💡Okay, so I have this problem here: "What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides 15!?" The answer choices are from A to E, with numbers 8 to 12. Hmm, that sounds a bit complicated, but let me try to break it down step by step.First, I need to understand what 15! is. 15! means 15 factorial, which is the product of all positive integers up to 15. So, 15! = 15 × 14 × 13 × ... × 2 × 1. That's a huge number, but I don't need to calculate it directly. Instead, I need to find its prime factors.The next part is about the largest perfect square that divides 15!. A perfect square is a number that can be expressed as some integer squared, like 4 (which is 2²), 9 (3²), 16 (4²), etc. So, the largest perfect square that divides 15! would be the product of the highest powers of all prime factors of 15! that are even exponents. Because in a perfect square, all the exponents in its prime factorization must be even.Once I find that largest perfect square, I need to take its square root. Taking the square root of a perfect square will give me an integer, and the exponents in its prime factorization will be half of those in the perfect square. Then, I have to sum those exponents to get the answer.Alright, let's start by finding the prime factors of 15! and their exponents. The primes less than or equal to 15 are 2, 3, 5, 7, 11, and 13. I need to find how many times each of these primes divides into 15!.To find the exponent of a prime p in n!, we can use the formula:Exponent of p in n! = floor(n/p) + floor(n/p²) + floor(n/p³) + ... until the division results in zero.Let me apply this formula for each prime:1. **Prime 2:** - floor(15/2) = 7 - floor(15/4) = 3 - floor(15/8) = 1 - floor(15/16) = 0 So, total exponent of 2 is 7 + 3 + 1 = 11.2. **Prime 3:** - floor(15/3) = 5 - floor(15/9) = 1 - floor(15/27) = 0 So, total exponent of 3 is 5 + 1 = 6.3. **Prime 5:** - floor(15/5) = 3 - floor(15/25) = 0 So, total exponent of 5 is 3.4. **Prime 7:** - floor(15/7) = 2 - floor(15/49) = 0 So, total exponent of 7 is 2.5. **Prime 11:** - floor(15/11) = 1 - floor(15/121) = 0 So, total exponent of 11 is 1.6. **Prime 13:** - floor(15/13) = 1 - floor(15/169) = 0 So, total exponent of 13 is 1.So, the prime factorization of 15! is:15! = 2¹¹ × 3⁶ × 5³ × 7² × 11¹ × 13¹Now, to find the largest perfect square that divides 15!, I need to adjust the exponents of each prime factor to the nearest lower even number because perfect squares require even exponents.Let's adjust each exponent:- For 2: 11 is odd, so the largest even exponent less than 11 is 10.- For 3: 6 is already even, so it stays 6.- For 5: 3 is odd, so the largest even exponent less than 3 is 2.- For 7: 2 is already even, so it stays 2.- For 11: 1 is odd, so the largest even exponent less than 1 is 0.- For 13: 1 is odd, so the largest even exponent less than 1 is 0.Therefore, the largest perfect square that divides 15! is:(2¹⁰ × 3⁶ × 5² × 7²)Now, I need to take the square root of this perfect square. Taking the square root of a number essentially halves all the exponents in its prime factorization.So, the square root will be:2^(10/2) × 3^(6/2) × 5^(2/2) × 7^(2/2) = 2⁵ × 3³ × 5¹ × 7¹Now, I need to find the sum of the exponents of the prime factors in this square root.The exponents are:- 2: 5- 3: 3- 5: 1- 7: 1Adding them up: 5 + 3 + 1 + 1 = 10Wait, but looking back at the answer choices, 10 is option C. But I thought the answer was 9. Did I make a mistake?Let me double-check my steps.1. Prime factorization of 15!: Correct, exponents are 11, 6, 3, 2, 1, 1 for primes 2,3,5,7,11,13 respectively.2. Adjusting exponents for perfect square: 10,6,2,2,0,0. That seems correct.3. Square root: 2⁵ × 3³ × 5¹ × 7¹. Exponents 5,3,1,1. Sum is 10.Hmm, but the answer choices include 9 as option B. Maybe I missed something.Wait, perhaps I included too many primes. Let me think again.The largest perfect square is 2¹⁰ × 3⁶ × 5² × 7². So, when taking the square root, it's 2⁵ × 3³ × 5¹ × 7¹. So, the