🧠:Okay, let's try to tackle this problem. So, we have a function f defined on the positive integers, and it satisfies two conditions: first, for any two positive integers m and n, f(m·n) = f(m) + f(n). Second, the function is non-decreasing, meaning that f(n+1) is always greater than or equal to f(n). We need to find what f(n) is.Hmm, let me start by recalling similar functions. The condition f(m·n) = f(m) + f(n) reminds me a lot of logarithms because logarithms convert multiplication into addition. But here, the function is defined on positive integers, not real numbers, and the codomain isn't specified. However, logarithms are usually the go-to example for such multiplicative functions. But wait, the problem also states that f is non-decreasing. That might be important because not all logarithmic functions (with different bases) are non-decreasing on the integers, but actually, logarithms with base greater than 1 are increasing functions. So maybe f(n) is related to the logarithm of n? But since we're dealing with integers, maybe it's a constant multiple of the logarithm? Let me check.But first, let's verify if such a function can indeed be logarithmic. Let's suppose f(n) = log_b(n) for some base b. Then, f(m·n) = log_b(m·n) = log_b(m) + log_b(n) = f(m) + f(n), so that condition is satisfied. Now, is this function non-decreasing? Yes, because logarithm functions are increasing if the base b is greater than 1. So for example, base 2, 10, e, etc. But since n is a positive integer, log_b(n) is defined and increasing as n increases. However, the problem states that f is defined on positive integers, but it doesn't specify the codomain. If we're allowed to have real numbers as outputs, then logarithms would work. But if the function f must map to integers, then logarithms might not work unless we use a base that results in integer outputs for integer inputs, which is not typical except for trivial cases.Wait, the problem says "function defined on the set of positive integers", but it doesn't specify the codomain. So perhaps the codomain is the real numbers? Or maybe integers? The problem doesn't say. Hmm, this is a bit ambiguous. But in mathematical problems like this, if the codomain isn't specified, it's often assumed to be real numbers unless stated otherwise. So maybe we can proceed under that assumption.But let's not jump to conclusions. Let's see if there are other functions that satisfy f(mn) = f(m) + f(n). Another example is the number of prime factors, but that's only additive over multiplication if you count with multiplicity. For example, the function Ω(n), which counts the total number of prime factors of n, including multiplicity, satisfies Ω(mn) = Ω(m) + Ω(n). But is this function non-decreasing? Let's check. Let's take n=1, Ω(1)=0; n=2, Ω(2)=1; n=3, Ω(3)=1; n=4, Ω(4)=2; n=5, Ω(5)=1; n=6, Ω(6)=2; n=7, Ω(7)=1; n=8, Ω(8)=3; etc. Wait, but Ω(n) isn't non-decreasing. For example, going from n=4 to n=5, Ω(n) decreases from 2 to 1. So that violates the second condition. Therefore, Ω(n) is not a candidate here.Another thought: maybe the function is related to the exponent in the prime factorization. For example, if we fix a prime p and define f(n) as the exponent of p in the prime factorization of n. Let's denote this as f(n) = v_p(n). Then, for m and n, v_p(mn) = v_p(m) + v_p(n), which satisfies the first condition. However, is this function non-decreasing? Let's check. Take p=2. Then, v_2(1)=0, v_2(2)=1, v_2(3)=0, v_2(4)=2, v_2(5)=0, etc. So this function is definitely not non-decreasing. For example, from n=2 to n=3, the function decreases from 1 to 0. Hence, this also doesn't satisfy the second condition.Hmm, so maybe the function isn't related to prime factors. Let me think again. The first condition is that f is additive over multiplication, which is similar to a logarithmic function. The second condition is that it's non-decreasing. So, if we consider logarithms, as I initially thought, they are increasing functions. But if we take f(n) = log(n), then f(n) is indeed non-decreasing. But the problem is about functions defined on positive integers. So, unless the function is a constant multiple of log(n), but scaled in some way.Wait, but even log(n) isn't integer-valued unless n is a power of the base. So if the codomain is supposed to be integers, then f(n) must be integer-valued, which complicates things. But the problem didn't specify, so maybe the answer is f(n) = log(n) for some base. However, the problem is presented in a way that expects a specific function, possibly with some constraints. Wait, but there's another possibility: the zero function. If f(n) = 0 for all n, then f(mn) = 0 = 0 + 0 = f(m) + f(n), and it's non-decreasing because all values are equal. But is this the only constant function? If f(n) is a constant function, say c, then f(mn) = c = c + c => c = 0. So the only constant function possible is f(n)=0. But is this the only solution?Wait, no. If we consider f(n) = 0 for all n, that works, but maybe there are other functions. For example, if f(n) is a logarithm function, which is non-constant. However, if we take f(n) = 0 for all n, then it's certainly a solution. But the problem says "Find f(n)", so perhaps there are multiple solutions? Or maybe the only solutions are f(n)=0 and f(n)=log_b(n) for some base b. But wait, if we need the function to be defined on positive integers, then f(n) can be log(n) scaled by a constant, but unless the log is integer-valued, which is only for n being powers of the base. Otherwise, log(n) would not be an integer, but since the codomain isn't specified, maybe real numbers are allowed. Hmm.But the problem is in a style that might expect a unique solution. Let me check the conditions again. The function is additive over multiplication and non-decreasing. Let's think step by step.First, let's analyze the functional equation: f(mn) = f(m) + f(n) for all m, n in N+. This is Cauchy's functional equation but in the multiplicative form. The solutions to this over positive integers (if we consider codomain as real numbers) are functions of the form f(n) = log(n) for some base, or equivalently f(n) = c * log(n) where c is a constant. However, over the positive integers, there might be more constraints.But in addition, the function is non-decreasing. So even if we have a logarithmic function, we need to check if it's non-decreasing. Since log(n) is increasing, that's fine. But if we take c * log(n), then if c is positive, it's increasing; if c is negative, it's decreasing. So, to maintain non-decreasingness, c must be non-negative. But if c=0, then f(n)=0 for all n, which is also a valid solution.Wait, so maybe the solutions are f(n)=c * log(n) for some constant c ≥ 0. But the problem says "Find f(n)". If c can be any non-negative real number, then there are infinitely many solutions. But the problem might be expecting a specific function, perhaps f(n)=0 or f(n)=log(n). Wait, but in the problem statement, it's possible that the only solution is f(n)=0. Wait, let's check that.Suppose there exists some n>1 such that f(n) >0. Then, since f is non-decreasing, for all k >=n, f(k) >= f(n) >0. But let's take a prime number p. Then f(p) must satisfy certain conditions. For example, if we take p=2, then f(2) is some value. Then f(4)=f(2)+f(2)=2f(2). Similarly, f(8)=3f(2), etc. Similarly, for other primes, like p=3: f(3) is some value. Now, if the function is non-decreasing, then f(3) must be >= f(2), because 3>2. Similarly, f(4)=2f(2) must be >= f(3). So 2f(2) >= f(3) >= f(2). Similarly, f(5) >= f(4)=2f(2). Then f(6)=f(2)+f(3) must be >= f(5). But f(6)=f(2)+f(3) >= f(2) + f(2)=2f(2). But f(5) >= 2f(2), so f(6) >= f(5). Continuing this way, we might get that the function is determined by its values at primes, but with constraints from the non-decreasing condition.Alternatively, if we suppose that f(n) = c log n, then to make it non-decreasing, c must be >=0. If c>0, then the function is strictly increasing, which is non-decreasing. If c=0, it's constant. So these are all possible solutions. But the problem might be expecting us to conclude that f(n) is proportional to log n, but since the problem is in Chinese or from an Olympiad, perhaps expecting f(n)=0 or f(n)=k log n where k is a constant.But wait, the problem says "Find f(n)", which might imply a unique solution. So maybe only the zero function is possible. Let's test this. Suppose f is not identically zero. Then there exists some n such that f(n) >0. Let's pick the smallest such n, say n=k. Since f(1*1)=f(1)+f(1) => f(1)=2f(1) => f(1)=0. So f(1)=0. Then, for n=1, f(1)=0. Suppose the smallest n where f(n) >0 is n=2. Then f(2) >0. Then f(2)=f(2*1)=f(2)+f(1)=f(2)+0, which is okay. Now, since f is non-decreasing, f(3) >= f(2). Also, f(4)=f(2*2)=2f(2). Then, since 3 <4, f(3) <= f(4)=2f(2). So f(2) <= f(3) <=2f(2). Similarly, f(5) >=f(4)=2f(2). Then f(6)=f(2*3)=f(2)+f(3). Since f(3) <=2f(2), f(6) <=f(2)+2f(2)=3f(2). But f(5) >=2f(2), and f(6) >=f(5)>=2f(2). So f(6) is between 2f(2) and 3f(2). But how does this continue?This seems complicated. Maybe if we assume f(n) = c log n, then the non-decreasing condition is automatically satisfied. But perhaps the problem constraints force c=0. Let's see. Suppose f(n) = c log n. Then for the function to be non-decreasing, c >=0. But unless c=0, the function is strictly increasing. However, can a non-zero c exist? For example, take c=1, so f(n)=log n. Then, f(n+1) >f(n) for all n, which satisfies f(n+1) >=f(n). So this is acceptable. But the problem doesn't specify any other conditions, so why would the only solution be f(n)=0?Wait, but maybe over the positive integers, the logarithm function isn't well-defined unless we choose a specific base. Wait, no. Logarithms are defined for any positive real number. But if we're considering f(n) = log n, then for each n, it's a real number, which is allowed if the codomain is the reals. But maybe the problem expects an integer-valued function. If the codomain is integers, then f(n) must be integer. In that case, log n can only be integer if n is a power of the base. For example, if base is 2, then f(n)=log2(n) is integer only when n is a power of 2. But for other n, it's non-integer. So unless the function is zero everywhere, which is integer, or unless the function is defined as zero, maybe that's the only solution.But the problem didn't specify that the function must be integer-valued. If it's allowed to be real-valued, then f(n)=c log n is a solution for any c>=0. But perhaps the problem is in a context where only integer values are considered. For example, in Olympiad problems, sometimes functions are expected to be integer-valued unless stated otherwise. But in the problem statement, there's no such specification. Hmm.Alternatively, maybe there's a unique solution regardless of codomain. Let's think differently. Let's analyze small n and see what f(n) must be.We know that f(1) = f(1*1) = f(1) + f(1), which implies f(1) = 0.Now, consider n=2. Let's denote f(2) = a. Then, f(4) = f(2*2) = 2a. Similarly, f(8)=3a, and so on. For n=3, let's denote f(3)=b. Then, f(9)=2b, etc.Now, since the function is non-decreasing, we have:For n=1: f(1)=0n=2: f(2)=an=3: f(3)=bn=4: f(4)=2an=5: f(5)=cn=6: f(6)=f(2)+f(3)=a + bn=7: f(7)=dn=8: f(8)=3aAnd so on.Now, the non-decreasing condition implies:f(2) >= f(1)=0 => a >=0f(3) >= f(2)=a => b >=af(4)=2a >= f(3)=b => 2a >= bf(5) >= f(4)=2a => c >=2af(6)=a + b >= f(5)=c => a + b >=cf(7)=d >= f(6)=a + b => d >=a + bf(8)=3a >= f(7)=d =>3a >= dAnd so on.Let's try to find relations between these variables. Let's suppose we can assign values step by step.From f(3) >=a and 2a >= f(3), so a <=b <=2a.Similarly, f(5) >=2a, and a + b >=f(5). So 2a <=c <=a + b.But since b <=2a, then a + b <=a +2a=3a. Therefore, c is between 2a and 3a.Similarly, f(7)=d >=a + b, and 3a >=d. So a + b <=d <=3a.But since b >=a and <=2a, a + b is between 2a and 3a. So d is between 2a and 3a.This seems quite flexible. Let's see if we can set a=0. If a=0, then f(2)=0. Then from b >=a=0 and 2a=0 >=b, so b=0. Similarly, c >=2a=0, but c <=a + b=0. So c=0. Similarly, d >=a + b=0 and d <=3a=0, so d=0. Continuing this way, all f(n)=0. So the zero function is a solution.Alternatively, suppose a>0. Let's set a=1 for simplicity. Then f(2)=1.From b >=1 and 2a=2 >=b, so 1 <=b <=2.Similarly, c >=2, and c <=a + b=1 + b.Since b <=2, 1 + b <=3, so c is between 2 and 3.But c must be an integer if we assume integer values, but the problem doesn't specify. If real values are allowed, c could be any real number between 2 and 3.But let's suppose we are allowed real numbers. Then, we can set b=1.5, which is between 1 and 2. Then c can be between 2 and 2.5 (since 1 +1.5=2.5). So c=2.3, for example. Then d must be between 2.5 and 3. Let's say d=2.8. And so on. However, this seems arbitrary. The problem is to find f(n) in general, not to construct an example. So perhaps there's a unique form.But considering that f(mn)=f(m)+f(n), this is similar to a logarithm function. If we define f(n)=k log n, then f(mn)=k log(mn)=k log m +k log n =f(m)+f(n), so this works. And if k >=0, then the function is non-decreasing. So the general solution would be f(n)=k log n for some constant k >=0. However, if the problem expects a specific function, maybe k=0, leading to f(n)=0, since otherwise, there are infinitely many solutions. But the problem says "Find f(n)", which could imply that all solutions are of the form f(n)=k log n with k >=0. But maybe in the context of the problem, the only possible solution is f(n)=0.Wait, let's check for n=2. Suppose f(2)=a>0. Then f(4)=2a. Since the function is non-decreasing, f(3) must be >=a and <=2a. Similarly, f(5)>=2a, but f(6)=f(2)+f(3)=a + f(3). Since f(3)<=2a, f(6)<=3a. But f(5)>=2a and f(6)>=f(5). So f(6)>=2a. So f(6)=a + f(3)>=2a => f(3)>=a. Which we already have. So f(3) is between a and 2a.But then f(4)=2a, f(5)>=2a, f(6)>=2a. Then f(8)=3a. Now, f(7)>=f(6)>=2a and <=3a. But how does this generalize? It seems that for primes greater than 2, their f-values must be at least the f-value of the previous number, which complicates things.However, if we assume that f(n) =k log n, then for primes p, f(p)=k log p. For example, f(2)=k log2, f(3)=k log3, etc. Since log3 > log2, and k>0, this would imply f(3) >f(2), which is okay. Similarly, f(4)=2k log2, which is greater than f(3)=k log3 if 2 log2 > log3. Since 2 log2 ≈1.386, and log3≈1.098, so 1.386>1.098, so yes, f(4)>f(3). Wait, but that contradicts because f(3)=k log3 and f(4)=2k log2. If k=1, then f(4)=2*0.693≈1.386, f(3)=1.098, so indeed f(4)>f(3). But the function needs to be non-decreasing. However, from n=4 to n=5, f(4)=1.386 and f(5)=log5≈1.609, so f(5)>f(4), which is okay. Similarly, n=6: f(6)=f(2)+f(3)=0.693+1.098≈1.791, which is greater than f(5)=1.609, so f(6)>f(5). Then n=7: f(7)=log7≈1.946>1.791, etc. So actually, f(n)=log n is increasing, hence non-decreasing. So if we take any k>=0, then f(n)=k log n is non-decreasing if k>=0. Therefore, all such functions are solutions.However, the problem states "Find f(n)". If multiple solutions exist, we need to characterize all of them. But maybe in the problem's context, the only solution is f(n)=0. Let's see why. Suppose there exists a prime p where f(p)>0. Then, for powers of p, f(p^k)=k f(p). As k increases, f(p^k) increases without bound. However, for numbers between p^k and p^{k+1}, since the function is non-decreasing, their f-values must be at least f(p^k)=k f(p). But for example, consider numbers like p^k +1. Their f-values must be >=k f(p). But p^k +1 is not divisible by p, so f(p^k +1) can't be directly related to f(p). However, in reality, with f(n)=k log n, this is satisfied because log(p^k +1) >k log p, but it's not linear in k. Wait, log(p^k +1) ≈k log p + log(1 +1/p^k) ≈k log p, so for large k, it's approximately k log p. But the exact value would still be less than log(p^{k+1})= (k+1) log p. So f(n)=log n grows logarithmically, which is slower than any linear function. Therefore, if we take n=p^k +1, f(n)=log(p^k +1) < log(p^{k}·p)=log p^{k+1}= (k+1) log p. So this is okay because the function is allowed to increase slowly.But maybe there's a contradiction if we assume a non-zero solution. Let's suppose f is non-zero. Then there exists some n with f(n)>0. Let's take the smallest such n, which must be a prime. Because if n is composite, say n=ab, then f(n)=f(a)+f(b). If a and b are smaller than n, and if n is the smallest with f(n)>0, then f(a) and f(b) would be zero, leading to f(n)=0, a contradiction. Therefore, the smallest n with f(n)>0 must be a prime. Let's call this prime p. Then f(p)=c>0. Now, consider the number p+1. Since p is the smallest prime (if p=2), then p+1=3, which is also prime. But if p=2, then p+1=3, which is prime. If we take p=3, p+1=4, which is composite. Wait, but the smallest prime is 2. So if the smallest n with f(n)>0 is 2, then f(2)=c>0. Then f(3)>=f(2)=c. But 3 is prime, so f(3)=d>=c. Then f(4)=f(2)+f(2)=2c. Since 3<4, f(3)=d<=2c. So c <=d<=2c. Similarly, f(5)>=f(4)=2c. Now, 5 is prime, so f(5)=e>=2c. Then f(6)=f(2)+f(3)=c + d >=c +c=2c. But f(6) must also be >=f(5)=e>=2c. So f(6)>=2c. Similarly, f(7)>=f(6)>=2c. And so on.This seems possible. For example, if we take c=1, then f(2)=1, f(3) can be between 1 and 2. Suppose f(3)=1.5. Then f(4)=2, f(5)>=2. Suppose f(5)=2. Then f(6)=f(2)+f(3)=1+1.5=2.5 >=f(5)=2. Then f(7)>=2.5, say 3, and so on. However, this requires defining f(n) for primes in a way that they are non-decreasing and satisfy the multiplicative condition. But this seems arbitrary unless there's a pattern.However, if we follow the logarithmic function, then f(n)=log n. For primes p, f(p)=log p, which increases as p increases. This would automatically satisfy the non-decreasing condition because primes are increasing and their logarithms are increasing. For composite numbers, their f-values are sums of the logs of their prime factors, which also maintain the non-decreasing property because when you multiply numbers, their logarithms add up. So as n increases, the total sum (log n) increases, hence f(n) is non-decreasing.But the problem doesn't specify any constraints other than the two given. Therefore, the solutions are all functions of the form f(n)=k log n for some constant k>=0. However, the problem might expect the general form. But in Chinese mathematical competitions, sometimes they expect the answer to be the zero function if additional constraints aren't given. Let's check with n=2 and n=3.Suppose f(2)=a and f(3)=b. From the non-decreasing condition, a <=b <=2a. Then f(4)=2a, f(5)>=2a, f(6)=a + b >=2a => b >=a. But we already have that. Now, consider f(5). Since 5 is prime, f(5)=c >=2a. Then f(10)=f(2)+f(5)=a + c >=c >=2a. Also, f(9)=2b, and since 9>8, f(9)>=f(8)=3a. So 2b >=3a => b >=1.5a. So combining with previous b <=2a, we have 1.5a <=b <=2a.Similarly, for f(7)=d >=f(6)=a + b. Since 7 is prime, f(7)=d >=a + b. Then f(14)=f(2)+f(7)=a + d >=a + (a + b)=2a + b. But f(14) must also be >=f(13)>=f(12)=f(3)+f(4)=b +2a. So f(14)>=b +2a. Which matches.This seems to allow for a non-zero solution. For example, let's set a=1. Then b must be between 1.5 and 2. Let's choose b=1.5. Then f(3)=1.5. Then f(4)=2, f(5)>=2, let's say f(5)=2. Then f(6)=1 +1.5=2.5, f(7)>=2.5, say f(7)=3. Then f(8)=3, f(9)=2*1.5=3, f(10)=1 +2=3, f(11)>=3, etc. But this construction requires choosing values arbitrarily, and there's no guarantee that these choices would be consistent for all n. For example, consider n=12=3*4. Then f(12)=f(3)+f(4)=1.5 +2=3.5. But f(12) must also be >=f(11)>=3. So f(12)=3.5 is okay. Then f(13)>=3.5, and so on.However, this approach requires defining f(n) for each prime in a way that maintains the non-decreasing condition. But unless there's a systematic way to assign these values, it's hard to see how this could be done without ending up with a logarithmic function. In fact, if we enforce that the function is additive over multiplication and non-decreasing, the only functions that satisfy both conditions are the constant zero function and the logarithm functions scaled by a non-negative constant.But why can't there be other functions? Suppose we try to define f(n) in a piecewise manner. For example, f(n)=0 for n<5, and f(n)=1 for n>=5. But then f(2*3)=f(6)=1, but f(2)+f(3)=0+0=0, which doesn't equal 1. So that doesn't work. Therefore, such a piecewise function won't satisfy the additive condition.Another attempt: suppose f(n) counts the number of digits in n. But f(mn) is not equal to f(m)+f(n). For example, f(10*10)=f(100)=3, but f(10)+f(10)=2+2=4 ≠3. So that doesn't work.Alternatively, suppose f(n) is the identity function, f(n)=n. Then f(mn)=mn ≠m +n =f(m)+f(n). Doesn't work.Alternatively, f(n) is the exponent in the prime factorization for a specific prime, but as we saw before, that's not non-decreasing.Therefore, the only viable solutions seem to be f(n)=k log n for k>=0 or f(n)=0.But the problem asks to "Find f(n)". If there are multiple solutions, we need to state all of them. But in the absence of additional constraints, both the zero function and logarithmic functions are valid. However, in many mathematical problems, especially in Olympiads, if a functional equation allows for a family of solutions, they might expect you to characterize all solutions. But the way the problem is phrased, "Find f(n)", might suggest that there's a unique solution. Let's check again.If we consider that f is defined on positive integers and the codomain is also positive integers, then the only solution is f(n)=0 for all n. Because if we assume f maps to integers, then for f(n)=k log n to be integer, n must be a power of the base. But since n can be any positive integer, the only way this works is if k=0, leading to f(n)=0. Otherwise, for non-integer logs, f(n) would not be an integer. Therefore, if the codomain is integers, the only solution is f(n)=0.However, the problem doesn't specify the codomain. If the codomain is real numbers, then there are infinitely many solutions of the form f(n)=k log n. But in Olympiad problems, unless specified, functions often map to real numbers. However, the fact that the function is non-decreasing might interact with the multiplicative condition in a way that forces k=0.Wait, let's suppose f(n)=k log n is a solution with k>0. Then it's strictly increasing, which satisfies non-decreasing. So why would k=0 be forced? There's no contradiction here. So perhaps both the zero function and logarithmic functions are valid solutions. But the problem says "Find f(n)", which might expect all possible solutions. Therefore, the answer would be that f(n)=c log n for some constant c>=0. But to write this, we need to confirm.Alternatively, maybe the only solution is f(n)=0. Let's consider that if f is non-decreasing and satisfies f(mn)=f(m)+f(n), then f(n)=0 for all n.Suppose there exists some n with f(n) >0. Let's take the smallest such n, which must be prime, as we discussed earlier. Let's call this prime p. Then f(p)=c>0. Now, consider the number p+1. Since p is the smallest prime with f(p)=c>0, and since f is non-decreasing, f(p+1)>=c. However, p+1 is either a prime or composite. If it's composite, it can be factored into primes smaller than p, but since p is the smallest prime with f(p)=c>0, all smaller primes have f(q)=0. Therefore, if p+1 is composite, its factors are smaller than p, so f(p+1)=sum of f of its factors, which would be sum of zeros, so f(p+1)=0. But f(p+1)>=c>0, which is a contradiction. Therefore, p+1 must be prime. But primes are not always consecutive. For example, if p=2, then p+1=3, which is prime. Then f(3)>=f(2)=c>0. Then 3 is the next prime. Then 4 is composite, and f(4)=f(2)+f(2)=2c. Since 4>3, f(4)>=f(3)>=c. So 2c >=f(3)>=c. Then, 5 is prime. f(5)>=f(4)=2c. But 5 is prime, so f(5)=d>=2c. Then 6=2*3, so f(6)=f(2)+f(3)=c + f(3). Since f(3)>=c, f(6)>=2c, which is okay because f(5)>=2c and f(6)>=f(5). However, if we go to 7, which is prime, f(7)>=f(6)>=2c. But 7 is prime, so f(7)=e>=2c. Then 8=2*4, f(8)=f(2)+f(4)=c +2c=3c. Since 8>7, f(8)=3c >=f(7)=e>=2c. This is okay.However, when we get to p=2, p+1=3 is prime, which is okay. Then p=3, p+1=4, which is composite. But f(4)=2c. However, 4's factors are 2 and 2, which have f(2)=c>0. So f(4)=2c. Now, f(4)=2c >=f(3)>=c. If f(3)=c, then f(4)=2c >=c, which is okay. But then f(5)>=2c. Since 5 is prime, f(5)=d>=2c. Then f(6)=f(2)+f(3)=c +c=2c. But f(6)>=f(5)=d>=2c. So f(6)=2c >=d>=2c => d=2c. So f(5)=2c.Next, f(7)>=f(6)=2c. Since 7 is prime, f(7)=e>=2c. Then f(8)=3c >=f(7)=e>=2c. So e can be between 2c and 3c. But then f(9)=f(3*3)=f(3)+f(3)=c +c=2c. But 9>8, so f(9)=2c >=f(8)=3c? No, that would require 2c >=3c, which implies c<=0. But c>0, which is a contradiction. Therefore, our assumption leads to a contradiction.Ah! Here's the contradiction. If f(9)=2c, but 9>8, so f(9)>=f(8)=3c. Therefore, 2c >=3c => c<=0. But c>0, which is a contradiction. Therefore, our initial assumption that there exists a prime p with f(p)=c>0 leads to a contradiction. Hence, the only possible solution is f(n)=0 for all n.Therefore, the answer is f(n)=0 for all n.This conclusion comes from the fact that assuming a non-zero solution leads to a contradiction when considering n=9. Because f(9) would have to be both 2c (from being 3*3) and at least 3c (from being greater than 8, which has f(8)=3c). This forces c=0, hence f(n)=0.

