of integers is 2012 -1024 +1.2012 -1024 = 988, so 988 +1=989. Correct.Another check: let's compute the total number of terms. The sum from k=0 to 9 of 2^k is 2^10 -1=1023 (since sum_{k=0}^n 2^k = 2^{n+1}-1). So, for k=0 to 9, the number of terms is 1+2+4+8+16+32+64+128+256+512=1023. Then, for k=10, 989 terms. Total terms=1023 +989=2012. Which matches the problem's range from 1 to 2012. So the count is correct.Therefore, the total sum is indeed 18084.But just to be thorough, let me check with a different approach.Alternative approach: the value [log₂n] = k for n in [2^k, 2^{k+1}-1]. So, for each k, the numbers contributing k are 2^k numbers. Except for the last k where it might be less.So, the maximum k is floor(log₂2012). Let's compute log₂2012.Since 2^10=1024, 2^11=2048. 2012 is between 2^10 and 2^11. So, floor(log₂2012)=10. Hence, the maximum k is 10.So, for k=0 to 10, each contributes except k=10, which contributes fewer terms.Number of terms for each k:For k=0: 1 term (1)For k=1: 2 terms (2,3)...For k=10: 2012 - 2^10 +1=2012-1024+1=989 terms.Sum is Σ_{k=0}^{9} k*2^k +10*989=8194+9890=18084.Same result.Alternatively, we can use the formula for the sum Σ_{k=0}^n k*2^k = (n-1)*2^{n+1} +2.Wait, let me check that formula. For example, if n=1:Σ_{k=0}^1 k*2^k=0+2=2. According to formula: (1-1)*2^{2}+2=0+2=2. Correct.For n=2: 0+2+8=10. Formula: (2-1)*2^3 +2=1*8 +2=10. Correct.For n=3: Σ=0+2+8+24=34. Formula: (3-1)*2^4 +2=2*16 +2=34. Correct.So, formula: Σ_{k=0}^n k*2^k = (n -1)2^{n+1} +2.Testing with n=9:Σ_{k=0}^9 k*2^k = (9 -1)*2^{10} +2=8*1024 +2=8192 +2=8194. Which matches our previous calculation.Thus, this formula gives us the same result. So, using this formula, for n=9, sum=8194.Then, adding 10*989=9890, total sum=8194+9890=18084.Therefore, the answer is 18084.I think this is correct. Let me just make sure there are no calculation errors.Wait, 2012 -1024=988, right? 1024+988=2012. So, 988 terms from 1024 to 2011, and 2012 is the 989th term. Correct.Yes, 1024 +988=2012, but actually, 1024 +988=2012, but since we count both endpoints, the number of terms is 2012 -1024 +1=989. Correct.Thus, everything checks out. The final answer should be 18084.Final Answerboxed{18084}

✅:To find the sum (left[log_{2} 1right] + left[log_{2} 2right] + left[log_{2} 3right] + cdots + left[log_{2} 2012right]), we need to consider the intervals where the floor function (left[log_{2} nright]) is constant. For each integer (k), the value (left[log_{2} nright] = k) when (2^k leq n < 2^{k+1}). The number of integers in this interval is (2^{k+1} - 2^k = 2^k), except for the last interval which may be incomplete.1. Identify the intervals and their contributions: - For (k = 0) to (k = 9), the intervals are complete and each contributes (k cdot 2^k). - For (k = 10), the interval is from (1024) to (2012), which contains (2012 - 1024 + 1 = 989) terms, contributing (10 cdot 989).2. Calculate the sum for complete intervals: - The sum (sum_{k=0}^{9} k cdot 2^k) can be computed using the formula for the sum of (k cdot r^k), which results in (8194).3. Calculate the contribution from the last interval: - The contribution from (k = 10) is (10 cdot 989 = 9890).4. Combine the results: - The total sum is (8194 + 9890 = 18084).Thus, the final answer is (boxed{18084}).

🔑:Given the problem:[ leftlfloor log_2 1 rightrfloor + leftlfloor log_2 2 rightrfloor + leftlfloor log_2 3 rightrfloor + cdots + leftlfloor log_2 2012 rightrfloor = , ? ]# Step-by-Step Detailed Solution:1. Understanding the Floor Function: [ leftlfloor log_2 x rightrfloor text{ gives the largest integer } k text{ such that } 2^k leq x < 2^{k+1}. ] 2. Determine Range for Logs: Note that (1024 = 2^{10}) and (2012 < 2^{11} = 2048), hence: [ leftlfloor log_2 x rightrfloor text{ for } 1024 leq x < 2012 text{ is } 10. ] 3. Sum for (leftlfloor log_2 x rightrfloor) from 1024 to 2012: [ leftlfloor log_2 1024 rightrfloor + leftlfloor log_2 1025 rightrfloor + cdots + leftlfloor log_2 2012 rightrfloor = 10 times (2012 - 1024 + 1) ] The number of terms from 1024 to 2012 is (2012 - 1024 + 1 = 989): [ 10 times 989 = 9890 ] 4. Summation for (leftlfloor log_2 x rightrfloor) from 1 to 1023: [ leftlfloor log_2 1 rightrfloor + leftlfloor log_2 2 rightrfloor + leftlfloor log_2 3 rightrfloor + cdots + leftlfloor log_2 1023 rightrfloor ] For this part, we recognize that: [ S = 1 times 2 + 2 times 2^2 + 3 times 2^3 + cdots + 9 times 2^9 ]5. Calculate Modified Summation (2S): For easier computation, we multiply each term in ( S ) by 2: [ 2S = 1 times 2^2 + 2 times 2^3 + 3 times 2^4 + cdots + 9 times 2^{10} ]6. Subtract Original (S) from (2S): [ 2S - S = (1 times 2^2 + 2 times 2^3 + 3 times 2^4 + cdots + 9 times 2^{10}) - (1 times 2 + 2 times 2^2 + 3 times 2^3 + cdots + 9 times 2^9) ] Collect terms: [ S = 2 + 2^2 + 2^3 + cdots + 2^9 - 9 times 2^{10} ] The summation ((2^1 + 2^2 + cdots + 2^9) = 2^{10} - 2): [ S = 2^{10} - 2 - 9 times 2^{10} ] Simplify: [ S = 2^{10} - 2 - 9 times 2^{10} = -8 times 2^{10} - 2 = -8192 - 2 = -8194 ]7. Combine Results: [ S = 8194 ]8. Final Summation: The total sum is given by combining both parts: [ 8194 + 9890 = 18084 ] Conclusion:[ boxed{18084} ]

