💡Alright, let's tackle this problem step by step. So, we have a group of monkeys divided into three subgroups, each with an equal number of monkeys. They go to pick peaches, combine them, and then distribute them equally. First, let's try to understand the problem. When each monkey gets 5 peaches, there are 27 left over. That means if we take the total number of peaches and subtract 27, it should be exactly divisible by the number of monkeys. On the other hand, when each monkey tries to get 7 peaches, at least one monkey ends up with fewer than 7. This implies that if we try to distribute 7 peaches per monkey, we don't have enough to give everyone exactly 7, so at least one monkey gets less. Let me denote the number of monkeys as ( n ) and the total number of peaches as ( m ). From the first scenario, if each monkey gets 5 peaches and 27 are left over, we can write the equation:[ m = 5n + 27 ]From the second scenario, if each monkey gets 7 peaches, but at least one monkey gets fewer, this means that the total number of peaches is not enough to give 7 to each monkey. So, the total peaches can be expressed as:[ m = 7n - k ]where ( k ) is the number of peaches short, and ( k ) must be at least 1 because at least one monkey gets fewer than 7.Now, we have two expressions for ( m ):1. ( m = 5n + 27 )2. ( m = 7n - k )Since both equal ( m ), we can set them equal to each other:[ 5n + 27 = 7n - k ]Let's solve for ( k ):[ 5n + 27 = 7n - k ][ 27 + k = 7n - 5n ][ 27 + k = 2n ][ k = 2n - 27 ]Now, since ( k ) is the number of peaches short, it must be a positive integer. So, ( 2n - 27 > 0 ), which implies:[ 2n > 27 ][ n > 13.5 ]Since the number of monkeys must be an integer, ( n geq 14 ).Also, since ( k ) is the number of peaches short, and each monkey can only be short by up to 6 peaches (because if they were short by 7, that would mean another monkey could have gotten 7), we have:[ k < 7 ][ 2n - 27 < 7 ][ 2n < 34 ][ n < 17 ]So, combining the two inequalities:[ 14 leq n < 17 ]Thus, ( n ) can be 14, 15, or 16.Let's test these values:1. If ( n = 14 ): [ m = 5(14) + 27 = 70 + 27 = 97 ] Check the second condition: [ 97 = 7(14) - k ] [ 97 = 98 - k ] [ k = 1 ] This works because ( k = 1 ) is less than 7.2. If ( n = 15 ): [ m = 5(15) + 27 = 75 + 27 = 102 ] Check the second condition: [ 102 = 7(15) - k ] [ 102 = 105 - k ] [ k = 3 ] This also works because ( k = 3 ) is less than 7.3. If ( n = 16 ): [ m = 5(16) + 27 = 80 + 27 = 107 ] Check the second condition: [ 107 = 7(16) - k ] [ 107 = 112 - k ] [ k = 5 ] This works as well because ( k = 5 ) is less than 7.Hmm, so all three values of ( n ) seem to satisfy the conditions. But the problem mentions that the monkeys are divided into three subgroups with an equal number of monkeys. That means the total number of monkeys ( n ) must be divisible by 3.Let's check which of our ( n ) values are divisible by 3:- 14 divided by 3 is not an integer.- 15 divided by 3 is 5, which is an integer.- 16 divided by 3 is not an integer.So, only ( n = 15 ) fits the condition that the monkeys are divided into three equal subgroups.Therefore, the total number of peaches is:[ m = 5(15) + 27 = 75 + 27 = 102 ]Let me double-check:- If each monkey gets 5 peaches: ( 15 times 5 = 75 ). Total peaches are 102, so ( 102 - 75 = 27 ) left over. That matches.- If each monkey gets 7 peaches: ( 15 times 7 = 105 ). But we only have 102 peaches, so ( 105 - 102 = 3 ) peaches short. That means 3 monkeys get 6 peaches instead of 7, which satisfies the condition that at least one monkey gets fewer than 7.Everything checks out!