✅:The function ( f ) satisfies two key properties: it is additive over multiplication (( f(mn) = f(m) + f(n) )) and non-decreasing (( f(n+1) geq f(n) )). To determine ( f(n) ):1. Functional Equation Analysis: The equation ( f(mn) = f(m) + f(n) ) resembles logarithmic functions, which convert multiplication to addition. However, we must verify if logarithmic functions or other solutions meet the non-decreasing condition.2. Non-Decreasing Constraint: If ( f ) were logarithmic (( f(n) = c log n )), it would satisfy the additive property. For ( c > 0 ), ( f(n) ) is strictly increasing, hence non-decreasing. However, this leads to contradictions when analyzing specific values. For instance, ( f(9) = 2c ) (from ( 3 times 3 )) must also satisfy ( f(9) geq f(8) = 3c ), implying ( 2c geq 3c ), which forces ( c = 0 ).3. Zero Function Verification: The constant function ( f(n) = 0 ) trivially satisfies both conditions. Testing for non-zero solutions leads to inconsistencies, as shown in the step-by-step analysis, confirming that no non-zero ( c ) is valid.Final Answer: The only function satisfying the given conditions is [boxed{f(n) = 0}] for all positive integers ( n ).

🔑:To determine ( f(n) ) where ( f ) is a function defined on the set of positive integers ( mathbf{N}^{+} ) such that for all ( m, n in mathbf{N}^{+} ), the following properties hold:1. ( f(m cdot n) = f(m) + f(n) )2. ( f(n+1) geqslant f(n) )we proceed as follows:1. Initial Setup: ( r, s, n in mathbf{N} ) where ( n geqslant 1 ), ( s geqslant r geqslant 2 ), and ( m in mathbf{N}^{+} ). From the properties of exponential functions, we can deduce that there exists an appropriate ( m ) satisfying: [ r^{m} leqslant s^{m} leqslant r^{m+1} ]2. Taking the Natural Logarithm: By taking the natural logarithm of the inequality, we get: [ m ln r leqslant n ln s leqslant (m+1) ln r ] Dividing the entire inequality by ( n ): [ frac{m}{n} leqslant frac{ln s}{ln r} leqslant frac{m}{n} + frac{1}{n} ]3. Applying the Function ( f ): By applying ( f ) to the exponential terms, we use the property ( f(r^n) = n f(r) ) and generate the following inequalities: [ f(r^m) leqslant f(s^n) leqslant f(r^{m+1}) ]4. Substituting ( f ) Values: Since ( f(r^m) = m f(r) ) and ( f(r^{m+1}) = (m+1) f(r) ), we have: [ m f(r) leqslant n f(s) leqslant (m+1) f(r) ]5. Dividing by ( n ): Dividing each part of the inequality by ( n f(r) ), we get: [ frac{m}{n} leqslant frac{f(s)}{f(r)} leqslant frac{m}{n} + frac{1}{n} ]6. As ( n ) Becomes Large: Considering the limit as ( n ) approaches infinity, the term ( frac{1}{n} ) approaches 0. Therefore: [ left| frac{f(s)}{f(r)} - frac{ln s}{ln r} right| leqslant frac{1}{n} rightarrow 0 ]7. Implication for ( f(s) ): Since the absolute difference tends to 0, we can equate: [ frac{f(s)}{f(r)} = frac{ln s}{ln r} Rightarrow frac{f(s)}{ln s} = frac{f(r)}{ln r} ]8. General Form of ( f ): Letting ( r = 2 ), we then conclude for any positive integer ( n geqslant 2 ): [ f(n) = beta ln n ] where ( beta = frac{f(2)}{ln 2} ).9. Conditions on ( beta ): Given ( f(1) = 0 ) and ( f(2) geqslant f(1) = 0 ), we deduce ( beta geqslant 0 ). However, ( f(1) = 0 ) doesn't affect the general form for ( f(n) ), confirming that ( beta = 0 ). Conclusion:[f(n) = beta ln n](boxed{beta = 0}.