🧠:Okay, let's see. The user wants me to calculate the change in oxygen concentration, but the problem given here is about finding the temperature using the ideal gas law. Hmm, maybe there's a follow-up question related to oxygen concentration after finding the temperature? Let me check the problem again.Alright, the given equation is the ideal gas law: PV = vRT. They solved for T and plugged in the numbers to get approximately 206 K. The task mentions "Calculate the change in oxygen concentration," but the provided solution is for temperature. Maybe there was a typo or a missing part in the problem statement? Let me try to figure out how temperature relates to oxygen concentration.Oxygen concentration could refer to the concentration in terms of moles per volume (n/V) or maybe partial pressure? Using the ideal gas law, concentration (n/V) is equal to P/(RT). So if temperature changes, the concentration would change inversely with temperature, assuming pressure is constant. Alternatively, if volume changes, it would affect concentration as well.Wait, the problem statement here doesn't mention any change, though. It just calculates the temperature. Maybe the original problem had a part where, after finding T, there's a scenario where pressure or volume changes, leading to a change in oxygen concentration? But the information provided only shows the calculation for T. Let me re-examine the given values:p = 9e5 Pa, V = 95e-5 m³, v = 0.5 moles, R = 8.31 J/(mol·K). Calculating T gives 206 K. If the task is to find the change in oxygen concentration, perhaps we need to compare two different states? But only one state is given here. Maybe the initial and final states are implied, but the user hasn't provided both. Alternatively, maybe "change in oxygen concentration" refers to a miscalculation in the provided solution, but the answer seems correct.Wait, let me verify the calculation step-by-step. The numerator is pV: 9e5 Pa * 95e-5 m³. Let's compute that:9e5 is 900,000, and 95e-5 is 0.00095. Multiplying them: 900,000 * 0.00095 = 900,000 * 9.5e-4 = 900,000 * (1e-4 * 9.5) = 900,000 * 1e-4 = 90, so 90 * 9.5 = 855. So the numerator is 855.Denominator: 0.5 moles * 8.31 J/(mol·K) = 0.5 * 8.31 = 4.155. Then T = 855 / 4.155 ≈ 205.776 K, which rounds to 206 K. That calculation checks out.But where does oxygen concentration come into play? Unless the original problem had more parts, like changing temperature and finding how concentration changes, but in the given info, only one state is present. Maybe the question is actually to compute the concentration (n/V) from the given data, which would be 0.5 moles / 95e-5 m³. Let's compute that:0.5 / 95e-5 = 0.5 / 0.00095 ≈ 526.3158 mol/m³. But the problem statement mentions "change in oxygen concentration," which implies a difference between two concentrations. Without a second state, it's impossible to calculate a change.Alternatively, maybe the problem is misstated, and the task was to find the temperature, not the concentration change. Or perhaps the user intended to ask about the concentration instead of the temperature. Let me check the initial problem statement again: "Calculate the change in oxygen concentration." But the solution provided is for temperature. There's a disconnect here.Wait, maybe the problem was to first find the temperature and then use that to find the concentration change in another step? For example, if oxygen is at this temperature and pressure, then changing one variable affects the concentration. But without additional information on another state (like before and after a change), we can't compute a change. Alternatively, perhaps the problem is about concentration in terms of partial pressure. If oxygen is part of a gas mixture, the concentration could relate to partial pressure. But again, with the given data, only the temperature is calculated. Given the provided information, there's insufficient data to compute a change in oxygen concentration. The solution presented correctly calculates the temperature using the ideal gas law, but the task mentions a different quantity. This might be an error in the problem statement, or perhaps the user intended to ask for temperature but wrote concentration by mistake. If we proceed under the assumption that the task was indeed to calculate temperature, then the solution is correct. However, if the task is truly about oxygen concentration, we need more information such as initial and final states (pressure, volume, temperature) to determine how concentration changes. Without that, we can't proceed. In conclusion, either there's a typo in the task description, or additional information is missing. The provided solution accurately calculates the temperature as 206 K using the given values and the ideal gas law.

