answer：Okay, so I have this problem where I need to find the last two digits of the sum: 3! + 5! + 10! + 15! + ... + 95! + 100!. Hmm, let me think about how to approach this.First, I remember that factorials can get really big very quickly. For example, 10! is already 3,628,800. But since I'm only interested in the last two digits of the entire sum, maybe I don't need to calculate all these huge numbers. Instead, I can focus on the last two digits of each factorial and then add them up.Let me start by recalling that the last two digits of a number are its remainder when divided by 100. So, if I can find each factorial modulo 100, then add those results together modulo 100, I should get the last two digits of the sum.Alright, let's list out the factorials given in the problem: 3!, 5!, 10!, 15!, ..., 95!, 100!. I notice that the factorials are increasing by 5 each time after 5!, so it's 3!, 5!, 10!, 15!, ..., 95!, 100!.Now, I should calculate each of these factorials modulo 100. But wait, calculating 100! directly is going to be a massive number. However, I remember that for n ≥ 10, n! will have at least two factors of 5 and two factors of 2, which means n! will be divisible by 100. Therefore, for n ≥ 10, n! ≡ 0 mod 100. That simplifies things a lot because all the factorials from 10! onwards will contribute 0 to the last two digits of the sum.So, I only need to calculate 3! and 5! modulo 100 and then add them together. Let's do that.First, 3! = 6. So, 6 mod 100 is just 6.Next, 5! = 120. So, 120 mod 100 is 20.Now, adding these two results together: 6 + 20 = 26. Therefore, the last two digits of the entire sum should be 26.Wait, let me double-check that. I know that 10! and above are all multiples of 100, so they don't affect the last two digits. So, only 3! and 5! contribute. 3! is 6, which is 06 in two digits, and 5! is 120, which is 20 in two digits. Adding 06 and 20 gives 26. Yep, that seems right.I don't think I missed anything here. The key insight was recognizing that factorials beyond 9! have at least two trailing zeros, making them congruent to 0 modulo 100. So, only the smaller factorials, 3! and 5!, contribute to the last two digits of the sum.Just to be thorough, let me write out the factorials:3! = 65! = 12010! = 3,628,800 (ends with two zeros)15! = 1,307,674,368,000 (ends with three zeros)...95! and 100! both end with many zeros.So, adding 6 and 120 gives 126, whose last two digits are 26. All the other terms add zero in the last two digits, so the total sum's last two digits are 26.I think that's solid. I don't see any mistakes in my reasoning. The main takeaway is that beyond a certain point, factorials have enough factors of 2 and 5 to make them divisible by 100, thus contributing nothing to the last two digits. So, only the smaller factorials matter here.Another thing I could consider is whether there are any other factorials between 3! and 10! that might contribute. But looking at the sequence given, it's 3!, 5!, 10!, 15!, etc. So, after 5!, the next factorial is 10!, which we've already established contributes 0. So, no, there are no other factorials in between that would affect the last two digits.Therefore, I'm confident that the last two digits of the sum are 26.

answer：First, I need to determine the initial cost of 9 liters of milk. Since 1 liter used to cost 80 rubles, multiplying 80 rubles by 9 liters gives the total initial cost.Next, I calculate the current cost of 9 liters with the new pricing. Each pack now contains 0.9 liters and costs 99 rubles. To find out how many packs are needed for 9 liters, I divide 9 liters by 0.9 liters per pack. Then, I multiply the number of packs by the new price to get the total current cost.After finding both the initial and current costs, I determine the increase in cost by subtracting the initial cost from the current cost.Finally, to find the percentage increase, I divide the increase in cost by the initial cost and multiply by 100%. This gives the percentage by which the manufacturer's revenue has increased.