exponents are 5,3,1,1. Sum is 10.But maybe the question is only considering the prime factors, not all primes. Wait, the square root is 2⁵ × 3³ × 5¹ × 7¹. So, the exponents are 5,3,1,1. Sum is 10.But the answer choices are A)8, B)9, C)10, D)11, E)12.So, 10 is an option, it's C. But the initial thought was 9. Maybe I miscalculated somewhere.Wait, let's check the exponents again.15! = 2¹¹ × 3⁶ × 5³ × 7² × 11¹ × 13¹Largest perfect square: 2¹⁰ × 3⁶ × 5² × 7²Square root: 2⁵ × 3³ × 5¹ × 7¹Exponents: 5,3,1,1. Sum: 10.So, the answer should be 10, which is C.But in the initial problem statement, the user had written the same question and the assistant had concluded the answer was B)9. So, maybe I made a mistake.Wait, perhaps I included 7 in the exponents, but in the square root, 7 is only to the power of 1, so it's included. So, the sum is 5+3+1+1=10.Alternatively, maybe the question is asking for the sum of the exponents in the prime factorization of the square root, but only considering the primes that were in the original perfect square. Wait, no, the square root includes all primes from the perfect square.Alternatively, maybe the question is asking for the sum of the exponents in the prime factorization of the square root, but the square root is an integer, so it's 2⁵ × 3³ × 5¹ × 7¹, so the exponents are 5,3,1,1, sum is 10.Alternatively, maybe the question is considering only the exponents of the primes in the original 15!, but that doesn't make sense.Alternatively, maybe the largest perfect square is different.Wait, let me check the exponents again.15! has exponents:2:11, 3:6,5:3,7:2,11:1,13:1.To make the largest perfect square, we take the floor division by 2 of each exponent:For 2: floor(11/2)=5, but wait, no, to get the exponent in the perfect square, we take the largest even number less than or equal to the exponent.So, for 2:11, the largest even exponent is 10.Similarly, 3:6 is even, so 6.5:3, largest even is 2.7:2, even, so 2.11:1, largest even is 0.13:1, largest even is 0.So, the perfect square is 2¹⁰ × 3⁶ × 5² × 7².Square root is 2⁵ × 3³ × 5¹ × 7¹.Exponents:5,3,1,1. Sum:10.So, the answer should be 10, which is C.But the initial answer was 9. Maybe the initial answer was wrong.Alternatively, perhaps I made a mistake in the exponents.Wait, let me recalculate the exponents in 15!.For 2:floor(15/2)=7floor(15/4)=3floor(15/8)=1floor(15/16)=0Total:7+3+1=11. Correct.For 3:floor(15/3)=5floor(15/9)=1floor(15/27)=0Total:5+1=6. Correct.For 5:floor(15/5)=3floor(15/25)=0Total:3. Correct.For 7:floor(15/7)=2floor(15/49)=0Total:2. Correct.For 11:floor(15/11)=1floor(15/121)=0Total:1. Correct.For 13:floor(15/13)=1floor(15/169)=0Total:1. Correct.So, exponents are correct.Then, largest perfect square exponents:2:10, 3:6,5:2,7:2.Square root exponents:5,3,1,1.Sum:10.So, the answer should be C)10.But the initial answer was 9. Maybe the initial answer was incorrect.Alternatively, perhaps the question is asking for the sum of the exponents in the prime factorization of the largest perfect square, not the square root. Let me check.No, the question is: "the sum of the exponents of the prime factors of the square root of the largest perfect square that divides 15!".So, it's the square root, so exponents are halved.So, 5,3,1,1. Sum is 10.Therefore, the correct answer is C)10.But the initial answer was 9. Maybe the initial answer was wrong.Alternatively, perhaps I made a mistake in the exponents.Wait, let me check the exponents again.Wait, when taking the square root, the exponents are halved, but if the exponent was even, it becomes half, if it was odd, it becomes (exponent-1)/2.Wait, no, the largest perfect square has exponents adjusted to even numbers, so when taking the square root, it's just exponent/2.So, for 2:10 becomes 5.3:6 becomes 3.5:2 becomes 1.7:2 becomes 1.So, exponents are 5,3,1,1. Sum is 10.Therefore, the answer is 10, which is C.So, I think the initial answer was incorrect, and the correct answer is C)10.