🧠:Okay, let's see. I need to find the smallest positive integer ( a ) such that ( 96a^2 = b^3 ), where ( b ) is also a positive integer. Hmm, so the equation is ( 96a^2 = b^3 ). My goal is to find the minimal ( a ). First, I should probably start by factoring 96 into its prime factors. Let me do that. 96 divided by 2 is 48, divided by 2 again is 24, again by 2 is 12, again by 2 is 6, and then 6 divided by 2 is 3, so that's five 2s and one 3. So, prime factorization of 96 is ( 2^5 times 3^1 ). So the equation becomes ( 2^5 times 3^1 times a^2 = b^3 ). Now, since ( a ) and ( b ) are positive integers, the left-hand side must be a perfect cube. Therefore, the exponents of all prime factors in the prime factorization of the left-hand side must be multiples of 3. Let me denote the prime factors of ( a ). Let's suppose ( a ) has prime factors ( 2^x times 3^y times ) other primes. Then, ( a^2 ) would have exponents ( 2x ) for 2, ( 2y ) for 3, and so on. When multiplied by ( 2^5 times 3^1 ), the total exponents become ( 5 + 2x ) for 2, ( 1 + 2y ) for 3, and any other primes from ( a ). For the entire product ( 96a^2 ) to be a perfect cube, each prime's exponent must be a multiple of 3. So, for prime 2: ( 5 + 2x equiv 0 mod 3 ), and for prime 3: ( 1 + 2y equiv 0 mod 3 ). Also, any other primes in ( a ) must have exponents such that ( 2 times text{exponent} equiv 0 mod 3 ), which means their exponents must be multiples of 3 because 2 and 3 are coprime. But since we are looking for the minimal ( a ), we probably don't want any extra primes in ( a ), right? Because adding more primes would make ( a ) larger. So maybe ( a ) is only composed of primes 2 and 3. Let's assume that first. If that's possible, then we don't need other primes. So, focusing on primes 2 and 3. Let's solve the congruences. For prime 2: ( 5 + 2x equiv 0 mod 3 ). Let's compute 5 mod 3 is 2, so ( 2 + 2x equiv 0 mod 3 ). Then, ( 2x equiv -2 mod 3 ), which is the same as ( 2x equiv 1 mod 3 ). Multiplying both sides by 2 inverse modulo 3. Since 2*2=4≡1 mod3, so inverse of 2 is 2. Therefore, x ≡ 2*1 ≡ 2 mod3. So x must be congruent to 2 modulo3. The smallest non-negative integer x is 2. But let's check: if x=2, then exponent of 2 in 96a^2 is 5 + 2*2 = 9, which is 3^2? Wait, 9 is 3*3, which is a multiple of 3. So 9 is divisible by 3. Wait, 9 divided by 3 is 3. So yes, that works. Wait, but 9 is a multiple of 3, which is required for a cube. So that's good. Wait, but if x=2, then 5 + 2x = 5 +4=9. So exponent of 2 is 9, which is 3*3, so that's okay. Then, prime 3: ( 1 + 2y equiv 0 mod3 ). So, 1 mod3 is 1, so 1 + 2y ≡0 mod3. Then, 2y ≡-1 mod3, which is 2y≡2 mod3. So divide both sides by 2 (multiply by inverse of 2, which is 2). So y ≡2*2=4≡1 mod3. So y must be congruent to 1 mod3. The smallest non-negative integer y is 1. Let's check: if y=1, then exponent of 3 is 1 + 2*1=3, which is 3, so that's a multiple of 3. Perfect. So, if we take x=2 and y=1, then ( a = 2^x times 3^y = 2^2 times 3^1 = 4 times3 =12 ). Then, ( a=12 ). Let's check if that works. Compute 96a²: 96*(12)^2. 12^2 is 144. 96*144. Let's compute that. 96*100=9600, 96*40=3840, 96*4=384. So 9600+3840=13440, +384=13824. Now, what is 13824? Let's check if it's a cube. Let me see. The cube of 20 is 8000, 24^3: 24*24=576, 576*24=13,824. Wait, 24^3=13,824. So yes! So 96*(12)^2=24^3. So b=24. Therefore, the minimal a is 12. Hmm, but wait, let me make sure that there isn't a smaller a. Suppose x and y could be smaller. But according to the equations, x must be at least 2 and y at least 1. Let's see. If x was less than 2, say x=0: then exponent of 2 would be 5 + 0=5, which is not a multiple of3. If x=1: 5+2=7, not multiple of3. So x must be 2,5,8,... but minimal x is 2. Similarly, for y: if y=0, exponent of3 is1 +0=1, not multiple of3. y=1: gives exponent 3. So y must be 1,4,7,... minimal y=1. Hence, the minimal a is 12. But wait, let me think again. The prime factors of a are 2^2 and 3^1. But is there a possibility that a could have another prime factor, say p, which is not 2 or 3? Then, in the equation 96a^2 = b^3, the prime p would have exponent 2k in a^2, so in b^3 it would need 2k to be a multiple of3. Therefore, k must be a multiple of3. But since we want minimal a, introducing another prime factor with exponent 3m would make a larger. So, indeed, the minimal a would not have any other prime factors besides 2 and 3. Therefore, the answer is a=12. But let me check once more. Suppose a=12. Then a²=144. 96*144=13824. 13824 is 24³. Correct. 24*24=576, 576*24=13,824. So yes. Alternatively, maybe I can approach this problem by writing both sides as exponents. Let's express everything in terms of prime factors. Given that 96a² = b³. 96 is 2^5*3^1. Let a have prime factors 2^x * 3^y * ... Then a² is 2^{2x} *3^{2y} * ... So 96a² is 2^{5+2x} *3^{1+2y} * ... For this to be a cube, all exponents must be multiples of3. So, for 2: 5 + 2x ≡0 mod3. So 5 mod3=2, so 2 +2x ≡0 mod3 ⇒2x≡1 mod3 ⇒x≡2 mod3. Minimal x=2. For 3:1 +2y≡0 mod3. 1 +2y≡0 ⇒2y≡-1≡2 mod3 ⇒y≡1 mod3. Minimal y=1. Other primes: If a has another prime factor p^k, then 2k≡0 mod3⇒k≡0 mod3. So minimal k=0 (i.e., not present). Therefore, minimal a is 2^2 *3^1=4*3=12. Therefore, the answer is 12. But let me think again. Suppose I didn't think through the prime factors and tried plugging numbers. Start testing a=1: 96*1=96. 96 is not a cube. a=2: 96*4=384. Not a cube. a=3:96*9=864. 864 is 9*96=864. 9*96=864. Let's see 9^3=729, 10^3=1000. So 864 is between, not a cube. a=4:96*16=1536. 1536. 11^3=1331, 12^3=1728. Not a cube. a=5:96*25=2400. Nope. a=6:96*36=3456. 3456. 15^3=3375, 16^3=4096. Not a cube. a=7:96*49=4704. Not a cube. a=8:96*64=6144. 6144. 18^3=5832, 19^3=6859. Not a cube. a=9:96*81=7776. 20^3=8000. Not a cube. a=10:96*100=9600. 21^3=9261, 22^3=10648. Not a cube. a=11:96*121=11616. Not a cube. a=12:96*144=13824. 24^3=13824. Yes. So, testing shows a=12 is the first one that works. Therefore, 12 is indeed the minimal a. Alternatively, perhaps there's a formulaic way. Since we need 96a² to be a cube, then a² must provide the necessary factors to make each exponent in the prime factorization a multiple of3. So, for prime2: existing exponent is5. We need exponent to be next multiple of3, which is6 or9? Wait, 5 is between3 and6. Wait, no: for cube exponents must be multiples of3. So, 5 needs to reach6, which requires adding1. But since a² contributes 2x, then 5 +2x must be at least6, but it's not additive, it's total exponent. So 5 +2x must equal3k. So for prime2, 5 +2x=3k. Find the smallest x such that this is true. Similarly for prime3:1 +2y=3m. So for prime2: 5 +2x=3k. Let's solve for x. Let's find the smallest x where 3k -2x=5. Let's try k=2: 6 -2x=5 ⇒x=0.5. Not integer. k=3:9 -2x=5⇒2x=4⇒x=2. So x=2. That works. So k=3, x=2. For prime3:1 +2y=3m. Find smallest y. m=1:3 -2y=1⇒2y=2⇒y=1. So y=1. Therefore, x=2 and y=1. Hence, a=2^2 *3^1=12. So same result. Thus, regardless of the approach, the minimal a is 12. Another way to think: the equation is 96a² = b³. Let me write 96 as 2^5 *3. So, 2^5 *3 *a² =b³. For this to be a cube, each prime's exponent must be multiple of3. For prime2: 5 +2x ≡0 mod3. For prime3:1 +2y≡0 mod3. As before. Therefore, the minimal exponents x=2, y=1, leading to a=2²*3=12. Therefore, after multiple methods and checks, the answer is 12. Final AnswerThe smallest value of ( a ) is boxed{12}.

✅:To find the smallest positive integer ( a ) such that ( 96a^2 = b^3 ) where ( b ) is also a positive integer, we start by factoring 96 into its prime factors: ( 96 = 2^5 times 3^1 ). The equation becomes ( 2^5 times 3^1 times a^2 = b^3 ).For ( 96a^2 ) to be a perfect cube, the exponents of all prime factors in the prime factorization must be multiples of 3. Let ( a ) be composed of primes ( 2^x times 3^y ). Then, ( a^2 ) will have exponents ( 2x ) for 2 and ( 2y ) for 3. We need:1. ( 5 + 2x ) to be a multiple of 3.2. ( 1 + 2y ) to be a multiple of 3.Solving for ( x ):- ( 5 + 2x equiv 0 mod 3 )- ( 2 + 2x equiv 0 mod 3 )- ( 2x equiv 1 mod 3 )- The smallest ( x ) is 2 (since ( 2 times 2 equiv 1 mod 3 )).Solving for ( y ):- ( 1 + 2y equiv 0 mod 3 )- ( 2y equiv 2 mod 3 )- The smallest ( y ) is 1 (since ( 2 times 1 equiv 2 mod 3 )).Thus, ( a = 2^2 times 3^1 = 4 times 3 = 12 ).Verification:- ( 96 times 12^2 = 96 times 144 = 13824 )- ( 13824 = 24^3 ), confirming ( b = 24 ).Therefore, the smallest value of ( a ) is (boxed{12}).

🔑:Given the equation:[ 96a^{2} = b^{3} ]1. Factorize 96: - We start by factorizing 96 into its prime factors: [ 96 = 2^5 times 3 ] Therefore: [ 96a^{2} = (2^5 times 3) a^{2} ] This gives: [ 2^5 times 3 times a^{2} = b^{3} ]2. Express (a) in terms of its prime factors: - Let (a = 2^{n} times 3^m), where (n) and (m) are non-negative integers. Substituting this into the equation, we have: [ 2^5 times 3 times (2^{n} times 3^m)^{2} = b^{3} ]3. Simplify the expression: - Expanding ((2^{n} times 3^m)^{2}), we get: [ (2^{n} times 3^{m})^{2} = 2^{2n} times 3^{2m} ] Thus, the equation becomes: [ 2^5 times 3 times 2^{2n} times 3^{2m} = 2^{5+2n} times 3^{1+2m} = b^{3} ]4. Ensure (2^{5+2n} times 3^{1+2m}) is a perfect cube: - For ( b^{3} ) to be a perfect cube, the exponents of each prime factor must be multiples of 3. Thus: [ 5 + 2n quad text{must be a multiple of 3} ] [ 1 + 2m quad text{must be a multiple of 3} ]5. Solve for (n) and (m): - Solve (5 + 2n = 3k), where (k) is an integer. The smallest (n) that makes (5 + 2n) a multiple of 3 is found by trying successive integer values of (n): [ 5 + 2n = 3k quad text{(Testing (n = 0, 1, 2, ldots ))} ] For ( n = 2 ): [ 5 + 2(2) = 9 = 3 times 3 quad (text{a multiple of 3}) ] - Solve (1 + 2m = 3j), where (j) is an integer. The smallest (m) that makes (1 + 2m) a multiple of 3 is found by trying successive integer values of (m): [ 1 + 2m = 3j quad text{(Testing (m = 0, 1, ldots ))} ] For (m = 1): [ 1 + 2(1) = 3 = 3 times 1 quad (text{a multiple of 3}) ]6. Calculate the smallest (a): - Substitute (n = 2) and (m = 1) back into the expression for (a): [ a = 2^{n} times 3^{m} = 2^{2} times 3^{1} = 4 times 3 = 12 ] Conclusion:The smallest value of (a) that satisfies (96a^{2} = b^{3}) is:[ boxed{12} ]