💡Okay, so I have this problem about a man with a seven-candle chandelier. Each evening, he lights an increasing number of candles: one on the first evening, two on the second, and so on until he lights all seven on the seventh evening. After burning for one hour each evening, all candles finish simultaneously on the seventh evening. The question is, how did he choose which candles to light each evening?Alright, let me try to wrap my head around this. First, there are seven candles, and each night he lights a different number of them. Each candle burns for one hour each night it's lit. By the end of the seventh night, all candles have burned out completely. So, each candle must have been lit for a total of four hours because 7 nights minus the first three nights when fewer candles were lit... Wait, no, that doesn't quite make sense.Let me think again. If he starts with one candle on the first night, then two on the second, up to seven on the seventh, each candle must have been lit on multiple nights such that the total burning time adds up to four hours. So, each candle is lit on four different evenings, burning one hour each time, totaling four hours.Okay, so the challenge is to figure out which combination of candles he lit each evening so that each candle is lit exactly four times. That way, by the seventh evening, all candles have been burning for four hours and finish together.Maybe I can approach this by trying to assign each candle to four different evenings. There are seven candles and seven evenings, with the number of candles lit each evening increasing from one to seven. So, on the first evening, only one candle is lit; on the second, two; and so on until the seventh evening, where all seven are lit.I need to make sure that each candle is included in exactly four of these evenings. Let me list the evenings and the number of candles lit each night:1. Evening 1: 1 candle2. Evening 2: 2 candles3. Evening 3: 3 candles4. Evening 4: 4 candles5. Evening 5: 5 candles6. Evening 6: 6 candles7. Evening 7: 7 candlesNow, I need to assign each candle to four of these evenings. Each candle must appear in four different evenings, and the total number of candle-lightings should be the sum from 1 to 7, which is 28. Since there are seven candles, each being lit four times, that also totals 28 (7 candles * 4 lightings each). So, the numbers match up.But how do I assign them so that each candle is lit exactly four times without overlapping in a way that causes some candles to be lit more or less than four times?Maybe I can think of it as a combinatorial problem. I need to select subsets of the seven candles for each evening, with the size of the subset matching the evening number, such that each candle is included in exactly four subsets.This sounds like a design problem, perhaps similar to a block design in combinatorics where each element is in a certain number of blocks. Specifically, a (v, k, λ) design where v is the number of elements, k is the block size, and λ is the number of blocks each element is in. But in this case, the block sizes vary from 1 to 7, and each element needs to be in exactly four blocks.This might be more complex than a standard block design. Maybe I can approach it by trying to construct the subsets step by step.Let's denote the candles as C1, C2, C3, C4, C5, C6, C7.Starting with Evening 1: Only one candle is lit. Let's say C1.Evening 2: Two candles. Maybe C2 and C3.Evening 3: Three candles. Perhaps C4, C5, C6.Evening 4: Four candles. Now, I need to include C7 and maybe some from previous evenings. Let's say C7, C1, C2, C3.Evening 5: Five candles. I need to include C4, C5, C6, C7, and maybe one more. Let's include C1.Evening 6: Six candles. I need to include almost all candles. Let's include C2, C3, C4, C5, C6, C7.Evening 7: All seven candles.Now, let's check how many times each candle is lit:C1: Evenings 1, 4, 5, 7 → 4 timesC2: Evenings 2, 4, 6, 7 → 4 timesC3: Evenings 2, 4, 6, 7 → 4 timesC4: Evenings 3, 5, 6, 7 → 4 timesC5: Evenings 3, 5, 6, 7 → 4 timesC6: Evenings 3, 5, 6, 7 → 4 timesC7: Evenings 4, 5, 6, 7 → 4 timesPerfect! Each candle is lit exactly four times, so they all burn out after four hours, finishing simultaneously on the seventh evening.Wait, did I just come up with a solution? Let me double-check.Evening 1: C1 → 1 candleEvening 2: C2, C3 → 2 candlesEvening 3: C4, C5, C6 → 3 candlesEvening 4: C7, C1, C2, C3 → 4 candlesEvening 5: C4, C5, C6, C7, C1 → 5 candlesEvening 6: C2, C3, C4, C5, C6, C7 → 6 candlesEvening 7: C1, C2, C3, C4, C5, C6, C7 → 7 candlesYes, that seems to work. Each candle is lit four times, so they all burn out after four hours, finishing together on the seventh evening.I think this makes sense. By carefully selecting which candles to light each evening, ensuring that each candle is included in exactly four evenings, the man can have all candles finish simultaneously on the seventh evening.