💡Okay, so I have this problem here: If ( n ) is 2 more than a multiple of 40, what is the remainder when ( n^2 - 3n + 5 ) is divided by 40? Hmm, let me try to figure this out step by step.First, I need to understand what it means for ( n ) to be 2 more than a multiple of 40. That means ( n ) can be written as ( n = 40k + 2 ) where ( k ) is some integer. So, ( n ) is like 2, 42, 82, 122, and so on. Basically, when you divide ( n ) by 40, the remainder is 2.Now, I need to find the remainder when ( n^2 - 3n + 5 ) is divided by 40. Since ( n ) is expressed in terms of 40, maybe I can substitute ( n = 40k + 2 ) into the expression and simplify it modulo 40.Let me write that out:( n^2 - 3n + 5 = (40k + 2)^2 - 3(40k + 2) + 5 )Okay, let's expand ( (40k + 2)^2 ). That should be ( (40k)^2 + 2 times 40k times 2 + 2^2 ), right? So:( (40k)^2 = 1600k^2 )( 2 times 40k times 2 = 160k )( 2^2 = 4 )So, ( (40k + 2)^2 = 1600k^2 + 160k + 4 )Now, let's expand the other part: ( -3(40k + 2) ). That would be:( -3 times 40k = -120k )( -3 times 2 = -6 )So, putting it all together:( n^2 - 3n + 5 = 1600k^2 + 160k + 4 - 120k - 6 + 5 )Now, let's combine like terms:First, the ( k^2 ) term: 1600k^2Then the ( k ) terms: 160k - 120k = 40kConstant terms: 4 - 6 + 5 = 3So, the expression simplifies to:( 1600k^2 + 40k + 3 )Now, I need to find the remainder when this is divided by 40. Since 1600 and 40 are both multiples of 40, their terms will be divisible by 40, right? So, when we take modulo 40, those terms will vanish, leaving only the constant term.Let me see:( 1600k^2 ) modulo 40: Well, 1600 divided by 40 is 40, so 1600k^2 is 40 times 40k^2, which is obviously a multiple of 40. So modulo 40, this term is 0.Similarly, ( 40k ) modulo 40: 40k is 40 times k, so again, a multiple of 40. Modulo 40, this term is also 0.So, the entire expression modulo 40 is just the constant term, which is 3.Therefore, the remainder when ( n^2 - 3n + 5 ) is divided by 40 is 3.Wait, let me double-check to make sure I didn't make any mistakes. Maybe I can test it with a specific value of ( n ). Let's take ( k = 0 ), so ( n = 2 ).Plugging into the expression:( 2^2 - 3 times 2 + 5 = 4 - 6 + 5 = 3 ). Divided by 40, the remainder is 3. That checks out.Let's try another one, say ( k = 1 ), so ( n = 42 ).( 42^2 = 1764 )( 3 times 42 = 126 )So, ( 1764 - 126 + 5 = 1764 - 126 = 1638; 1638 + 5 = 1643 )Now, divide 1643 by 40:40 times 41 is 1640, so 1643 - 1640 = 3. The remainder is 3. That also checks out.Maybe one more, ( k = 2 ), so ( n = 82 ).( 82^2 = 6724 )( 3 times 82 = 246 )So, ( 6724 - 246 + 5 = 6724 - 246 = 6478; 6478 + 5 = 6483 )Divide 6483 by 40:40 times 162 is 6480, so 6483 - 6480 = 3. Again, remainder 3.Okay, so it seems consistent. The remainder is always 3 when ( n ) is 2 more than a multiple of 40. So, I think my answer is correct.**Final Answer**The remainder is boxed{3}.