✅:The problem presented involves calculating the temperature ( T ) using the ideal gas law ( mathrm{pV} = vmathrm{RT} ). The solution provided correctly computes ( T ) as follows:Given values:- Pressure: ( mathrm{p} = 9 cdot 10^5 , text{Pa} ),- Volume: ( mathrm{V} = 95 cdot 10^{-5} , text{m}^3 ),- Moles: ( v = 0.5 , text{moles} ),- Gas constant: ( mathrm{R} = 8.31 , text{J/(mol·K)} ).Calculation:[mathrm{T} = frac{mathrm{pV}}{vmathrm{R}} = frac{(9 cdot 10^5 , text{Pa}) cdot (95 cdot 10^{-5} , text{m}^3)}{0.5 , text{moles} cdot 8.31 , text{J/(mol·K)}} = frac{855}{4.155} approx 205.776 , text{K} approx 206 , text{K}.]Conclusion:The computed temperature is 206 K. However, the task mentions "calculating the change in oxygen concentration," which is not addressed by the given data. To determine a change in concentration, additional information about two distinct states (e.g., initial and final pressure, volume, or temperature) would be required. As posed, the problem only solves for temperature using the ideal gas law. If the task was intended to compute ( T ), the solution is correct. If there was an error in the task description, further clarification is needed. Final Answer:[boxed{206 , text{K}}]

🔑:1. Start with the ideal gas law equation: [ pV = nRT ] where: - ( p ) is the pressure, - ( V ) is the volume, - ( n ) is the number of moles, - ( R ) is the ideal gas constant, and - ( T ) is the temperature in Kelvin.2. Rearrange the equation to solve for the temperature ( T ): [ T = frac{pV}{nR} ]3. Substitute the given values into the equation: - Pressure (( p )) = ( 9 times 10^5 ) Pa - Volume (( V )) = ( 95 times 10^{-5} ) m(^3) - Number of moles (( n )) = 0.5 mol - Ideal gas constant (( R )) = 8.31 J/(mol·K)4. Perform the substitution: [ T = frac{left(9 times 10^5 , text{Pa} right) times left(95 times 10^{-5} , text{m}^3 right)}{0.5 , text{mol} times 8.31 , text{J/mol·K}} ]5. Calculate the numerator: [ (9 times 10^5) times (95 times 10^{-5}) = 9 times 95 times 10^0 = 855 ]6. Calculate the denominator: [ 0.5 times 8.31 = 4.155 ]7. Divide the numerator by the denominator to find ( T ): [ T = frac{855}{4.155} approx 205.77617 text{ K} ]8. Round the temperature to the nearest whole number: [ T approx 206 text{ K} ] Conclusion:[boxed{approx 206 text{ K}}]

🧠:Alright, so I need to find all integer pairs (k, m) where m > k > 0, such that the sequence x_n defined by x₀ = Fₖ/Fₘ and the recursive relation x_{n+1} = (2x_n - 1)/(1 - x_n) if x_n ≠ 1, and x_{n+1} = 1 otherwise, includes the number 1. First, let me recall that the Fibonacci sequence starts with F₀ = 0, F₁ = 1, and each subsequent term is the sum of the two previous ones. So F₂ = 1, F₃ = 2, F₄ = 3, F₅ = 5, etc. The ratio Fₖ/Fₘ will be a fraction where both numerator and denominator are Fibonacci numbers, with the denominator larger since m > k.Now, the sequence x_n is defined recursively. Starting with x₀ = Fₖ/Fₘ, each subsequent term is obtained by applying the transformation (2x - 1)/(1 - x) unless x is 1, in which case it stays 1. The problem states that the sequence must contain the number 1. That means at some point, applying this transformation to x_n will result in 1. So I need to find all k and m such that iterating this function starting from Fₖ/Fₘ eventually leads to 1.First thought: Maybe there's a pattern or a property of the transformation that can help here. Let's analyze the function f(x) = (2x - 1)/(1 - x). Let's see what happens when we apply it multiple times.Let me compute the first few terms manually for some small k and m to get an idea. Let's pick k=1, m=2. Then F₁=1, F₂=1, so x₀ = 1/1 = 1. But since m must be greater than k, m=2 and k=1 is allowed, but x₀ is already 1. So in this case, the sequence immediately contains 1. Wait, but the problem states m > k > 0. So k must be at least 1, m at least 2.Wait, but if k=1 and m=2, then F₁=1, F₂=1, so x₀=1/1=1. Then the sequence starts at 1, so 1 is present. So (1,2) is a solution. But wait, the problem says m > k > 0. So m must be strictly greater than k. If m=2 and k=1, then m is greater than k. So that's okay. But in this case, x₀=1, so the sequence already contains 1. So (1,2) is a valid pair. But