💡First, I need to determine the volume of the entire cake. The cake is cylindrical, so its volume can be calculated using the formula for the volume of a cylinder: ( V = pi r^2 h ), where ( r ) is the radius and ( h ) is the height.The diameter of the cake is 16 inches, so the radius ( r ) is half of that, which is 8 inches. The thickness of the cake, which serves as the height ( h ), is ( frac{1}{2} ) inch.Plugging these values into the formula:[V = pi (8)^2 left(frac{1}{2}right) = pi times 64 times frac{1}{2} = 32pi text{ cubic inches}.]Next, since the cake is cut into 8 equal pieces, the volume of one piece is the total volume divided by 8:[frac{32pi}{8} = 4pi text{ cubic inches}.]Therefore, the volume of one piece of cake is ( 4pi ) cubic inches.

💡Alright, so I have this problem here: Let ( a geq 2 ) be an integer, and let ( b ) be a composite number (( b > 0 )). If there are ( r ) different positive integers that can divide ( b ), then there are at least ( r ) different positive integers that can divide ( a^b - 1 ). Hmm, okay, let me try to make sense of this.First, let me parse the problem. We have two integers, ( a ) and ( b ). ( a ) is at least 2, and ( b ) is composite, meaning it has more than two positive divisors. The number of positive divisors of ( b ) is ( r ). The claim is that ( a^b - 1 ) has at least ( r ) positive divisors.So, my goal is to show that if ( b ) has ( r ) divisors, then ( a^b - 1 ) has at least ( r ) divisors. I need to find a relationship between the divisors of ( b ) and the divisors of ( a^b - 1 ).Let me recall some number theory concepts. I know that if ( d ) divides ( b ), then ( a^d - 1 ) divides ( a^b - 1 ). Is that correct? Let me verify. Suppose ( d ) divides ( b ), so ( b = d cdot k ) for some integer ( k ). Then, ( a^b - 1 = a^{d cdot k} - 1 ). I can factor this as ( (a^d)^k - 1 ), which can be written as ( (a^d - 1)(a^{d(k-1)} + a^{d(k-2)} + dots + a^d + 1) ). So, yes, ( a^d - 1 ) is a factor of ( a^b - 1 ).Therefore, each divisor ( d ) of ( b ) gives rise to a divisor ( a^d - 1 ) of ( a^b - 1 ). Since ( b ) has ( r ) divisors, this would imply that ( a^b - 1 ) has at least ( r ) divisors. But wait, is that necessarily true? Are all these ( a^d - 1 ) distinct?Hmm, that's a good question. If ( d_1 ) and ( d_2 ) are distinct divisors of ( b ), does that mean ( a^{d_1} - 1 ) and ( a^{d_2} - 1 ) are distinct? Let's think about it. Suppose ( a^{d_1} - 1 = a^{d_2} - 1 ). Then, ( a^{d_1} = a^{d_2} ). Since ( a geq 2 ), this would imply ( d_1 = d_2 ). So, yes, each distinct divisor ( d ) of ( b ) gives a distinct divisor ( a^d - 1 ) of ( a^b - 1 ).Therefore, since ( b ) has ( r ) divisors, ( a^b - 1 ) must have at least ( r ) divisors. That seems to make sense.But let me test this with an example to make sure I'm not missing something. Let's take ( a = 2 ) and ( b = 4 ). The divisors of ( b = 4 ) are 1, 2, and 4, so ( r = 3 ). Now, ( a^b - 1 = 2^4 - 1 = 16 - 1 = 15 ). The divisors of 15 are 1, 3, 5, and 15, so that's 4 divisors. Since 4 is greater than ( r = 3 ), it satisfies the condition.Another example: Let ( a = 3 ) and ( b = 6 ). The divisors of 6 are 1, 2, 3, and 6, so ( r = 4 ). Then, ( a^b - 1 = 3^6 - 1 = 729 - 1 = 728 ). The divisors of 728 are 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, and 728, which is 16 divisors. Again, 16 is greater than ( r = 4 ).Wait, but what if ( a ) and ( b ) are such that ( a^d - 1 ) is the same for different ( d )? Is that possible? Let's see. Suppose ( a = 2 ), ( d_1 = 2 ), ( d_2 = 3 ). Then ( 2^2 - 1 = 3 ) and ( 2^3 - 1 = 7 ). These are different. Another example: ( a = 5 ), ( d_1 = 2 ), ( d_2 = 4 ). Then ( 5^2 - 1 = 24 ) and ( 5^4 - 1 = 624 ). These are different as well.So, it seems that for ( a geq 2 ), ( a^{d_1} - 1 ) and ( a^{d_2} - 1 ) are distinct if ( d_1 ) and ( d_2 ) are distinct. Therefore, each divisor ( d ) of ( b ) gives a unique divisor ( a^d - 1 ) of ( a^b - 1 ). Hence, the number of divisors of ( a^b - 1 ) is at least the number of divisors of ( b ), which is ( r ).But let me think about whether ( a^d - 1 ) could potentially share factors with other terms or if there's any overlap that I'm not considering. For example, could ( a^{d_1} - 1 ) and ( a^{d_2} - 1 ) share a common factor other than 1? Yes, they could. For instance, if ( d_1 ) and ( d_2 ) are not coprime, ( a^{d_1} - 1 ) and ( a^{d_2} - 1 ) might share a common factor. However, in terms of the number of distinct divisors, each ( a^d - 1 ) is a distinct divisor, even if they share common factors. So, the count of distinct divisors would still be at least ( r ).Another angle: The number of divisors of ( a^b - 1 ) is related to its prime factorization. Each ( a^d - 1 ) contributes to this factorization, potentially introducing new prime factors or increasing the exponents of existing ones. However, regardless of how the factors interact, each ( a^d - 1 ) is a distinct divisor, so the total number of divisors must be at least ( r ).Wait, but is ( a^d - 1 ) necessarily a proper divisor of ( a^b - 1 )? Let me see. If ( d ) is a proper divisor of ( b ), meaning ( d < b ), then ( a^d - 1 ) divides ( a^b - 1 ) and is less than ( a^b - 1 ), so it's a proper divisor. If ( d = b ), then ( a^d - 1 = a^b - 1 ), which is the number itself, so it's a trivial divisor. Therefore, all ( a^d - 1 ) for ( d ) dividing ( b ) are divisors of ( a^b - 1 ), including the number itself.So, in terms of counting, each ( d ) gives a distinct divisor, and since ( d ) ranges over all divisors of ( b ), including 1 and ( b ) itself, ( a^d - 1 ) ranges over at least ( r ) distinct divisors of ( a^b - 1 ). Therefore, the number of divisors of ( a^b - 1 ) is at least ( r ).I think this makes sense. Each divisor of ( b ) gives a unique divisor of ( a^b - 1 ), and since ( b ) has ( r ) divisors, ( a^b - 1 ) must have at least ( r ) divisors. I can't think of a case where this wouldn't hold, given that ( a geq 2 ) and ( b ) is composite.Just to be thorough, let me consider a case where ( b ) is a prime power. Wait, no, ( b ) is composite, so it can't be a prime. It has to have at least three divisors: 1, itself, and at least one other. So, for example, ( b = 4 ), which has divisors 1, 2, 4. Then ( a^4 - 1 ) would have divisors ( a^1 - 1 ), ( a^2 - 1 ), and ( a^4 - 1 ). As we saw earlier, these are distinct, so ( a^4 - 1 ) has at least 3 divisors, which is the number of divisors of ( b ).Another example: ( b = 6 ), which has divisors 1, 2, 3, 6. Then ( a^6 - 1 ) would have divisors ( a^1 - 1 ), ( a^2 - 1 ), ( a^3 - 1 ), and ( a^6 - 1 ). These are all distinct, so ( a^6 - 1 ) has at least 4 divisors.Wait, but in the case where ( a = 1 ), ( a^d - 1 = 0 ), which isn't a positive integer. But the problem states ( a geq 2 ), so that's fine.Also, if ( b = 1 ), but ( b ) is composite, so ( b ) must be at least 4, since 1 is neither prime nor composite, and 2 and 3 are primes. So, ( b ) starts from 4 upwards.I think I've covered the bases here. Each divisor of ( b ) gives a distinct divisor of ( a^b - 1 ), so the number of divisors of ( a^b - 1 ) is at least ( r ). Therefore, the statement is true.