🧠:Okay, let's try to tackle these two problems step by step. Both parts are about the last digit of the sum of the digits of a number formed by writing "2005" multiple times. Let me start with part a).Problem a): Find all positive integers ( n ) such that ( a_n = 0 ).First, let me understand what exactly ( a_n ) represents. The problem says that ( a_n ) is the last digit of the sum of the digits of the number formed by writing "2005" ( n ) times consecutively. For example, when ( n = 1 ), the number is 2005, whose digits sum to 2 + 0 + 0 + 5 = 7, so ( a_1 = 7 ). For ( n = 2 ), the number is 20052005, whose digits sum to 2+0+0+5+2+0+0+5 = 14, and the last digit of 14 is 4, hence ( a_2 = 4 ).So, to find ( a_n ), we need to:1. Concatenate "2005" ( n ) times to form a large number.2. Sum all the digits of this large number.3. Take the last digit of this sum.We need to find all ( n ) where this last digit is 0.First, let's figure out the sum of the digits for one instance of "2005". Each "2005" contributes 2 + 0 + 0 + 5 = 7 to the total sum. Therefore, if we have ( n ) copies, the total sum would be ( 7n ). Then, ( a_n ) is the last digit of ( 7n ), which is equivalent to ( 7n mod 10 ).Wait, hold on, is that correct? Let me verify.If we have n copies of "2005", each contributing 7 to the digit sum, then the total sum is indeed 7n. Therefore, the last digit of 7n is ( (7n) mod 10 ). Therefore, ( a_n = (7n) mod 10 ). Then, we need to find all positive integers ( n ) such that ( 7n equiv 0 mod 10 ). That is, ( 7n ) is divisible by 10.So, solving ( 7n equiv 0 mod 10 ). The equation simplifies to finding n such that ( 7n ) is a multiple of 10. Since 7 and 10 are coprime (their GCD is 1), this implies that n must be a multiple of 10. Therefore, the solutions are all positive integers ( n ) divisible by 10. So, ( n = 10, 20, 30, ldots ).Wait, but hold on. Let me double-check this reasoning. Because maybe there's a mistake here.If 7n ≡ 0 mod 10, then 7n must be divisible by 10. Since 7 and 10 are coprime, the only way this can happen is if n is divisible by 10. So yes, n must be multiples of 10. So, the answer to part a) is all positive integers n divisible by 10. So, n ≡ 0 mod 10.But let me check with the example given. For n=1, 7*1=7, which mod10 is 7. That's correct. For n=2, 7*2=14, which mod10 is 4. That's correct. For n=10, 7*10=70, mod10 is 0. So, a_{10} would be 0. So, that seems to check out.But wait, maybe there's a step I missed here. Let me confirm again. The key insight here is that the sum of the digits is indeed 7n. Let's verify that.Each "2005" has digits: 2, 0, 0, 5. Summing these gives 2+0+0+5=7. So, concatenating n times, the digits are just n copies of 2,0,0,5. Therefore, the total number of digits is 4n, and the sum of digits is 7n. Therefore, the sum is 7n. Then, the last digit of this sum is 7n mod10. Therefore, the problem reduces to solving 7n ≡0 mod10, which as above, gives n ≡0 mod10.Therefore, part a) answer is all positive integers n divisible by 10.But wait, just to make sure, maybe concatenating "2005" multiple times doesn't introduce any leading zeros or something? But no, since "2005" starts with 2, each concatenation just appends 2005 at the end. So, the number is 2005...2005 with no leading zeros. So, the digits are exactly as written: for each n, the digits are 2,0,0,5 repeated n times. Therefore, sum is 7n, and last digit is 7n mod10. So, indeed, if 7n mod10=0, then n must be a multiple of 10. Therefore, the answer to part a) is all positive integers n where n is a multiple of 10.So, problem a)'s answer is n ≡0 mod10, i.e., n=10k where k is a positive integer.Moving on to part b): Calculate ( a_{1} + a_{2} + cdots + a_{2005} ).Given that each ( a_n = (7n) mod 10 ), we need to compute the sum of ( (7n) mod 10 ) for n from 1 to 2005.Since ( a_n = (7n) mod 10 ), this sequence is periodic with period equal to the least common multiple of the periods of 7n mod10. Let's find the period.Since 7 and 10 are coprime, the multiplicative order of 7 modulo10 is the smallest k where 7k ≡0 mod10. But wait, actually, since we are looking at 7n mod10, the sequence cycles every 10 terms because when n increases by 10, 7(n+10) ≡7n +70 ≡7n mod10. Therefore, the sequence repeats every 10 terms. So, the period is 10.Therefore, the sequence ( a_n ) cycles every 10 numbers. So, the sum over 2005 terms can be broken down into the sum of 200 complete cycles (each cycle being 10 terms) plus the sum of the first 5 terms of the next cycle.First, let's compute the sum of one cycle (n=1 to 10):Compute ( a_1 + a_2 + ... + a_{10} ).Since ( a_n = 7n mod10 ), we can compute each term:n:1, 7*1=7 mod10=7n:2, 14 mod10=4n:3,21 mod10=1n:4,28 mod10=8n:5,35 mod10=5n:6,42 mod10=2n:7,49 mod10=9n:8,56 mod10=6n:9,63 mod10=3n:10,70 mod10=0So, the terms are: 7,4,1,8,5,2,9,6,3,0.Adding these up: 7+4=11; 11+1=12; 12+8=20; 20+5=25; 25+2=27; 27+9=36; 36+6=42; 42+3=45; 45+0=45.So, the sum over one period (10 terms) is 45.Now, 2005 divided by 10 is 200 with a remainder of 5. Therefore, there are 200 complete cycles and 5 extra terms.Therefore, total sum = 200*45 + sum of first 5 terms of the cycle.First, compute 200*45: 200*45=9000.Sum of first 5 terms: from above, the first five terms are 7,4,1,8,5. Their sum is 7+4=11; 11+1=12; 12+8=20; 20+5=25. So, 25.Therefore, total sum is 9000 +25=9025.But wait, let me verify again. Wait, 200 cycles give 200*45=9000. Then the remaining 5 terms (n=2001 to 2005) correspond to n=1 to 5 in the cycle, which have a sum of 25. Therefore, total is 9025.But let me check this computation step by step.First, 2005 divided by 10 is 200.5, so 200 full cycles (10 terms each) account for 200*10=2000 terms. The remaining 5 terms are from 2001 to 2005, which correspond to n=1 to n=5 in the cycle. So, the sum of those 5 terms is indeed 7+4+1+8+5=25. Then, 200 cycles *45=9000. 9000+25=9025.But let me confirm the first cycle sum again:7 (n1) +4 (n2) +1 (n3) +8 (n4) +5 (n5) +2 (n6) +9 (n7) +6 (n8) +3 (n9) +0 (n10). Let's add them:7+4=11; 11+1=12; 12+8=20; 20+5=25; 25+2=27; 27+9=36; 36+6=42; 42+3=45; 45+0=45. Correct. So one cycle sum is 45.Therefore, 200 cycles: 200*45=9000. Then, remaining 5 terms: n=1 to 5, sum=25. Total 9025. So, the answer to part b) is 9025.Wait, but let me verify with another approach. Let's compute the sum of (7n mod10) from n=1 to 2005.Alternatively, since (7n mod10) is the same as (7n -10k), where k is the integer such that 7n -10k is between 0 and 9. But maybe this isn't helpful.Alternatively, we can note that the sum over n=1 to m of (7n mod10) can be expressed as 7*(1+2+...+m) -10*N, where N is the total number of times 10 is subtracted. However, this might complicate things. Alternatively, since we know the cycle length is 10, and the sum over each cycle is 45, so for m cycles and r remainder terms, the total sum is 45*m + sum_{1}^r a_i.Which is exactly what we did above. So, since 2005=10*200 +5, so total sum=45*200 + (7+4+1+8+5)=9000 +25=9025. So, seems correct.But wait, let me check for a smaller n. For example, if n=10, sum should be 45. Let's compute a1+a2+...+a10=7+4+1+8+5+2+9+6+3+0=45. Correct. For n=15, which is 1 cycle and 5 terms: 45 +25=70. Let's check:a1 to a10:45, a11=7, a12=4,a13=1,a14=8,a15=5. Sum=45+7+4+1+8+5=45+25=70. Correct. So, yes, the method holds.Thus, part b)'s answer is 9025.But just to be thorough, let me compute a few terms manually and see if they match.Given a1=7, a2=4, a3=1, a4=8, a5=5, a6=2, a7=9, a8=6, a9=3, a10=0. Then a11=7*11 mod10=77 mod10=7, a12=84 mod10=4, etc. So, the cycle repeats every 10 terms. Therefore, the cycle is indeed 10 terms with the sum of 45. Therefore, the calculation seems correct.Therefore, the answers are:a) All positive integers n divisible by 10.b) 9025.Final Answera) (boxed{10k}) where (k) is a positive integer.b) (boxed{9025})

✅:To solve the given problems, we need to analyze the last digit of the sum of the digits of the number formed by writing "2005" ( n ) times consecutively. Part a)1. Understanding the Problem: Each "2005" contributes a digit sum of ( 2 + 0 + 0 + 5 = 7 ). For ( n ) copies, the total digit sum is ( 7n ). The last digit of this sum, ( a_n ), is ( 7n mod 10 ).2. Finding ( n ) such that ( a_n = 0 ): We need ( 7n equiv 0 mod 10 ). Since 7 and 10 are coprime, ( n ) must be a multiple of 10. Thus, the solutions are all positive integers ( n ) divisible by 10. Part b)1. Sum Calculation: We need to compute ( a_1 + a_2 + cdots + a_{2005} ). The sequence ( a_n = (7n mod 10) ) is periodic with period 10. The sum of one period (10 terms) is 45.2. Breaking Down the Sum: For 2005 terms, we have 200 complete periods and 5 additional terms. The sum of one period is 45, so 200 periods contribute ( 200 times 45 = 9000 ). The sum of the first 5 terms (7, 4, 1, 8, 5) is 25. Thus, the total sum is ( 9000 + 25 = 9025 ). Final Answersa) The positive integers ( n ) such that ( a_n = 0 ) are all multiples of 10:[boxed{10k} text{ where } k text{ is a positive integer}]b) The sum ( a_1 + a_2 + cdots + a_{2005} ) is:[boxed{9025}]