💡Okay, so I have this problem where a teacher offers candy to her class of 30 students, and the average number of pieces each student takes is 5. I need to find out the greatest number of pieces one student could have taken, given that every student takes at least one piece. Hmm, let's see how to approach this.First, I know that the mean number of candies taken is 5. Since there are 30 students, I can calculate the total number of candies by multiplying the mean by the number of students. So, 5 times 30 is 150. That means there are 150 pieces of candy in total.Now, I need to figure out how to maximize the number of candies one student could take. To do that, I should minimize the number of candies taken by the other students because if they take as few as possible, that leaves more candies for one student. The problem says every student takes some candy, so the minimum each student can take is 1 piece.If 29 students each take 1 piece, that's 29 pieces in total. Subtracting that from the total number of candies, 150 minus 29 equals 121. So, the one student could take 121 pieces.Wait a second, does this make sense? Let me double-check. If 29 students take 1 piece each, that's 29 pieces. Adding the 121 pieces taken by the last student gives us 150, which matches the total number of candies. Yeah, that seems right.Is there any other way to distribute the candies that would allow one student to take more than 121 pieces? If I tried to have some students take more than 1 piece, that would leave fewer candies for the one student, which isn't what I want. So, to maximize one student's candies, I need to minimize the others as much as possible.Therefore, the greatest number of pieces one student could have taken is 121.

💡Okay, so Jeremy has 10 identical stickers and 3 identical sheets of paper. He wants to distribute all the stickers onto the sheets, but each sheet must have at least one sticker. And the only thing that matters is the number of stickers on each sheet, not which specific sheet gets which stickers. Hmm, okay.First, since the sheets are identical, the order doesn't matter. So, if I give 2 stickers to one sheet and 3 to another, it's the same as giving 3 to the first and 2 to the second. That means I need to find the number of distinct ways to partition 10 stickers into 3 groups, each group having at least one sticker.I remember something called the "stars and bars" theorem from combinatorics. It's used to find the number of ways to distribute identical items into distinct groups. But wait, in this case, the sheets are identical, not distinct. So, does that change things?Let me think. If the sheets were distinct, like different colors, then I could use the stars and bars formula directly. The formula is C(n + k - 1, k - 1), where n is the number of items and k is the number of groups. So for 10 stickers and 3 sheets, it would be C(10 + 3 - 1, 3 - 1) = C(12, 2) = 66 ways. But since the sheets are identical, this counts each distinct distribution multiple times, once for each permutation of the sheets.Wait, no, actually, if the sheets are identical, then we need to count the number of integer partitions of 10 into 3 parts, each at least 1. That's different from the stars and bars approach, which assumes distinguishable groups.So, maybe I should approach this by finding all the partitions of 10 into exactly 3 positive integers, where the order doesn't matter. Let me list them out.Starting with the largest possible number on a sheet, which would be 8, but then the other two sheets would have 1 each, so 8 + 1 + 1. But since the sheets are identical, 8 + 1 + 1 is the same as 1 + 8 + 1 or 1 + 1 + 8. So that's one unique distribution.Next, 7 + 2 + 1. Again, since the sheets are identical, any permutation of these numbers counts as the same distribution. So that's another unique distribution.Then, 6 + 3 + 1. That's another one.6 + 2 + 2. That's another unique distribution because two sheets have the same number of stickers.5 + 4 + 1. Another one.5 + 3 + 2. Another unique distribution.4 + 4 + 2. That's another one.4 + 3 + 3. Another unique distribution.Wait, let me check if I've missed any.Starting from the highest possible number:1. 8 + 1 + 12. 7 + 2 + 13. 6 + 3 + 14. 6 + 2 + 25. 5 + 4 + 16. 5 + 3 + 27. 4 + 4 + 28. 4 + 3 + 3Is that all? Let me see. After 4 + 3 + 3, the next would be 3 + 3 + 4, which is the same as 4 + 3 + 3. Similarly, 3 + 4 + 3 is the same. So, I think I have all the unique partitions.So, counting them up: 1, 2, 3, 4, 5, 6, 7, 8. That's 8 ways.Wait, but earlier I thought stars and bars gave 66 ways when sheets are distinct, but since sheets are identical, it's only 8 ways. That seems like a big difference. Maybe I'm missing something.Alternatively, maybe I should adjust the stars and bars approach for identical groups. I remember that when groups are identical, the problem becomes finding the number of integer partitions, which is more complicated.In this case, since the number of groups is small (3), listing them out is feasible. So, based on my earlier list, there are 8 unique ways to distribute 10 identical stickers onto 3 identical sheets, with each sheet having at least one sticker.But wait, let me double-check my list to make sure I didn't miss any partitions.Starting with the largest part:1. 8 + 1 + 12. 7 + 2 + 13. 6 + 3 + 14. 6 + 2 + 25. 5 + 4 + 16. 5 + 3 + 27. 4 + 4 + 28. 4 + 3 + 3Is there a 9 + something? 9 + 1 + 0, but each sheet must have at least one sticker, so 0 is not allowed. So, 9 is too big because the remaining two sheets would have to have at least 1 each, which would require 9 + 1 + 0, which is invalid.Similarly, 7 + 3 + 0 is invalid. So, all partitions must have each part at least 1.I think I have all the possible partitions. So, the answer should be 8 ways.Wait, but I recall that the number of partitions of 10 into exactly 3 positive integers is actually 8. Let me confirm that.Yes, according to integer partition tables, the number of partitions of 10 into exactly 3 parts is 8. So, that matches my count.Therefore, Jeremy can distribute the stickers in 8 different ways.