maybe the problem wants pairs where m > k >0, but maybe m=2 is allowed? Let me check.But hold on, Fibonacci numbers: F₀=0, F₁=1, F₂=1, F₃=2, etc. So F₂ is 1, same as F₁. So x₀=F₁/F₂=1/1=1. So in that case, x₀=1, so the sequence immediately has 1. So (1,2) is a solution. But maybe the problem wants 1 to appear in the sequence after some steps, not necessarily at the start? The problem says "contains the number 1". So even if x₀=1, it's included. So (1,2) is a solution. Similarly, if k=1 and m=1, but m must be greater than k, so m=1 is invalid. So the next one is k=1, m=2.But maybe there are other pairs. Let's try k=1, m=3. Then F₁=1, F₃=2, so x₀=1/2. Let's compute the sequence:x₀=1/2.x₁=(2*(1/2) - 1)/(1 - 1/2) = (1 -1)/(1/2)=0/(1/2)=0.x₂=(2*0 -1)/(1 -0)=(-1)/1=-1.x₃=(2*(-1)-1)/(1 - (-1))=(-2 -1)/2=-3/2.x₄=(2*(-3/2)-1)/(1 - (-3/2))=(-3 -1)/(1 + 3/2)=(-4)/(5/2)=-8/5.Hmm, this seems to be diverging. So in this case, starting from 1/2, the sequence goes to 0, -1, -3/2, -8/5, etc., never reaching 1. So (1,3) is not a solution.How about k=1, m=4? F₁=1, F₄=3. So x₀=1/3.Compute x₁=(2*(1/3) -1)/(1 -1/3)=(2/3 -1)/(2/3)=(-1/3)/(2/3)=-1/2.x₂=(2*(-1/2)-1)/(1 - (-1/2))=(-1 -1)/(3/2)=(-2)/(3/2)=-4/3.x₃=(2*(-4/3)-1)/(1 - (-4/3))=(-8/3 -1)/(7/3)=(-11/3)/(7/3)=-11/7.Continuing, it's getting more negative. Not reaching 1. So (1,4) is not a solution.How about k=2, m=3. F₂=1, F₃=2. x₀=1/2. Then same as above, leading to 0, -1, etc. So same as before, not reaching 1.k=2, m=4. F₂=1, F₄=3. x₀=1/3. Then same as k=1, m=4. Doesn't reach 1.How about k=3, m=4. F₃=2, F₄=3. x₀=2/3.Compute x₁=(2*(2/3)-1)/(1 -2/3)=(4/3 -1)/(1/3)=(1/3)/(1/3)=1.So x₁=1. Therefore, the sequence contains 1. Therefore, (3,4) is a solution.Interesting. So starting with 2/3, applying the function once gives 1. So (3,4) works.Similarly, let's check k=1, m=2: x₀=1, so already 1. That's allowed. So (1,2) is a solution. But (3,4) is another solution.Wait, so maybe there's a pattern here. Let's check k=5, m=6. F₅=5, F₆=8. So x₀=5/8.Compute x₁=(2*(5/8)-1)/(1 -5/8)=(10/8 -8/8)/(3/8)=(2/8)/(3/8)=2/3.Then x₂=(2*(2/3)-1)/(1 -2/3)=(4/3 -3/3)/(1/3)=(1/3)/(1/3)=1. So x₂=1. Therefore, starting with 5/8, after two steps we reach 1. So (5,6) is a solution.Similarly, check k=4, m=5. F₄=3, F₅=5. x₀=3/5.x₁=(2*(3/5)-1)/(1 -3/5)=(6/5 -5/5)/(2/5)=(1/5)/(2/5)=1/2.x₂=(2*(1/2)-1)/(1 -1/2)=0/(1/2)=0.x₃=(0 -1)/(1 -0)=-1/1=-1.Continues to negative numbers, not reaching 1. So (4,5) is not a solution.Wait, but k=5, m=6 works. Let's check k=2, m=5. F₂=1, F₅=5. x₀=1/5.x₁=(2*(1/5)-1)/(1 -1/5)=(2/5 -5/5)/(4/5)=(-3/5)/(4/5)=-3/4.x₂=(2*(-3/4)-1)/(1 - (-3/4))=(-6/4 -4/4)/(7/4)=(-10/4)/(7/4)=-10/7.Not reaching 1.Another example, k=5, m=7. F₅=5, F₇=13. x₀=5/13.x₁=(2*(5/13)-1)/(1 -5/13)=(10/13 -13/13)/(8/13)=(-3/13)/(8/13)=-3/8.x₂=(2*(-3/8)-1)/(1 - (-3/8))=(-6/8 -8/8)/(11/8)=(-14/8)/(11/8)=-14/11.Not reaching 1.Hmm. So seems like some pairs where m = k +1 might work. For instance, (1,2), (3,4), (5,6). Let's check (5,6). As we saw, it worked. Let's check k=5, m=6: 5/8. Then x₁=2/3, x₂=1. So two steps. Similarly, k=3, m=4: 2/3, x₁=1. One step. k=1, m=2: x₀=1. Zero steps. So perhaps the pairs where k is odd and m = k +1. Let me check k=7, m=8.F₇=13, F₈=21. x₀=13/21.Compute x₁=(2*(13/21)-1)/(1 -13/21)=(26/21 -21/21)/(8/21)=(5/21)/(8/21)=5/8.Then x₂=(2*(5/8)-1)/(1 -5/8)=(10/8 -8/8)/(3/8)=2/8 / 3/8=2/3.Then x₃=(2*(2/3)-1)/(1 -2/3)= same as before, leads to 1. So x₃=1. Therefore, starting from 13/21, after three steps, reaches 1. So (7,8) is a solution.So the pattern seems that for k=2n-1 and m=k+1, i.e., m=k+1 where k is odd. But in the examples, k=1 (odd), m=2; k=3 (odd), m=4; k=5 (odd), m=6; k=7 (odd), m=8. So all these pairs where k is odd and m=k+1. Then their ratio Fₖ/F_{k+1} leads to a sequence that eventually reaches 1.But wait, let's test k=2, m=3. k=2 even, m=3. x₀=1/2. As before, leads to 0, -1, etc. Doesn't reach 1. Similarly, k=4, m=5: x₀=3/5. Leads to 1/2, 0, -1, etc. Doesn't reach 1. So only when k is odd and m=k+1?But wait, let's check k=1, m=3. k=1 (odd), m=3 which is k+2. x₀=1/2. As before, doesn't reach 1. So maybe only when m =k+1.So conjecture: all pairs where m=k+1 and k is odd.Wait, but k=1, m=2: k=1 (odd), m=k+1=2. Works. k=3, m=4: works. k=5, m=6: works. k=7, m=8: works. So seems like the pattern is k odd and m=k+1.But let's check k=9, m=10. F₉=34, F_{10}=55. x₀=34/55.Compute x₁=(2*(34/55)-1)/(1 -34/55)=(68/55 -55/55)/(21/55)=13/55 /21/55=13/21.x₂=(2*(13/21)-1)/(1 -13/21)=(26/21 -21/21)/(8/21)=5/21 /8/21=5/8.x₃=(2*(5/8)-1)/(1 -5/8)= same as before, leading to 2/3, then 1. So x₄=1. Therefore, (9,10) is a solution.So indeed, this pattern continues. So starting with Fₖ/F_{k+1} where k is odd, the sequence x_n decreases by steps until it reaches 1. For example, k=1: F₁/F₂=1/1=1. So already at 1. k=3: 2/3, which leads to 1 in one step. k=5:5/8 leads to 2/3, then 1. k=7:13/21 leads to 5/8, then 2/3, then 1. Etc. Each time, the ratio Fₖ/F_{k+1} where k is odd leads to a sequence that reduces down to 1.But wait, why does this happen? Let me think about the transformation f(x) = (2x -1)/(1 -x). Let's analyze this function.Let me compute f(f(x)) to see if there's a cycle. Let's compute f(f(x)):First, f(x) = (2x -1)/(1 -x).Then f(f(x)) = [2*( (2x -1)/(1 -x) ) -1 ] / [1 - ( (2x -1)/(1 -x) ) ]Simplify numerator:2*(2x -1)/(1 -x) -1 = [4x -2 - (1 -x)] / (1 -x) = [4x -2 -1 +x]/(1 -x) = (5x -3)/(1 -x)Denominator:1 - (2x -1)/(1 -x) = [ (1 -x) - (2x -1) ] / (1 -x) = [1 -x -2x +1 ]/(1 -x) = (2 -3x)/(1 -x)Therefore, f(f(x)) = [ (5x -3)/(1 -x) ] / [ (2 -3x)/(1 -x) ] = (5x -3)/(2 -3x)Similarly, compute f^3(x) = f(f(f(x))):f(f(f(x))) = f( (5x -3)/(2 -3x) ) = [2*( (5x -3)/(2 -3x) ) -1 ] / [1 - ( (5x -3)/(2 -3x) ) ]Numerator:2*(5x -3)/(2 -3x) -1 = [10x -6 - (2 -3x)] / (2 -3x) = [10x -6 -2 +3x]/(2 -3x) = (13x -8)/(2 -3x)Denominator:1 - (5x -3)/(2 -3x) = [ (2 -3x) - (5x -3) ] / (2 -3x) = [2 -3x -5x +3 ]/(2 -3x) = (5 -8x)/(2 -3x)Thus, f^3(x) = (13x -8)/(5 -8x)Similarly, f^4(x) can be computed, but this might get tedious. However, noticing a pattern here. Each iteration, the coefficients seem to follow Fibonacci numbers? Let's see:First, f(x) numerator: 2x -1, denominator: -x +1f(f(x)) numerator:5x -3, denominator: -3x +2f^3(x) numerator:13x -8, denominator: -8x +5Wait, coefficients: 2,5,13,... which are Fibonacci numbers. Similarly denominators:1,3,8,...Wait:2,5,13: these are F₃=2, F₅=5, F₇=13. Similarly denominators 1,3,8: F₂=1, F₄=3, F₆=8. So seems like f^n(x) = (F_{2n+1}x - F_{2n}) ) / ( -F_{2n}x + F_{2n-1} )Let me check:For n=1: f(x) = (F₃x - F₂)/( -F₂x + F₁ ) = (2x -1)/(-1x +1). Yes, matches.For n=2: f(f(x)) = (F₅x - F₄)/( -F₄x + F₃ ) = (5x -3)/(-3x +2). Yes, matches.For n=3: f^3(x) = (F₇x - F₆)/( -F₆x + F₅ ) = (13x -8)/(-8x +5). Yes. So the pattern holds.Therefore, the nth iteration of f is given by:f^n(x) = (F_{2n+1}x - F_{2n}) / (-F_{2n}x + F_{2n-1})Assuming this pattern continues, which seems to hold based on the first few computations.Therefore, if after n iterations, we reach 1, then f^n(x₀) =1. Let's set f^n(x₀) =1 and solve for x₀.So:(F_{2n+1}x₀ - F_{2n}) / (-F_{2n}x₀ + F_{2n-1}) ) =1Multiply both sides by denominator:F_{2n+1}x₀ - F_{2n} = -F_{2n}x₀ + F_{2n-1}Bring all terms to left:F_{2n+1}x₀ - F_{2n} + F_{2n}x₀ - F_{2n-1} =0Factor x₀:x₀(F_{2n+1} + F_{2n}) ) - (F_{2n} + F_{2n-1})=0But F_{2n+1} + F_{2n} = F_{2n+2} (by Fibonacci recurrence)Similarly, F_{2n} + F_{2n-1} = F_{2n+1}Therefore:x₀ * F_{2n+2} - F_{2n+1} =0Thus:x₀ = F_{2n+1}/F_{2n+2}But x₀ is given as Fₖ/Fₘ. So Fₖ/Fₘ = F_{2n+1}/F_{2n+2} for some integer n ≥0.Therefore, the pairs (k, m) must satisfy Fₖ/Fₘ = F_{2n+1}/F_{2n+2} for some n ≥0.But since k and m are positive integers with m >k, and Fₖ/Fₘ must equal F_{2n+1}/F_{2n+2}, then we must have k=2n+1 and m=2n+2. Because the Fibonacci sequence is strictly increasing for n ≥2, and the ratios Fₖ/Fₘ are unique for k <m when considering consecutive terms. Wait, but wait: for example, F₃/F₄=2/3, which is unique. Similarly, F₅/F₆=5/8, etc. So unless there's another pair where Fₖ/Fₘ equals F_{2n+1}/F_{2n+2}, but since Fibonacci numbers are distinct and the ratios between consecutive terms are unique (since each ratio is in lowest terms? Not sure, but for Fibonacci numbers, consecutive terms are coprime. Let me verify.It's known that consecutive Fibonacci numbers are coprime. Proof: gcd(Fₙ, Fₙ₊₁) = gcd(Fₙ, Fₙ₊₁ - Fₙ) = gcd(Fₙ, Fₙ₋₁) = ... = gcd(F₁, F₂)=1. Therefore, Fₖ and Fₘ are coprime if m =k+1. Therefore, the fraction Fₖ/Fₘ is in lowest terms when m=k+1. Therefore, the only way Fₖ/Fₘ = F_{2n+1}/F_{2n+2} is if k=2n+1 and m=2n+2. Because if two fractions a/b and c/d are in lowest terms and equal, then a=c and b=d.Therefore, the only pairs (k, m) where m=k+1 and k is odd. Because 2n+1 is odd for any integer n ≥0. Thus, k must be odd, and m=k+1.Therefore, all solutions are pairs where k is an odd positive integer and m=k+1.But wait, let's check if there are other possibilities. Suppose that Fₖ/Fₘ = F_{2n+1}/F_{2n+2}. Since Fₖ and Fₘ are coprime when m=k+1, but if m≠k+1, maybe Fₖ/Fₘ can still equal F_{a}/F_{a+1} for some a. For example, is there a pair (k, m) with m≠k+1 where Fₖ/Fₘ = F_{a}/F_{a+1}?Suppose Fₖ/Fₘ = F_{a}/F_{a+1}. Then cross-multiplying: Fₖ F_{a+1} = Fₘ F_{a}. But since consecutive Fibonacci numbers are coprime, unless k=a and m=a+1. Wait, but maybe not. Let's test with numbers.For example, F₃=2, F₄=3, F₅=5. So F₃/F₅=2/5. Is there a pair F_a/F_{a+1}=2/5? Let's see:F₄/F₅=3/5, F₃/F₄=2/3, F₅/F₆=5/8. No, so 2/5 is not equal to any consecutive ratio. Similarly, F₃/F₅=2/5 is not equal to any F_a/F_{a+1}.Another example: F₂=1, F₄=3. So F₂/F₄=1/3. Is 1/3 equal to any F_a/F_{a+1}? F₂/F₃=1/2, F₃/F₄=2/3, F₁/F₂=1/1=1. No, so 1/3 is not a consecutive ratio. Therefore, it's likely that the only pairs where Fₖ/Fₘ is equal to F_a/F_{a+1} are when k=a and m=a+1. Hence, the only solutions are (k, m)=(2n+1, 2n+2) for n ≥0. But since k >0, n can be 0,1,2,...But m must be greater than k, so for n ≥0, k=2n+1 ≥1, m=2n+2 ≥2. So all pairs where k is odd and m=k+1.But the problem states m >k >0, so these pairs are all valid. Thus, the solutions are all pairs of integers (k, m) where m=k+1 and k is a positive odd integer.But wait, when n=0, k=1, m=2. Which is allowed. Then n=1, k=3, m=4; n=2, k=5, m=6, etc.Therefore, the answer should be all pairs where k is odd and m=k+1.But let's confirm with another example. Take k=11, m=12. F₁₁=89, F₁₂=144. x₀=89/144.Applying f(x):x₁=(2*(89/144)-1)/(1 -89/144)=(178/144 -144/144)/(55/144)=(34/144)/(55/144)=34/55.x₂=(2*(34/55)-1)/(1 -34/55)=(68/55 -55/55)/(21/55)=13/55 /21/55=13/21.x₃=(2*(13/21)-1)/(1 -13/21)= similar to previous steps, leads to 5/8, then 2/3, then 1. So after several iterations, it reaches 1. Hence, (11,12) is a solution.Thus, the pattern holds. Therefore, all pairs where k is an odd positive integer and m=k+1 satisfy the condition. Therefore, the solution is all pairs (k, m) with m=k+1 and k odd.But wait, need to make sure there are no other solutions. Suppose there's a pair (k, m) where m ≠k+1, but Fₖ/Fₘ equals F_{2n+1}/F_{2n+2} for some n. But since Fₖ and Fₘ are coprime only if m=k+1, but if m≠k+1, then Fₖ and Fₘ might share a common factor. However, even if they do, F_{2n+1}/F_{2n+2} is in lowest terms (since consecutive Fibonacci numbers are coprime), so Fₖ/Fₘ must reduce to F_{2n+1}/F_{2n+2}. Therefore, Fₖ= d*F_{2n+1} and Fₘ= d*F_{2n+2} for some integer d ≥1. However, Fibonacci numbers have the property that Fₐ divides F_b if and only if a divides b. So if F_{2n+1} divides Fₖ and F_{2n+2} divides Fₘ, then 2n+1 divides k and 2n+2 divides m. But since m >k, and 2n+2 >2n+1, but unless k=2n+1 and m=2n+2, which are consecutive integers. However, Fibonacci numbers are not multiples of each other unless the index is a multiple. For example, F_6=8, which is divisible by F_3=2 (since 6 is a multiple of 3). Similarly, F_9=34, which is divisible by F_3=2 (34/2=17), but F_3 divides F_9. However, if we suppose that Fₖ = d*F_{2n+1} and Fₘ= d*F_{2n+2}, then d must be a common divisor. But since F_{2n+1} and F_{2n+2} are coprime, the only common divisor d can have is a divisor of both F_{2n+1} and F_{2n+2}, which are coprime, so d=1. Therefore, Fₖ=F_{2n+1} and Fₘ=F_{2n+2}, implying k=2n+1 and m=2n+2. Therefore, the only solutions are indeed when m=k+1 and k is odd.Hence, the answer is all pairs of integers (k, m) with m =k+1 and k an odd positive integer.Therefore, the final answer is all pairs where m =k+1 and k is odd, i.e., boxed{(k, m) text{ where } m = k + 1 text{ and } k text{ is an odd positive integer}}.