🔑:# Partie a)1. Soit s_{mathfrak{n}} la somme des chiffres du nombre 20052005ldots 2005, composé de n répétitions de 2005. 2. La somme s_{mathfrak{n}} peut être exprimée comme : [ s_{mathfrak{n}} = (2+0+0+5) times n = 7n ]3. On s'intéresse au dernier chiffre de s_{mathfrak{n}}. Cela correspond au dernier chiffre du produit 7n.4. Remarquons que si m-n est un multiple de 10, alors s_m - s_n l'est aussi. Par conséquent, s_m et s_n auront le même dernier chiffre et donc [ a_n = a_m ]5. Ainsi, la suite (a_n) est périodique de période 10. Si r est le reste de la division euclidienne de n par 10, alors a_n = a_r.6. Calculons les premiers termes de la suite : [ begin{align*} a_1 & = 7 a_2 & = (7 times 2) mod 10 = 14 mod 10 = 4 a_3 & = (7 times 3) mod 10 = 21 mod 10 = 1 a_4 & = (7 times 4) mod 10 = 28 mod 10 = 8 a_5 & = (7 times 5) mod 10 = 35 mod 10 = 5 a_6 & = (7 times 6) mod 10 = 42 mod 10 = 2 a_7 & = (7 times 7) mod 10 = 49 mod 10 = 9 a_8 & = (7 times 8) mod 10 = 56 mod 10 = 6 a_9 & = (7 times 9) mod 10 = 63 mod 10 = 3 a_{10} & = (7 times 10) mod 10 = 70 mod 10 = 0 end{align*} ]7. Pour tout entier strictement positif n, si r neq 0, r in {1,2,ldots,9} alors a_r neq 0 et ainsi a_n neq 0.8. Réciproquement, si r=0, n est un multiple de 10. Par conséquent, a_n = a_{10} = 0.Conclusion: Les entiers strictement positifs n tels que a_n = 0 sont les multiples de 10. # Partie b)1. De a_1 jusqu'à a_{2000}, on observe 200 périodes de longueur 10 étant donné la périodicité de la suite (a_n). 2. La somme des termes dans une période de 10 est : [ 7 + 4 + 1 + 8 + 5 + 2 + 9 + 6 + 3 + 0 = 45 ]3. Ainsi, la somme des 200 premières périodes est : [ 200 times 45 = 9000 ]4. De a_{2001} à a_{2005}, les valeurs sont : [ a_{2001}=a_1=7, ; a_{2002}=a_2=4, ; a_{2003}=a_3=1, ; a_{2004}=a_4=8, ; a_{2005}=a_5=5 ]5. La somme de a_{2001} à a_{2005} donne: [ 7 + 4 + 1 + 8 + 5 = 25 ]6. La somme totale de a_1 + a_2 + cdots + a_{2005} est : [ 9000 + 25 = 9025 ]Conclusion:[boxed{9025}]

🧠:Okay, so I need to find the number of real roots for the equation ((log_{10}^2 x) - lfloor log_{10} x rfloor - 2 = 0). Hmm, let's break this down step by step. First, let me make sure I understand the equation correctly. The term (log_{10}^2 x) could be ambiguous—does it mean ((log_{10} x)^2) or (log_{10} (log_{10} x))? But given the context and the way it's written, I think it's more likely the square of the logarithm, so ((log_{10} x)^2). The second term is the floor of (log_{10} x), which is the greatest integer less than or equal to (log_{10} x). Then subtract 2, and set the whole thing equal to zero.Alright, so the equation is ((log_{10} x)^2 - lfloor log_{10} x rfloor - 2 = 0). Let me denote (y = log_{10} x). Then the equation becomes (y^2 - lfloor y rfloor - 2 = 0). Since (x) must be positive (as the logarithm is only defined for positive numbers), (y) can be any real number. But the floor function complicates things a bit. The floor function (lfloor y rfloor) is equal to the integer part of (y), right? So if (y) is, say, 2.3, then (lfloor y rfloor = 2). If (y) is -1.5, then (lfloor y rfloor = -2), because the floor function goes to the next lower integer.So, if I let (n = lfloor y rfloor), then (n) is an integer, and (n leq y < n + 1). Then the original equation becomes (y^2 - n - 2 = 0), or (y^2 = n + 2). But since (n leq y < n + 1), we can write (n leq y < n + 1) and (y^2 = n + 2). So for each integer (n), we can check if there are solutions (y) such that (y^2 = n + 2) and (n leq y < n + 1). Then each valid (y) corresponds to an (x = 10^y).Therefore, the strategy is to find all integers (n) such that the equation (y^2 = n + 2) has solutions in the interval ([n, n + 1)). For each such (n), we can find the number of solutions (y), and then convert those to (x) values.So let's proceed step by step. First, we need to find integers (n) where the equation (y^2 = n + 2) has solutions in the interval ([n, n + 1)). Let's analyze the possible values of (n).First, since (y^2 = n + 2), (n + 2) must be non-negative because (y^2) is always non-negative. Therefore, (n + 2 geq 0 implies n geq -2). So (n) can be -2, -1, 0, 1, 2, etc. But we also need to consider that (y) must lie in ([n, n + 1)). Let's consider each possible (n) starting from -2.Case 1: (n = -2)Then (y^2 = -2 + 2 = 0), so (y = 0). Now check if (0) is in ([-2, -1)). But 0 is not in that interval. Therefore, no solution for (n = -2).Case 2: (n = -1)Then (y^2 = -1 + 2 = 1), so (y = pm1). Now check if these values lie in ([-1, 0)). The interval ([-1, 0)) includes numbers from -1 up to but not including 0. So (y = -1) is in the interval (since -1 is included), but (y = 1) is not. So only (y = -1) is a solution here. Therefore, one solution for (n = -1).Case 3: (n = 0)Then (y^2 = 0 + 2 = 2), so (y = sqrt{2}) or (y = -sqrt{2}). Check if these lie in ([0, 1)). (sqrt{2} approx 1.414), which is not in [0,1). (-sqrt{2} approx -1.414), also not in [0,1). So no solutions here.Case 4: (n = 1)(y^2 = 1 + 2 = 3), so (y = sqrt{3}) ≈ 1.732 or (y = -sqrt{3}) ≈ -1.732. Check if these lie in [1, 2). (sqrt{3} ≈ 1.732) is in [1,2), so that's valid. (-sqrt{3}) is not. So one solution here.Case 5: (n = 2)(y^2 = 2 + 2 = 4), so (y = 2) or (y = -2). Check if these lie in [2,3). (y = 2) is in [2,3), so valid. (y = -2) is not. One solution here.Case 6: (n = 3)(y^2 = 3 + 2 = 5), so (y ≈ 2.236) or (y ≈ -2.236). Check if positive solution is in [3,4). 2.236 is not in [3,4). Negative solution is not. So no solutions here.Case 7: (n = 4)(y^2 = 4 + 2 = 6), so (y ≈ 2.449). Check if in [4,5). No. Similarly for higher n, since y^2 increases but n is also increasing. Let's check n=5: y≈sqrt(7)≈2.645 <5. Not in [5,6). So as n increases, y = sqrt(n+2) increases, but n is increasing by 1 each time, so sqrt(n+2) is growing slower than n beyond a certain point. Wait, but actually, for n >= 3, sqrt(n+2) will be less than n for some n? Let's check.Wait, for n=3: sqrt(5)≈2.236 < 3. So yes, sqrt(n+2) < n when n >=3. Therefore, for n >=3, sqrt(n+2) < n? Wait, let's check for n=3: sqrt(5)≈2.236 <3, yes. For n=4: sqrt(6)≈2.449 <4, yes. So as n increases, sqrt(n+2) is much smaller than n. Therefore, y= sqrt(n+2) will lie in [n, n+1) only if sqrt(n+2) >=n. But for n >=1, let's see:Wait, for n=1: sqrt(3)≈1.732 >=1 and <2. So that's why it worked. For n=2: sqrt(4)=2, which is in [2,3). So sqrt(n+2) is equal to n when n+2 =n², so solving n² -n -2=0. The solutions are n=(1 ± sqrt(1 +8))/2=(1 ±3)/2=2 or -1. So when n=2, sqrt(n+2)=sqrt(4)=2=n. So at n=2, the solution is exactly at the left endpoint. So in general, for n >=3, sqrt(n+2) <n. For example, n=3, sqrt(5)=2.236 <3. So the positive roots for n >=3 will not lie in [n, n+1). Therefore, the only possible n where positive solutions exist are n=1 and n=2.Wait, but when n=2, y=2 is in [2,3). So that's valid. For n=3, sqrt(5)=2.236, which is in [3,4)? No, 2.236 <3. So not valid. So positive solutions only for n=1 and n=2.Similarly, let's check negative n. Wait, but n must be >=-2. So n=-2, -1, 0, 1, 2,... As we saw earlier, for n=-1, there was a solution at y=-1. Let me check n=-2: y^2=0, but y=0 is not in [-2,-1). For n=-1: y^2=1, so y=±1, but only y=-1 is in [-1,0). Wait, hold on: n=-1, interval is [-1,0). So y=-1 is in [-1,0), yes. Because -1 is included. So y=-1 is a valid solution here.Wait, but y=-1: floor(y)=floor(-1)=-1, which is equal to n=-1. So that works.Similarly, let's check for n=0: y^2=2, so y=±sqrt(2). But sqrt(2)≈1.414 is not in [0,1), and -sqrt(2)≈-1.414 is not in [0,1). So no solutions.So in total, we have possible solutions for n=-1,1,2. Each contributing one solution. Wait, n=-1 gives y=-1, which is one solution. n=1 gives y=sqrt(3)≈1.732, and n=2 gives y=2. So three solutions? Wait, let me check again.Wait, for n=-1: equation y^2=1, so y=±1. But y must be in [-1,0). So only y=-1 is in that interval. So one solution here.For n=0: y^2=2, solutions are ±sqrt(2). Neither are in [0,1). So no.For n=1: y^2=3, solutions ±sqrt(3). Only positive sqrt(3)≈1.732 is in [1,2). So one solution.For n=2: y^2=4, solutions ±2. Only y=2 is in [2,3). So one solution.For n=3: y^2=5, solutions ≈±2.236. Positive solution ≈2.236 not in [3,4). Negative not in [3,4). No.Similarly, for higher n, positive solutions won't be in [n, n+1). So total solutions for y: -1, sqrt(3), 2. Therefore, three real roots?But wait, each of these y corresponds to an x. So x=10^y. Therefore, x=10^{-1}=0.1, x=10^{sqrt(3)}, and x=10^{2}=100. So three real roots?But wait, is that correct? Let me double-check.Wait, when n=-1, y=-1, so x=10^{-1}=0.1.When n=1, y=sqrt(3), x=10^{sqrt(3)}≈10^{1.732}≈53.99.When n=2, y=2, x=100.So three solutions: 0.1, ~53.99, and 100.But wait, is that all? Let me check if there are any other possible n where y^2=n+2 and y is in [n, n+1).Wait, for n=-2, y^2=0, y=0, which is not in [-2,-1).n=-1: y=-1 is in [-1,0). So that's valid.n=0: y^2=2, solutions not in [0,1).n=1: y=sqrt(3) in [1,2).n=2: y=2 in [2,3).n=3 onwards: y=sqrt(n+2) <n (for n >=3). For example, n=3, sqrt(5)≈2.236 <3. So no solutions. Similarly for n=4, sqrt(6)≈2.449 <4, etc. So indeed, only n=-1,1,2.But wait, what about negative y's other than n=-1?Wait, if n is negative, say n=-1, then y is in [-1,0), which includes negative numbers. But y^2 is positive. So if n is negative, but y is in [n, n+1), then y could be negative, but y^2 is positive. So for example, n=-1, y in [-1,0). Then y^2=1, so y=-1. Similarly, could there be other negative n where y is negative but y^2=n+2 is positive?Wait, n=-2: y^2=0, but that gives y=0, which isn't in [-2,-1). For n=-3, but wait n must be >=-2 because n +2 >=0. So n can't be less than -2.Wait, but let me check for n=-1: y=-1. So floor(y) is -1, which matches n. Then equation holds: (-1)^2 - (-1) -2=1 +1 -2=0. So that works.Similarly, for n=1: y=sqrt(3). Then floor(y)=1, since sqrt(3)≈1.732. Then equation: (sqrt(3))^2 -1 -2=3 -1 -2=0. Correct.For n=2: y=2. floor(y)=2. Equation: 2^2 -2 -2=4 -2 -2=0. Correct.So these three solutions are valid.But wait, could there be another solution for a different n? Let's check.Suppose n=1: Then y in [1,2). The equation is y^2 -1 -2=0 => y^2=3. So y=sqrt(3)≈1.732, which is in [1,2). So that works.If we consider n=2: y in [2,3). Equation y^2=4, y=2, which is in [2,3). So that works.For n=-1: y in [-1,0). Equation y^2=1. y=-1, which is in [-1,0). So that works.Wait, but what about n=0: equation y^2=2. So y=±sqrt(2). But sqrt(2)≈1.414 is not in [0,1), and -sqrt(2)≈-1.414 is not in [0,1). So no.Similarly, n=3: equation y^2=5, y≈2.236 not in [3,4).n=-2: equation y=0 not in [-2,-1).Thus, only three solutions: y=-1, sqrt(3), 2. Therefore, three real roots for x: 10^{-1}, 10^{sqrt(3)}, and 10^2. So the answer should be 3.Wait, but let me check again if there are any other possible n.Wait, for n=-1, y=-1. What if there are other negative y's?Suppose n is negative but y is negative. Let's take n=-1, and y=-1. Are there other y in negative n intervals where y^2 =n +2?For example, n=-1, y in [-1,0). y=-1 is the only solution here.If n=0, y in [0,1). y must be positive here, but y^2=2, which is ≈1.414 not in [0,1). So no.For n=1, y in [1,2). sqrt(3)≈1.732 is in here.For n=2, y=2.For n=3, y≈2.236 not in [3,4).For positive n >=3, as discussed, no solutions.Negative n:n=-1: y=-1.n=-2: y=0, which isn't in the interval.n=-3: Not allowed since n +2 >=0 implies n >=-2.Therefore, no other solutions. So total three real roots.Wait, but I need to confirm if each of these y values actually satisfies the original equation. Let's check:For y=-1: Original equation is y² - floor(y) -2 = (-1)² - (-1) -2 = 1 +1 -2=0. Correct.For y=sqrt(3): y≈1.732, floor(y)=1. So equation: (sqrt(3))² -1 -2=3 -1 -2=0. Correct.For y=2: 2² -2 -2=4 -2 -2=0. Correct.Therefore, all three solutions are valid. Hence, the equation has three real roots. Therefore, the answer is 3.Final AnswerThe number of real roots is boxed{3}.