💡Alright, so I've got this problem here about functions satisfying something called the "inverse negative" transformation. The definition given is that a function ( f(x) ) satisfies this property if ( fleft( frac{1}{x} right) = -f(x) ). I need to check which of the four given functions meet this condition.Let me go through each function one by one.**Function 1: ( f(x) = log_{a}x ) where ( a > 0 ) and ( a neq 1 ).**Okay, so I need to compute ( fleft( frac{1}{x} right) ) and see if it equals ( -f(x) ).Calculating ( fleft( frac{1}{x} right) ):[fleft( frac{1}{x} right) = log_{a}left( frac{1}{x} right)]I remember that ( log_{a}left( frac{1}{x} right) ) is the same as ( -log_{a}x ) because ( frac{1}{x} = x^{-1} ), and logarithm of ( x^{-1} ) is ( -1 times log_{a}x ).So,[fleft( frac{1}{x} right) = -log_{a}x = -f(x)]That's exactly what we needed! So function 1 satisfies the "inverse negative" transformation.**Function 2: ( f(x) = a^{x} ) where ( a > 0 ) and ( a neq 1 ).**Again, compute ( fleft( frac{1}{x} right) ):[fleft( frac{1}{x} right) = a^{frac{1}{x}}]Now, I need to see if this equals ( -f(x) ), which would be ( -a^{x} ).But ( a^{frac{1}{x}} ) is not the same as ( -a^{x} ) unless ( a^{frac{1}{x}} + a^{x} = 0 ), which isn't generally true because exponentials are always positive (since ( a > 0 )). So, this doesn't satisfy the condition.Therefore, function 2 does not satisfy the "inverse negative" transformation.**Function 3: ( y = x - frac{1}{x} ).**Let me write this as ( f(x) = x - frac{1}{x} ).Compute ( fleft( frac{1}{x} right) ):[fleft( frac{1}{x} right) = frac{1}{x} - frac{1}{frac{1}{x}} = frac{1}{x} - x]Simplify that:[fleft( frac{1}{x} right) = frac{1}{x} - x = -left( x - frac{1}{x} right) = -f(x)]Perfect, that's exactly the condition we need. So function 3 satisfies the "inverse negative" transformation.**Function 4: A piecewise function defined as:**[f(x) = begin{cases} x, & text{if } 0 < x < 1 0, & text{if } x = 1 -frac{1}{x}, & text{if } x > 1 end{cases}]This one is a bit more involved because it's piecewise. I need to check the condition ( fleft( frac{1}{x} right) = -f(x) ) for different intervals of ( x ).Let's break it down into cases:1. **Case 1: ( 0 < x < 1 )** In this interval, ( f(x) = x ). Now, ( frac{1}{x} ) will be greater than 1 because ( x ) is between 0 and 1. So, ( fleft( frac{1}{x} right) ) falls into the third case of the piecewise function, which is ( -frac{1}{x} ). So, [ fleft( frac{1}{x} right) = -frac{1}{x} ] Now, let's compute ( -f(x) ): [ -f(x) = -x ] Wait, ( -frac{1}{x} ) is not equal to ( -x ) unless ( x = 1 ), which it's not in this interval. Hmm, that seems contradictory. Did I make a mistake? Wait, no. Let me re-examine. If ( x ) is between 0 and 1, then ( frac{1}{x} ) is greater than 1, so ( fleft( frac{1}{x} right) = -frac{1}{x} ). On the other hand, ( f(x) = x ), so ( -f(x) = -x ). So, ( fleft( frac{1}{x} right) = -frac{1}{x} ) and ( -f(x) = -x ). These are not equal unless ( x = 1 ), which is not in this interval. So, does this mean function 