answer：First, substitute the given values of (x) and (y) into the equation (2x + 3z = 5y).Start with the equation:[2x + 3z = 5y]Substitute (x = 11) and (y = 8):[2(11) + 3z = 5(8)]Simplify the expressions:[22 + 3z = 40]Isolate (3z) by subtracting (22) from both sides of the equation:[3z = 40 - 22]Simplify the right-hand side:[3z = 18]Solve for (z) by dividing both sides by 3:[z = frac{18}{3} = 6]### Conclusion:[boxed{6}]

answer：Okay, so I have this problem here where I need to calculate the product of a series of fractions: (5/8) multiplied by (8/11), then (11/14), and so on, all the way up to (2002/2005). Hmm, that seems like a lot of fractions to multiply together. I wonder if there's a pattern or a shortcut I can use instead of multiplying each one step by step.Let me write out the first few terms to see if I can spot a pattern:(5/8) * (8/11) * (11/14) * (14/17) * ... * (2002/2005)Looking at this, I notice that each numerator after the first fraction seems to be the same as the denominator of the previous fraction. For example, the numerator of the second fraction is 8, which is the same as the denominator of the first fraction. Similarly, the numerator of the third fraction is 11, which matches the denominator of the second fraction. This pattern continues all the way through the series.So, if I think about multiplying these fractions together, a lot of terms should cancel out. Let me try multiplying the first two fractions:(5/8) * (8/11) = (5 * 8) / (8 * 11) = 5/11Oh, look at that! The 8 in the numerator cancels out with the 8 in the denominator. So, after multiplying the first two fractions, I'm left with 5/11.Now, let's multiply that result by the next fraction:(5/11) * (11/14) = (5 * 11) / (11 * 14) = 5/14Again, the 11 in the numerator cancels out with the 11 in the denominator. So now I have 5/14.Continuing this pattern, if I multiply by the next fraction:(5/14) * (14/17) = (5 * 14) / (14 * 17) = 5/17Once more, the 14 cancels out, leaving me with 5/17.I see a clear pattern here. Each time I multiply by the next fraction, the denominator of the previous result cancels out with the numerator of the new fraction, leaving me with just the numerator of the first fraction (which is 5) and the denominator of the last fraction I've multiplied so far.So, if I continue this process all the way to the last fraction, which is (2002/2005), the denominator of the previous fraction before that would be 2002, and it would cancel out with the numerator of the last fraction. That means the final result should be 5 divided by the denominator of the last fraction, which is 2005.Therefore, the entire product simplifies to 5/2005.Now, I should check if this fraction can be simplified further. To do that, I need to find the greatest common divisor (GCD) of 5 and 2005.Let's factorize 2005:2005 ÷ 5 = 401So, 2005 = 5 × 401Now, 401 is a prime number because it doesn't have any divisors other than 1 and itself. Therefore, the prime factors of 2005 are 5 and 401.Looking back at the fraction 5/2005, both the numerator and the denominator have a common factor of 5. So, I can divide both the numerator and the denominator by 5 to simplify the fraction:5 ÷ 5 = 12005 ÷ 5 = 401So, the simplified form of the fraction is 1/401.Let me just recap to make sure I didn't miss anything. The product is a telescoping product where each numerator cancels with the previous denominator, leaving only the first numerator and the last denominator. Then, simplifying that fraction gives me the final answer.I think that makes sense. I don't see any mistakes in my reasoning, so I'm confident that 1/401 is the correct answer.