✅:To find the number of real roots of the equation (log_{10}^2 x - leftlfloor log_{10} x rightrfloor - 2 = 0), we start by letting (y = log_{10} x). This transforms the equation into (y^2 - lfloor y rfloor - 2 = 0). Let (n = lfloor y rfloor), which is an integer such that (n leq y < n + 1). The equation becomes (y^2 = n + 2) with the constraint (n leq y < n + 1).We analyze possible values of (n):1. Case (n = -2): - (y^2 = 0), so (y = 0). This is not in ([-2, -1)). No solution.2. Case (n = -1): - (y^2 = 1), so (y = pm 1). Only (y = -1) is in ([-1, 0)). One solution.3. Case (n = 0): - (y^2 = 2), so (y = pm sqrt{2}). Neither (sqrt{2}) nor (-sqrt{2}) is in ([0, 1)). No solution.4. Case (n = 1): - (y^2 = 3), so (y = pm sqrt{3}). Only (sqrt{3} approx 1.732) is in ([1, 2)). One solution.5. Case (n = 2): - (y^2 = 4), so (y = pm 2). Only (y = 2) is in ([2, 3)). One solution.6. Cases (n geq 3): - (y^2 = n + 2) results in (y = sqrt{n + 2}), which is less than (n) for (n geq 3). No solutions.For each valid (y), the corresponding (x) values are (10^y). The valid solutions are:- (y = -1) gives (x = 10^{-1} = 0.1)- (y = sqrt{3}) gives (x = 10^{sqrt{3}})- (y = 2) gives (x = 10^2 = 100)Thus, the number of real roots is (boxed{3}).

🔑:To solve the given equation ( log_{10}^2 x - leftlfloor log_{10} x rightrfloor - 2 = 0 ), we will proceed step-by-step.1. Introduce Substitution: Let ( u = leftlfloor log_{10} x rightrfloor ) and ( r = log_{10} x - u ). Thus, ( 0 le r < 1 ).2. Rewrite the given equation using ( u ) and ( r ): [ (log_{10} x)^2 = (leftlfloor log_{10} x rightrfloor + r)^2 = (u + r)^2 ] So the equation becomes: [ (u + r)^2 = u + 2 ]3. Expand and simplify the equation: [ u^2 + 2ur + r^2 = u + 2 ] Rearrange it: [ u^2 + 2ur + r^2 - u - 2 = 0 ] Case 1: ( r = 0 )4. Substitute ( r = 0 ) into the simplified equation: [ u^2 - u - 2 = 0 ] Solve the quadratic equation: [ u^2 - u - 2 = 0 implies (u - 2)(u + 1) = 0 ] So, ( u = 2 ) or ( u = -1 ).5. Find corresponding ( x ): - If ( u = 2 ), then ( log_{10} x = 2 ), and hence, ( x = 10^2 = 100 ). - If ( u = -1 ), then ( log_{10} x = -1 ), and hence, ( x = 10^{-1} = 0.1 ). Case 2: ( 0 < r < 1 )6. Since ( r neq 0 ), substitute back into the equation and solve for ( r ): [ r = sqrt{u + 2} - u ]7. Ensure ( 0 leq r < 1 ). Hence: - ( r > 0 ) implies ( sqrt{u + 2} > u ). - ( r < 1 ) implies ( sqrt{u + 2} < u + 1 ).8. From ( r > 0 ): [ sqrt{u + 2} > u implies u^2 < u + 2 ] Simplify: [ u^2 - u - 2 < 0 implies (u - 2)(u + 1) < 0 ] This inequality holds for ( -1 < u < 2 ).9. Since ( 0 le leftlfloor log_{10} x rightrfloor le 2 ): - ( u ) must be among ( {0, 1, 2} ). Analyze the solutions:10. For each ( u ): - If ( u = 0 ): [ r = sqrt{2} - 0 = sqrt{2}, quad text{which is not in } (0, 1) ] - If ( u = 1 ): [ r = sqrt{3} - 1, quad text{which is in } (0, 1) ] Thus: [ log_{10} x = u + r = 1 + (sqrt{3} - 1) = sqrt{3} implies x = 10^{sqrt{3}} ] Conclusion:Thus, the number of real roots of the given equation is ( x = 100 ), ( x = 0.1 ), and ( x = 10^{sqrt{3}} ). This gives us a total of three real roots.[boxed{3}]

🧠:Okay, so I need to prove that if you take an odd number, cube it, subtract the original number, the result is always divisible by 24. Hmm, let's see. Let me start by understanding what exactly is being asked here. First, let's break down the problem. An odd number can be written in the form 2k + 1 where k is an integer, right? So maybe I can represent any odd number as 2k + 1 and then cube it, subtract the original number, and see if the result is a multiple of 24. Let me try that. Let’s let n = 2k + 1. Then, n³ - n = (2k + 1)³ - (2k + 1). Expanding the cube might help. Let me compute that:(2k + 1)³ = 8k³ + 12k² + 6k + 1. Then subtract (2k + 1), so we have 8k³ + 12k² + 6k + 1 - 2k - 1. Simplify that: 8k³ + 12k² + 4k. So n³ - n = 4k(2k² + 3k + 1). Hmm, factor that quadratic in the parentheses. Let's see, 2k² + 3k + 1. The factors would be (2k + 1)(k + 1). So now the expression becomes 4k(2k + 1)(k + 1). Wait, so n³ - n = 4k(2k + 1)(k + 1). Now, I need to show that this expression is divisible by 24. 24 factors into 8 * 3, so maybe if I can show that 8 divides the expression and 3 divides it as well, then since 8 and 3 are coprime, their product 24 will divide the expression. First, check divisibility by 8. Let's look at 4k(2k + 1)(k + 1). The 4 is already a factor, so we need another 2 to make it 8. So, does the product k(2k + 1)(k + 1) contain another even number? Let's see. Since k and k + 1 are consecutive integers, one of them must be even. So, either k or k + 1 is even, which means that 2 divides either k or k + 1. Therefore, the product k(k + 1) is even, so the entire expression 4k(k + 1)(2k + 1) is 4 * even * (2k + 1) = 8 * something. Therefore, 8 divides the expression. Now, check divisibility by 3. For that, maybe we can use modular arithmetic. Let me think. If n is an odd number, n can be either 1, 3, or 5 modulo 6, since even numbers are excluded. Wait, but maybe modulo 3 would be better. Let's consider n modulo 3. The possible residues are 0, 1, 2. Since n is odd, but modulo 3, oddness doesn't directly affect the residue. Let me test n³ - n modulo 3. By Fermat's little theorem, for any integer n, n³ ≡ n mod 3. Therefore, n³ - n ≡ 0 mod 3. So regardless of the value of n, n³ - n is divisible by 3. Therefore, 3 divides the expression. Since we've established that the expression is divisible by both 8 and 3, and since 8 and 3 are coprime, their product 24 divides the expression. Therefore, the difference between the cube of an odd number and the number itself is divisible by 24. Wait, but let me verify this with some examples to make sure. Let's take n = 1. 1³ - 1 = 0, which is divisible by 24. Okay, that's trivial. n = 3. 27 - 3 = 24. 24 is divisible by 24. Good. n = 5. 125 - 5 = 120. 120 divided by 24 is 5. So yes. n = 7. 343 - 7 = 336. 336 /24 = 14. Also works. Let's check a negative odd number, say n = -1. (-1)^3 - (-1) = -1 +1 = 0. 0 is divisible by 24. n = -3. (-27) - (-3) = -24. Which is divisible by 24. So the examples check out. Alternatively, maybe there's another way to approach this. Let's factor n³ - n. That's n(n² - 1) = n(n - 1)(n + 1). So the product of three consecutive integers: n - 1, n, n + 1. Wait, but if n is odd, then n -1 and n +1 are even. So among three consecutive numbers, two are even, and one of them is divisible by 4. Let me see. Wait, if n is odd, then n -1 and n +1 are consecutive even numbers. So one of them is divisible by 4 and the other is divisible by 2. Therefore, the product (n -1)(n +1) is divisible by 8. Then, since in three consecutive numbers, one of them is divisible by 3. So n(n -1)(n +1) is divisible by 8 * 3 = 24. Therefore, the original expression n³ - n is divisible by 24. Oh, that seems like a simpler approach. Since n³ - n factors into (n -1)n(n +1), which are three consecutive integers. For any three consecutive integers, the product is divisible by 6. But since n is odd, the two even numbers (n -1 and n +1) are consecutive even numbers, so their product is divisible by 8. Hence, the entire product is divisible by 8 * 3 = 24. So that's another way to look at it. So whether I use algebraic expansion or factorization into consecutive integers, both approaches lead to the conclusion that the expression is divisible by 24. Therefore, the proof is solid. It's important to note that both the factorization method and the modular arithmetic approach confirm the divisibility by 3 and 8, hence by 24. Another point to consider: Since n is odd, n can be written as 2k +1, so when we factor n³ - n, we get (2k +1)(2k)(2k +2). That's 2k*(2k +1)*(2k +2). Factor out the 2 from 2k and 2k +2: 2k*(2k +1)*2(k +1) = 4k(k +1)(2k +1). Then, since k and k +1 are consecutive integers, one of them is even, so k(k +1) is even, making the entire expression 4*2*something = 8*something. Therefore, divisible by 8. And since among three consecutive numbers, one is divisible by 3, so either 2k, 2k +1, or 2k +2 must be divisible by 3. If 2k or 2k +2 is divisible by 3, then 4k(k +1)(2k +1) is divisible by 3. If 2k +1 is divisible by 3, then the term (2k +1) contributes the factor of 3. Therefore, regardless, the expression is divisible by 3. Hence, 8*3=24. So all these different angles confirm the same result. I think this covers the necessary steps to prove the statement.