4 doesn't satisfy the condition? Wait, hold on. Maybe I need to think differently. Let me test with a specific value. Let's take ( x = frac{1}{2} ). Then, ( fleft( frac{1}{2} right) = frac{1}{2} ). ( frac{1}{x} = 2 ), so ( f(2) = -frac{1}{2} ). Now, ( -fleft( frac{1}{2} right) = -frac{1}{2} ). So, ( fleft( frac{1}{x} right) = -f(x) ) because ( -frac{1}{2} = -frac{1}{2} ). Wait, that's correct. Wait, but earlier, algebraically, it didn't seem to hold. Let me see: If ( f(x) = x ) for ( 0 < x < 1 ), then ( fleft( frac{1}{x} right) = -frac{1}{x} ), and ( -f(x) = -x ). So, ( -frac{1}{x} = -x ) implies ( frac{1}{x} = x ), which implies ( x^2 = 1 ), so ( x = 1 ) or ( x = -1 ). But in this interval, ( x ) is between 0 and 1, so ( x = 1 ) is the boundary, not inside the interval. But when I tested with ( x = frac{1}{2} ), it worked. So, maybe my algebraic approach was wrong. Wait, no. Let's think again. For ( 0 < x < 1 ), ( f(x) = x ), so ( -f(x) = -x ). But ( fleft( frac{1}{x} right) = -frac{1}{x} ). So, unless ( -frac{1}{x} = -x ), which would require ( frac{1}{x} = x ), which only happens at ( x = 1 ), which is not in the interval ( 0 < x < 1 ). But when I plugged in ( x = frac{1}{2} ), it worked because ( fleft( frac{1}{x} right) = -frac{1}{x} = -2 ), and ( -f(x) = -frac{1}{2} ). Wait, that's not equal. Wait, hold on, I think I made a mistake in my earlier test. Wait, if ( x = frac{1}{2} ), then ( f(x) = frac{1}{2} ), so ( -f(x) = -frac{1}{2} ). On the other hand, ( fleft( frac{1}{x} right) = f(2) = -frac{1}{2} ). So, ( fleft( frac{1}{x} right) = -frac{1}{2} ), which is equal to ( -f(x) = -frac{1}{2} ). So, in this case, it works. Wait, so algebraically, it seems like ( fleft( frac{1}{x} right) = -frac{1}{x} ) and ( -f(x) = -x ). So, unless ( -frac{1}{x} = -x ), which would imply ( x = 1 ), but in the interval ( 0 < x < 1 ), this isn't true. But when I plug in ( x = frac{1}{2} ), it works because ( fleft( frac{1}{x} right) = -frac{1}{x} = -2 ), but ( -f(x) = -frac{1}{2} ). Wait, that's not equal. Wait, I'm confused. Wait, no. If ( x = frac{1}{2} ), then ( f(x) = frac{1}{2} ), so ( -f(x) = -frac{1}{2} ). ( frac{1}{x} = 2 ), so ( f(2) = -frac{1}{2} ). Therefore, ( fleft( frac{1}{x} right) = -frac{1}{2} ), which is equal to ( -f(x) = -frac{1}{2} ). So, it works for ( x = frac{1}{2} ). Wait, but algebraically, ( fleft( frac{1}{x} right) = -frac{1}{x} ) and ( -f(x) = -x ). So, unless ( -frac{1}{x} = -x ), which is ( frac{1}{x} = x ), which is only true when ( x = 1 ). But in the interval ( 0 < x < 1 ), ( x neq 1 ), so this equality doesn't hold. But in the specific case of ( x = frac{1}{2} ), it does hold because ( fleft( frac{1}{x} right) = -frac{1}{x} = -2 ), but ( -f(x) = -frac{1}{2} ). Wait, that's not equal. Wait, I'm making a mistake here. Wait, no. If ( x = frac{1}{2} ), then ( f(x) = frac{1}{2} ), so ( -f(x) = -frac{1}{2} ). ( frac{1}{x} = 2 ), so ( f(2) = -frac{1}{2} ). Therefore, ( fleft( frac{1}{x} right) = -frac{1}{2} ), which equals ( -f(x) = -frac{1}{2} ). So, it works. Wait, but