✅:To determine all pairs of integers ((k, m)) satisfying (m > k > 0) for which the sequence (x_n) defined by (x_0 = frac{F_k}{F_m}) and the given recursive relation contains the number 1, we analyze the transformation function (f(x) = frac{2x - 1}{1 - x}) and its iterations.1. Function Analysis: - The function (f(x)) is applied iteratively starting from (x_0 = frac{F_k}{F_m}). - We need to find when this sequence reaches 1, i.e., when (f^n(x_0) = 1) for some (n).2. Iteration Pattern: - By computing iterations of (f(x)), we observe a pattern in the coefficients of the resulting fractions, which relate to Fibonacci numbers. - The (n)-th iteration of (f) results in (f^n(x) = frac{F_{2n+1}x - F_{2n}}{-F_{2n}x + F_{2n-1}}).3. Solving for 1: - Setting (f^n(x_0) = 1) leads to the equation (frac{F_{2n+1}x_0 - F_{2n}}{-F_{2n}x_0 + F_{2n-1}} = 1). - Solving this equation, we find (x_0 = frac{F_{2n+1}}{F_{2n+2}}).4. Fibonacci Ratios: - For (x_0 = frac{F_k}{F_m}) to equal (frac{F_{2n+1}}{F_{2n+2}}), we must have (k = 2n+1) and (m = 2n+2) due to the coprimality of consecutive Fibonacci numbers.5. Conclusion: - The pairs ((k, m)) must satisfy (m = k + 1) with (k) being an odd positive integer. This is because consecutive Fibonacci numbers are coprime, ensuring the ratio (frac{F_k}{F_{k+1}}) is in lowest terms and leads to the sequence reaching 1.Thus, the solution is all pairs ((k, m)) where (m = k + 1) and (k) is an odd positive integer. The final answer is:[boxed{(k, m) text{ where } m = k + 1 text{ and } k text{ is an odd positive integer}}]

🔑:1. Assume the existence of a specific (k,m) solution: - Suppose the couple (k, m) is a solution, and let n be the smallest index for which x_{n} = 1 . - For all s < n , we have x_{s+1} = frac{2 x_{s} - 1}{1 - x_{s}} .2. Express x_{s} in terms of x_{s+1} : - Solving x_{s+1} = frac{2 x_{s} - 1}{1 - x_{s}} for x_s gives: [ x_{s} = frac{1 + x_{s+1}}{2 + x_{s+1}} ]3. Define the new sequence y_t to match properties of x_n : - Let left( y_{t} right) be a sequence defined by y_{0} = 1 and y_{t+1} = frac{1 + y_{t}}{2 + y_{t}} . - Then x_{0} = y_{n} .4. Identify values for which x_n contains 1: - The values x_{0} for which the sequence ( ( x_{n} ) ) contains 1 are exactly the values taken by the terms of the sequence ( ( y_{t} ) ).5. Examine properties of ( y_t ) using Fibonacci ratios: - Using induction, show that: [ y_{t} = frac{F_{2 t + 1}}{F_{2 t + 2}} ] - We need to determine the pairs (k, m) for which there exists t geqslant 0 such that: [ frac{F_{2 t + 1}}{F_{2 t + 2}} = frac{F_{k}}{F_{m}} ]6. Analyze growth of Fibonacci sequence: - Using the identity F_{2 t + 2} = F_{2 t + 1} + F_{2 t} and properties of Fibonacci sequence growth, we deduce: [ 1 < frac{F_{2 t + 2}}{F_{2 t + 1}} < 2 ]7. Determine constraints on ( k ) and ( m ): - Similarly, from F_{k+2} = 2F_{k} + F_{k-1} , we find: [ frac{F_{k+2}}{F_{k}} > 2 text{ for } k > 1 ] - Thus, if m > k + 1 , we have frac{F_{m}}{F_{k}} > 2 , contradicting with ( frac{F_{2 t + 1}}{F_{2 t + 2}} ).8. Conclude k = m - 1 : - It must hold that m = k + 1 . - By properties of consecutive terms of Fibonacci sequence, they are relatively prime. - Thus, equality ( frac{F_{2 t + 1}}{F_{2 t + 2}} = frac{F_{k}}{F_{m}} ) implies k = 2t + 1 and m = 2t + 2 . Conclusion:- The solutions to the problem are the pairs of the form: [ (2t + 1, 2t + 2) text{ with } t geqslant 0 ] (boxed{(2t + 1, 2t + 2)})

🧠:Okay, let me try to solve this problem. The question is: If 496 equals 2^m minus 2^n, where m and n are integers, then what is m + n? The options are given as (A) 13, (B) 9, (C) 4, (D) 14, (E) 5. Hmm, interesting. So I need to find two integers m and n such that 2^m - 2^n = 496, and then add them together. Let's think about how to approach this.First, I know that 2^m - 2^n can be factored. Let's factor out 2^n from both terms. That would give 2^n(2^{m-n} - 1) = 496. So, 2^n multiplied by (2^{m-n} - 1) equals 496. Now, 496 is an even number, so it's divisible by 2. Let me check what 496 factors into. Let me divide 496 by 2: 496 ÷ 2 = 248. Then 248 ÷ 2 = 124, ÷2 again is 62, ÷2 again is 31. So 496 = 2^4 × 31. Wait, 2^4 is 16, and 16 × 31 is 496. So 496 factors into 16 × 31. Therefore, in the equation 2^n(2^{m-n} - 1) = 496, the left side is 2^n multiplied by another number (which is one less than a power of two), and the right side is 2^4 × 31. So, comparing the two sides, perhaps 2^n is one of the factors of 496 that's a power of two, and the other factor is 2^{m-n} - 1, which must be 31. Let's check.Since 496 is 2^4 × 31, then maybe 2^n is 2^4, which would make n = 4, and then the other factor would be 31. Then, 2^{m - n} - 1 = 31. So, 2^{m - 4} = 32. Because 32 is 2^5. So, 2^{m - 4} = 2^5, so m - 4 = 5, so m = 9. Therefore, m = 9 and n = 4, so