💡Alright, so I need to figure out how much Donald will pay for the laptop after a 15% discount. Okay, let's start by understanding what a 15% reduction means. Percentage means per hundred, so 15% is the same as 15 per hundred, or 0.15 in decimal form. First, I think I need to calculate what 15% of the original price is. The original price of the laptop is 800. So, if I take 15% of 800, that should give me the discount amount. To calculate 15% of 800, I can multiply 800 by 0.15. Let me do that: 800 times 0.15 equals... hmm, 800 times 0.1 is 80, and 800 times 0.05 is 40, so adding those together, 80 plus 40 is 120. So, the discount is 120.Now, to find out how much Donald will pay after the discount, I need to subtract the discount amount from the original price. That would be 800 minus 120. Let me calculate that: 800 minus 120 equals 680. So, Donald will pay 680 for the laptop after the 15% discount. Wait, let me double-check my calculations to make sure I didn't make any mistakes. First, 15% of 800: 800 times 0.15 is indeed 120. Then, subtracting that from 800 gives 680. Yep, that seems correct. Alternatively, I could think of it as paying 85% of the original price because if there's a 15% discount, you're paying 100% minus 15%, which is 85%. So, 85% of 800 is the same as 0.85 times 800. Let me calculate that: 0.85 times 800 equals 680. Okay, both methods give me the same result, so I'm confident that 680 is the correct amount Donald will pay for the laptop after the discount.

💡Alright, so I've got this problem here about a fraction where the numerator is 3x minus 2, and the denominator is 12 minus 3x. The question is asking for the values of x in the interval from -3 to 3 where the numerator is at least three times the denominator. Hmm, okay, let me try to unpack this step by step.First off, I need to understand what it means for the numerator to be at least three times the denominator. So, if I have a fraction, the numerator is on top, and the denominator is on the bottom. The problem is saying that the numerator should be greater than or equal to three times the denominator. So, in mathematical terms, that would translate to:3x - 2 ≥ 3(12 - 3x)Okay, so I need to solve this inequality for x, keeping in mind that x has to be between -3 and 3. Let me write that down:3x - 2 ≥ 3(12 - 3x)Alright, let's simplify the right side first. I'll distribute the 3 into the parentheses:3x - 2 ≥ 36 - 9xNow, I want to get all the x terms on one side and the constants on the other. So, I'll add 9x to both sides to move the x terms to the left:3x + 9x - 2 ≥ 36That simplifies to:12x - 2 ≥ 36Next, I'll add 2 to both sides to isolate the term with x:12x ≥ 38Now, to solve for x, I'll divide both sides by 12:x ≥ 38/12Simplifying that fraction, 38 divided by 12 is the same as 19/6, which is approximately 3.1667.Wait a minute, but the problem states that x must be in the interval [-3, 3]. So, x has to be less than or equal to 3. But according to my solution, x needs to be greater than or equal to 19/6, which is about 3.1667. That's outside the given interval because 19/6 is more than 3.Hmm, that seems like a problem. So, does that mean there are no solutions within the interval [-3, 3]? Because 19/6 is approximately 3.1667, which is greater than 3, and x can't exceed 3.Let me double-check my steps to make sure I didn't make a mistake. Starting from the inequality:3x - 2 ≥ 3(12 - 3x)Distribute the 3:3x - 2 ≥ 36 - 9xAdd 9x to both sides:12x - 2 ≥ 36Add 2 to both sides:12x ≥ 38Divide by 12:x ≥ 38/12 = 19/6 ≈ 3.1667Yeah, that seems correct. So, x needs to be at least 19/6, but since x can't be more than 3, there's no overlap between x ≥ 19/6 and x ≤ 3. Therefore, there are no values of x in the interval [-3, 3] that satisfy the inequality.Wait, but just to be thorough, let me test the boundary point x = 3. If I plug x = 3 into the numerator and denominator, what do I get?Numerator: 3(3) - 2 = 9 - 2 = 7Denominator: 12 - 3(3) = 12 - 9 = 3So, the numerator is 7, and three times the denominator is 3*3 = 9. Is 7 ≥ 9? No, it's not. So, even at x = 3, the numerator isn't three times the denominator.What about x = 19/6? Let me calculate that. 19 divided by 6 is approximately 3.1667, which is outside the interval. So, even if I plug that in, it's not within the allowed range.Therefore, it seems like there's no solution within the given interval. So, the answer should be that there's no valid solution.But just to make sure, let me think about the behavior of the function. The numerator is 3x - 2, which is a linear function increasing with x. The denominator is 12 - 3x, which is a linear function decreasing with x. So, as x increases, the numerator increases, and the denominator decreases. Therefore, the ratio of numerator to denominator increases as x increases.So, the point where the numerator equals three times the denominator is at x = 19/6, which is about 3.1667. Since our interval stops at x = 3, and at x = 3, the numerator is 7 and three times the denominator is 9, which is still less than 7. So, even at the upper bound of the interval, the numerator isn't three times the denominator.Therefore, there's no x in [-3, 3] that satisfies the condition. So, the correct answer is that there's no valid solution.