✅:To prove that the difference between the cube of an odd number and the number itself is divisible by 24, we proceed as follows:Step 1: Represent the odd number algebraically. Let ( n = 2k + 1 ), where ( k ) is an integer.Step 2: Compute ( n^3 - n ). [n^3 - n = (2k + 1)^3 - (2k + 1)] Expanding the cube: [(2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1] Subtract ( 2k + 1 ): [n^3 - n = 8k^3 + 12k^2 + 6k + 1 - 2k - 1 = 8k^3 + 12k^2 + 4k] Factor out ( 4k ): [n^3 - n = 4k(2k^2 + 3k + 1) = 4k(2k + 1)(k + 1)]Step 3: Prove divisibility by 8. The expression ( 4k(2k + 1)(k + 1) ) includes ( 4k ) and the product ( k(k + 1) ). Since ( k ) and ( k + 1 ) are consecutive integers, one must be even. Thus, ( k(k + 1) ) is divisible by 2, making the entire expression divisible by ( 4 times 2 = 8 ).Step 4: Prove divisibility by 3. Using modular arithmetic: For any integer ( n ), Fermat’s Little Theorem states ( n^3 equiv n pmod{3} ), so ( n^3 - n equiv 0 pmod{3} ). Thus, ( n^3 - n ) is divisible by 3.Step 5: Combine results. Since ( n^3 - n ) is divisible by both 8 and 3, and 8 and 3 are coprime, their product 24 divides ( n^3 - n ).Alternative Approach via Factorization: Factor ( n^3 - n ): [n^3 - n = n(n^2 - 1) = n(n - 1)(n + 1)] This represents the product of three consecutive integers ( n - 1, n, n + 1 ). Among three consecutive integers: - At least one is divisible by 3. - If ( n ) is odd, ( n - 1 ) and ( n + 1 ) are consecutive even numbers, with one divisible by 4. Thus, ( (n - 1)(n + 1) ) is divisible by 8. Hence, ( n(n - 1)(n + 1) ) is divisible by ( 8 times 3 = 24 ).Conclusion: For any odd integer ( n ), ( n^3 - n ) is divisible by 24. Final Answer The difference between the cube of an odd number and the number itself is divisible by 24, as proven by algebraic expansion, modular arithmetic, and factorization into consecutive integers. boxed{24}

🔑:To prove that the difference between the cube of an odd number and the number itself is divisible by 24, let's proceed step-by-step.1. Assume (2n + 1) is an odd number, where (n in mathbb{Z}). [ a = 2n + 1 ]2. Formulate the given problem using our assumption: [ (2n + 1)^3 - (2n + 1) ]3. Expand the expression ((2n + 1)^3). [ (2n + 1)^3 = (2n + 1)(2n + 1)(2n + 1) = (2n + 1)(4n^2 + 4n + 1) ]4. Multiply the expanded expressions: [ (2n + 1)(4n^2 + 4n + 1) = 8n^3 + 12n^2 + 6n + 1 ]5. Subtract ((2n + 1)) from the expanded cubic expression: [ 8n^3 + 12n^2 + 6n + 1 - (2n + 1) = 8n^3 + 12n^2 + 6n + 1 - 2n - 1 = 8n^3 + 12n^2 + 4n ]6. Factor out common terms: [ 8n^3 + 12n^2 + 4n = 4n(2n^2 + 3n + 1) ]7. Recognize that (n(n + 1)) is always an even number (as it is the product of two consecutive integers, one of which must be even): [ n(n + 1) text{ is even, i.e., divisible by } 2. ]8. Notice that since (2n + 1) is an odd number, (n(n + 1)) includes both factors of 2 and 3 (because it is the product of two consecutive integers, meaning one of them will contribute a 3 when (n = 3k) or (n+1 = 3k) for some integer (k)): [ n(n + 1) text{ is, however, divisible by } 6. ]9. Conclusively, if (n(n + 1)) is divisible by (6) and with the factor (4) in the overall expression: [ 4n(2n^2 + 3n + 1) ] This expression is divisible by (24). Now the term (4n(n(n + 1)) is divisible by (4) as well as by (6), making it divisible by (24).10. Thus, the given statement is proven: [ (2n + 1)^3 - (2n + 1) = 4n(n + 1)(2n + 1) text{ is divisible by } 24. ] Conclusion[boxed{(2n+1)^3 - (2n+1) text{ is divisible by } 24}]Which concludes our proof. blacksquare

🧠:Okay, let me try to figure out how to solve this problem. So, John is traveling from Glasgow to Dundee, which is 140 kilometers apart. He takes a bus for half an hour and then a train for two hours. The train is 20 km/h faster than the bus. Both the bus and the train are going at constant speeds. We need to find the speed of the bus.Alright, let me break this down. The total distance is 140 km. He spends part of the time on the bus and part on the train. The time he spends on the bus is half an hour, which is 0.5 hours, right? And the time on the train is two hours. So, the total distance covered by both the bus and the train should add up to 140 km.Let me denote the speed of the bus as 'b' km/h. Then the speed of the train would be 'b + 20' km/h because the train is 20 km/h faster. Now, distance equals speed multiplied by time. So, the distance covered by the bus would be bus speed multiplied by bus time, which is b * 0.5. Similarly, the distance covered by the train is train speed multiplied by train time, which is (b + 20) * 2. Adding these two distances together should equal the total distance of 140 km. So, the equation would be:0.5b + 2(b + 20) = 140Let me check if that makes sense. The bus distance is 0.5 hours times speed, and the train distance is 2 hours times its speed. Yeah, that seems right.Okay, now I need to solve this equation for 'b'. Let's expand the terms first.First, distribute the 2 into (b + 20):0.5b + 2b + 40 = 140Combine like terms:0.5b + 2b is 2.5b. So, 2.5b + 40 = 140Subtract 40 from both sides:2.5b = 140 - 402.5b = 100Now, divide both sides by 2.5:b = 100 / 2.5Let me compute that. 2.5 times 40 is 100, so 100 divided by 2.5 is 40. So, b = 40 km/h.Wait, so the bus speed is 40 km/h. Let me verify that. If the bus is going at 40 km/h for half an hour, that's 40 * 0.5 = 20 km. Then the train is going at 40 + 20 = 60 km/h for two hours, so 60 * 2 = 120 km. 20 + 120 = 140 km. Yep, that adds up to the total distance. So that checks out.Hmm, but wait, let me make sure I didn't make any mistakes in the algebra. Let me go through the steps again.Starting with the equation:0.5b + 2(b + 20) = 140Multiply out the second term:0.5b + 2b + 40 = 140Combine 0.5b and 2b. 0.5b is the same as (1/2)b, so (1/2)b + 2b = (1/2 + 4/2)b = (5/2)b, which is 2.5b. So, 2.5b + 40 = 140.Subtract 40: 2.5b = 100.Divide by 2.5: b = 100 / 2.5. Let me compute that. 2.5 * 40 = 100, so yes, 100 divided by 2.5 is 40. All steps are correct.So, the answer is 40 km/h for the bus. The train would then be 60 km/h. Let me check once more the total distance: 0.5 hours * 40 km/h = 20 km, and 2 hours * 60 km/h = 120 km. 20 + 120 = 140 km. Perfect, that's exactly the total distance given. So everything checks out.I don't think I missed anything here. The key was setting up the equation correctly by considering the distance each vehicle covers and summing them up to the total. The algebra steps were straightforward once the equation was set up. So, yeah, the speed of the bus is 40 km/h.

✅:The speed of the bus can be determined by analyzing the total distance traveled and the relationship between the bus and train speeds.Given:- Total distance = 140 km- Bus travel time = 0.5 hours- Train travel time = 2 hours- Train speed = Bus speed + 20 km/hLet:- ( b ) = speed of the bus (km/h)- ( b + 20 ) = speed of the train (km/h)Distance Equations:1. Bus distance: ( 0.5b )2. Train distance: ( 2(b + 20) )Total Distance Equation:[0.5b + 2(b + 20) = 140]Solve for ( b ):[0.5b + 2b + 40 = 140 implies 2.5b = 100 implies b = frac{100}{2.5} = 40]Verification:- Bus distance: ( 0.5 times 40 = 20 ) km- Train distance: ( 2 times (40 + 20) = 120 ) km- Total: ( 20 + 120 = 140 ) km (matches given total)Answer: The speed of the bus is (boxed{40}) km/h.

🔑:1. Define variables: Let the speed of the bus be ( v ) km/h. Therefore, the speed of the train will be ( v + 20 ) km/h (since the train travels 20 km/h faster than the bus).2. Set up the time-distance relationship: John travels by bus for half an hour (0.5 hours) and by train for two hours (2 hours), for a total distance of 140 km.3. Express the total distance: The distance traveled by bus is ( frac{1}{2} v ) km, and the distance traveled by train is ( 2 (v + 20) ) km. Therefore, the total distance is: [ frac{1}{2} v + 2 (v + 20) = 140 ]4. Simplify and solve the equation: First, distribute and combine like terms: [ frac{1}{2} v + 2v + 40 = 140 ] Combining the terms involving ( v ): [ frac{1}{2} v + 2v = frac{1}{2} v + frac{4}{2} v = frac{5}{2} v ] So, the equation becomes: [ frac{5}{2} v + 40 = 140 ]5. Isolate ( v ): Subtract 40 from both sides: [ frac{5}{2} v = 100 ] Multiply both sides by (frac{2}{5}) to solve for ( v ): [ v = 100 times frac{2}{5} ] Simplify it: [ v = 40 ]6. Conclusion: The speed of the bus is ( 40 ) km/h. [ boxed{40 text{ km/h}} ]