answer：Okay, so I have this problem here about logarithms and conditions. Let me try to understand it step by step. The function given is f(x) = log base 2 of x, which is written as f(x) = log₂x. The question is asking about the relationship between the statements "a > b" and "f(a) > f(b)". Specifically, it wants to know what kind of condition "a > b" is for "f(a) > f(b)". The options are whether it's sufficient but not necessary, necessary but not sufficient, both sufficient and necessary, or neither.First, I need to recall what logarithmic functions do. Log base 2 is a common logarithmic function, and I remember that logarithmic functions are only defined for positive real numbers. So, both a and b must be greater than zero for f(a) and f(b) to be defined. That's an important point because if a or b were zero or negative, the logarithm wouldn't make sense.Next, I should think about the properties of logarithmic functions. Log base 2 is an increasing function, which means that as x increases, log₂x also increases. So, if a is greater than b, then log₂a should be greater than log₂b. That seems straightforward. So, if a > b, then f(a) > f(b). That makes "a > b" a sufficient condition for "f(a) > f(b)" because knowing that a is greater than b guarantees that f(a) is greater than f(b).But is it a necessary condition? That is, does "f(a) > f(b)" necessarily mean that "a > b"? Well, since the logarithm is an increasing function, if f(a) is greater than f(b), then a must be greater than b. So, in that sense, "a > b" is also a necessary condition because you can't have f(a) > f(b) without having a > b.Wait, but hold on. The function is only defined for positive numbers, so if a and b are both positive, then yes, "a > b" is both sufficient and necessary for "f(a) > f(b)". But what if a or b are not positive? For example, if a is positive and b is negative, then f(b) wouldn't be defined. So, in that case, "f(a) > f(b)" wouldn't make sense because f(b) is undefined. Therefore, we have to assume that a and b are both positive for f(a) and f(b) to be defined.Given that, if we're only considering positive a and b, then "a > b" is both sufficient and necessary for "f(a) > f(b)". But the problem doesn't specify that a and b are positive. It just says "a > b" and "f(a) > f(b)". So, if a and b could be any real numbers, including negative ones, then "a > b" might not necessarily lead to "f(a) > f(b)" because f(a) or f(b) could be undefined.However, in the context of the function f(x) = log₂x, the domain is already restricted to positive real numbers. So, when we talk about f(a) and f(b), we're implicitly assuming that a and b are positive. Therefore, within the domain of the function, "a > b" is both a sufficient and necessary condition for "f(a) > f(b)".But wait, the original answer said it's necessary but not sufficient. That seems contradictory to what I just thought. Maybe I need to reconsider. If "a > b" is both sufficient and necessary, then the answer should be C. But the initial answer was B. Perhaps I missed something.Let me think again. If "a > b" is sufficient, meaning that if a > b, then f(a) > f(b), which is true because log is increasing. Is it necessary? That is, if f(a) > f(b), does that imply a > b? Yes, because log is strictly increasing, so f(a) > f(b) implies a > b. Therefore, it's both sufficient and necessary.But why did the initial answer say it's necessary but not sufficient? Maybe there's a misunderstanding. Perhaps the initial answer considered that "a > b" might not always lead to "f(a) > f(b)" if a and b are not in the domain. But in the context of the function, we already assume a and b are positive. So, within that domain, "a > b" is both sufficient and necessary.Alternatively, maybe the initial answer was considering that "a > b" is necessary for "f(a) > f(b)", but not sufficient on its own because you also need a and b to be in the domain. But in the problem statement, since f(a) and f(b) are given, a and b must be in the domain. So, "a > b" is both sufficient and necessary.I think the confusion arises from whether we're considering the condition within the domain or not. If we're strictly within the domain where a and b are positive, then "a > b" is both sufficient and necessary. If we're considering all real numbers, then "a > b" isn't sufficient because f(a) or f(b) might not be defined. But since the function f(x) = log₂x is only defined for positive x, the problem is implicitly assuming a and b are positive.Therefore, I think the correct answer should be C: Sufficient and necessary condition. But the initial answer was B. Maybe I need to check the properties again.Wait, another angle: If "a > b" is a condition for "f(a) > f(b)", then "a > b" is necessary because without it, "f(a) > f(b)" can't happen. But is it sufficient? If a > b, does that always mean f(a) > f(b)? Yes, because log is increasing. So, it's both necessary and sufficient.I think the initial answer might have made a mistake. So, I would conclude that the correct answer is C: Sufficient and necessary condition.

answer：First, I need to determine the price after the initial half-price sale. If the original price is considered as 100%, then the sale price would be 50% of the original price.Next, I apply the 30% coupon discount to the already discounted sale price. A 30% discount on the sale price means the customer pays 70% of the sale price.To find the final price after both discounts, I multiply the sale price by 70%. This gives me the final price in terms of the original price.Finally, to determine the total percentage discount from the original price, I subtract the final price percentage from 100%. This calculation reveals the overall discount percentage.