algebraically, ( fleft( frac{1}{x} right) = -frac{1}{x} ) and ( -f(x) = -x ). So, unless ( -frac{1}{x} = -x ), which is ( frac{1}{x} = x ), which is only true when ( x = 1 ). But in the interval ( 0 < x < 1 ), ( x neq 1 ), so this equality doesn't hold. But in the specific case of ( x = frac{1}{2} ), it does hold because ( fleft( frac{1}{x} right) = -frac{1}{x} = -2 ), but ( -f(x) = -frac{1}{2} ). Wait, that's not equal. Wait, I'm confused. Wait, maybe I'm mixing up the definitions. Let me clarify: For ( 0 < x < 1 ), ( f(x) = x ). So, ( fleft( frac{1}{x} right) ) is evaluated at ( frac{1}{x} ), which is greater than 1, so ( fleft( frac{1}{x} right) = -frac{1}{frac{1}{x}} = -x ). Wait, that's different from what I thought earlier. So, ( fleft( frac{1}{x} right) = -x ), and ( -f(x) = -x ). So, they are equal! Oh, I see where I went wrong earlier. I thought ( fleft( frac{1}{x} right) = -frac{1}{x} ), but actually, since ( frac{1}{x} > 1 ), ( fleft( frac{1}{x} right) = -frac{1}{frac{1}{x}} = -x ). So, ( fleft( frac{1}{x} right) = -x ), which is equal to ( -f(x) = -x ). So, it does satisfy the condition. Phew, that was a bit confusing, but I think I got it now.2. **Case 2: ( x = 1 )** Here, ( f(1) = 0 ). Compute ( fleft( frac{1}{1} right) = f(1) = 0 ). Now, ( -f(1) = -0 = 0 ). So, ( fleft( frac{1}{x} right) = 0 = -f(x) ). It works here as well.3. **Case 3: ( x > 1 )** In this interval, ( f(x) = -frac{1}{x} ). Now, ( frac{1}{x} ) will be between 0 and 1 because ( x > 1 ). So, ( fleft( frac{1}{x} right) ) falls into the first case of the piecewise function, which is ( frac{1}{x} ). So, [ fleft( frac{1}{x} right) = frac{1}{x} ] Now, ( -f(x) = -left( -frac{1}{x} right) = frac{1}{x} ). Therefore, ( fleft( frac{1}{x} right) = frac{1}{x} = -f(x) ). So, it satisfies the condition.So, after checking all cases, function 4 does satisfy the "inverse negative" transformation.**Summary:**- Function 1: Satisfies- Function 2: Does not satisfy- Function 3: Satisfies- Function 4: SatisfiesTherefore, the functions that satisfy the "inverse negative" transformation are 1, 3, and 4.

💡First, I need to determine the number of students in the Science club and the Math club, including the two students who are members of both clubs.Let ( x ) represent the number of students in the Science club, and ( 3x ) represent the number of students in the Math club since there are three times as many students in the Math club as in the Science club.The total number of students participating in either club is given as 30. This includes students who are only in the Science club, only in the Math club, and those in both. Therefore, the equation can be set up as:[x + 3x - 2 = 30]Simplifying the equation:[4x - 2 = 30][4x = 32][x = 8]So, there are 8 students in the Science club and ( 3 times 8 = 24 ) students in the Math club.Finally, to find the number of students who are in the Math club but not in the Science club, subtract the 2 students who are in both clubs from the total number of Math club students:[24 - 2 = 22]