m + n = 13. Wait, but 13 is one of the options, option A. But let me check if there could be other possibilities.Alternatively, maybe 2^n is a smaller power of two. Let's see. For example, 2^3 = 8. If n = 3, then the other factor would be 496 ÷ 8 = 62. So then, 2^{m - 3} - 1 = 62. Then, 2^{m - 3} = 63. But 63 is not a power of two. 2^6 is 64, which is 64, so 63 is one less than 64. But 2^{m - 3} would be 63, which is not a power of two. So that doesn't work. Similarly, if n = 2, then 2^2 = 4, 496 ÷ 4 = 124. So 2^{m - 2} - 1 = 124. Then 2^{m - 2} = 125. But 125 is 5^3, not a power of two. So that's invalid. If n = 1, 2^1 = 2, 496 ÷ 2 = 248. Then 2^{m - 1} - 1 = 248. So 2^{m - 1} = 249. Again, 249 is not a power of two. 2^7 is 128, 2^8 is 256. 249 is between them. So no. If n = 0, 2^0 = 1, then 496 ÷ 1 = 496. Then 2^{m - 0} - 1 = 496. So 2^m = 497. But 497 is not a power of two. 2^8 is 256, 2^9 is 512, so 497 is between them. So no. So that's invalid. So the only valid case seems to be when n = 4, leading to m = 9, so m + n = 13, which is option A. But wait, let me check if there are other factors.Wait, 496 is 16 × 31. But 31 is prime, right? Let me confirm. 31 is a prime number, yes. So 31 can't be broken down into smaller factors. So the only way to write 496 as 2^n times (something) is 16 × 31 or 8 × 62, 4 × 124, etc., but since 31 is prime and 62 is 2 × 31, 124 is 4 × 31, etc. So in all those cases, the (2^{m - n} - 1) term would have to be 31, 62, 124, etc., but only 31 is one less than a power of two. Because 31 + 1 = 32 = 2^5. Whereas 62 + 1 = 63, which is not a power of two, 124 + 1 = 125, which isn't a power of two either. So only when (2^{m - n} - 1) = 31, which gives 2^{m - n} = 32 = 2^5, so m - n = 5. Then, since 2^n = 16, which is 2^4, so n = 4, m = 4 + 5 = 9. Therefore, m + n = 13, which is option A.But let me check if maybe n is larger than 4? For example, if n = 5, then 2^5 = 32. Then 496 ÷ 32 = 15.5. Wait, but 15.5 is not an integer, so that's not possible. Similarly, n can't be higher than 4 because 2^5 is 32, which divides into 496 15.5 times, which is not integer. So n must be 4. Therefore, the only solution is m = 9 and n = 4, so m + n is 13, option A. So that should be the answer. Let me just verify by plugging the numbers back in. 2^9 is 512, 2^4 is 16. 512 - 16 is 496. Yes, that's correct. So the answer is 13, which is option A. Wait, but the options given are (A) 13, (B) 9, (C) 4, (D) 14, (E) 5. So yes, A is 13.But just to be thorough, let me check if there's another pair m and n that could satisfy 2^m - 2^n = 496. For example, maybe m and n are such that m is 10 and n is something else. Let's see. 2^10 is 1024. If we subtract 2^n from that to get 496, then 2^n would have to be 1024 - 496 = 528. But 528 is not a power of two. 512 is 2^9, 528 - 512 = 16, which is 2^4, but 528 = 512 + 16, which is 2^9 + 2^4, but that's not a single power of two, so no. How about m = 8? 2^8 = 256. Then 256 - 2^n = 496 would imply 2^n = 256 - 496 = -240, which is impossible since 2^n is positive. Similarly, m = 7: 128 - 2^n = 496 ⇒ 2^n = 128 - 496 = -368, which is invalid. So m has to be larger than n, and 2^m must be larger than 2^n. So the only possible solution is m = 9 and n = 4. Therefore, m + n = 13. So the answer is A.Final Answerboxed{A}

✅:Given the equation (496 = 2^m - 2^n) where (m) and (n) are integers, we need to find (m + n).First, we factorize the equation as (2^n(2^{m-n} - 1) = 496). The prime factorization of 496 is (2^4 times 31). This suggests that (2^n) must be a power of 2 factor of 496, and (2^{m-n} - 1) must be the remaining factor.Since 496 factors into (2^4 times 31), we consider (2^n = 2^4) (which gives (n = 4)) and (2^{m-n} - 1 = 31). Solving (2^{m-n} - 1 = 31) gives (2^{m-n} = 32), which is (2^5). Therefore, (m - n = 5), leading to (m = 4 + 5 = 9).Thus, (m = 9) and (n = 4), so (m + n = 9 + 4 = 13).Verification: (2^9 - 2^4 = 512 - 16 = 496), which confirms the solution is correct.The final answer is (boxed{A}).

🔑:To solve the equation ( 496 = 2^m - 2^n ) where ( m ) and ( n ) are integers, we need to express 496 as the difference between two powers of 2.1. Identify Closest Power of 2 Greater than 496: The first step is to identify a power of 2 that is greater than, but close to, 496. [ 2^8 = 256 quad text{(not close enough)} ] [ 2^9 = 512 quad text{(this works)} ]2. Compute the Subtraction to Match 496: We can see that: [ 496 = 512 - 16 = 2^9 - 2^4 ]3. Verify the Exponents: By expressing both 512 and 16 as powers of 2, we have: [ 2^9 = 512 quad text{and} quad 2^4 = 16 ] So: [ 496 = 2^9 - 2^4 ]4. Sum of the Exponents: The problem asks for the sum of ( m ) and ( n ) where ( m = 9 ) and ( n = 4 ): [ m + n = 9 + 4 = 13 ] Conclusion:The value of ( m + n ) is:[boxed{13}]