💡Okay, so I have this problem where a set of three real numbers can be represented in two different ways: one as {a, b/a, 1} and the other as {a², a + b, 0}. I need to find the value of a²⁰⁰³ + b²⁰⁰⁴. Hmm, let me try to figure this out step by step.First, since both representations are sets, the order doesn't matter, right? So, the elements in both sets must be the same, just possibly arranged differently. That means each element in the first set must be equal to one of the elements in the second set, and vice versa.Looking at the first set: {a, b/a, 1}. The second set is {a², a + b, 0}. So, one of the elements in the first set must be 0 because the second set has 0. But in the first set, the elements are a, b/a, and 1. So, either a = 0, b/a = 0, or 1 = 0. But 1 can't be 0, so that's not possible. If a were 0, then b/a would be undefined because we can't divide by zero. So, a can't be 0 either. Therefore, the only possibility is that b/a = 0. If b/a = 0, then b must be 0 because any number divided by a non-zero number is 0. So, b = 0. That simplifies things a bit. Now, let's substitute b = 0 into both sets.Substituting b = 0 into the first set, we get {a, 0/a, 1} which simplifies to {a, 0, 1}. The second set becomes {a², a + 0, 0}, which is {a², a, 0}. So now, both sets are {a, 0, 1} and {a², a, 0}. Since these sets are equal, their elements must be the same. So, the elements a, 0, and 1 must be the same as a², a, and 0. That means that a² must be equal to either a or 1 because those are the other elements in the first set.Let's consider the two possibilities:1. If a² = a, then we can solve for a: a² = a a² - a = 0 a(a - 1) = 0 So, a = 0 or a = 1. But earlier, we established that a can't be 0 because b/a would be undefined. So, a = 1 is the only solution here.2. If a² = 1, then: a² = 1 So, a = 1 or a = -1. Again, a = 1 is possible, but we also have a = -1 as another solution.Wait, so from the first case, we get a = 1, and from the second case, we get a = 1 or a = -1. So, combining these, the possible values for a are 1 and -1.But let's check if a = 1 works. If a = 1, then the first set becomes {1, 0, 1}, which simplifies to {0, 1} because sets don't have duplicate elements. The second set becomes {1², 1 + 0, 0} = {1, 1, 0}, which also simplifies to {0, 1}. So, both sets are {0, 1}, which is consistent. But wait, the original sets were supposed to have three elements each. If a = 1, then both sets only have two elements because 1 is repeated. That might be an issue because the problem states it's a set of three real numbers. So, maybe a = 1 isn't acceptable because it reduces the set size.Let me think about that. If a = 1, then the first set is {1, 0, 1} which is actually {0, 1}, a set with two elements. Similarly, the second set is {1, 1, 0}, which is also {0, 1}. But the problem mentions a set of three real numbers, so having only two elements might not satisfy the condition. Therefore, a = 1 might not be a valid solution.So, let's consider a = -1. If a = -1, then the first set becomes {-1, 0, 1}, which has three distinct elements. The second set becomes {(-1)², -1 + 0, 0} = {1, -1, 0}, which is also {-1, 0, 1}. So, both sets have three elements and are equal. That works perfectly.Therefore, a must be -1, and b is 0 as we found earlier.Now, we need to compute a²⁰⁰³ + b²⁰⁰⁴. Let's substitute the values we found:a = -1, so a²⁰⁰³ = (-1)²⁰⁰³. Since 2003 is an odd number, (-1) raised to an odd power is -1. So, a²⁰⁰³ = -1.b = 0, so b²⁰⁰⁴ = 0²⁰⁰⁴. Any number raised to any power is 0, so b²⁰⁰⁴ = 0.Adding them together: -1 + 0 = -1.So, the value of a²⁰⁰³ + b²⁰⁰⁴ is -1.**Final Answer**boxed{-1}

💡Okay, so I have this problem where I need to figure out the positional relationship between two circles defined by the equations (x^2 + y^2 - 6y = 0) and (x^2 + y^2 - 8x + 12 = 0). Hmm, I remember that to determine the relationship between two circles, I need to find their centers and radii first. Once I have those, I can figure out if they're separate, overlapping, tangent, or one inside the other.Let me start with the first circle: (x^2 + y^2 - 6y = 0). I think I need to rewrite this equation in the standard form of a circle, which is ((x - h)^2 + (y - k)^2 = r^2), where ((h, k)) is the center and (r) is the radius. Looking at the equation, I see that there's an (x^2) term and a (y^2) term, but the (y) term has a coefficient. I think I need to complete the square for the (y) terms. So, let's group the (y) terms together:(x^2 + (y^2 - 6y) = 0).To complete the square for (y^2 - 6y), I take half of the coefficient of (y), which is (-6), so half of that is (-3), and then square it, getting (9). So I add and subtract 9 inside the equation:(x^2 + (y^2 - 6y + 9 - 9) = 0).This simplifies to:(x^2 + (y - 3)^2 - 9 = 0).Adding 9 to both sides gives:(x^2 + (y - 3)^2 = 9).So, the center of the first circle is at ((0, 3)) and the radius is (sqrt{9} = 3).Alright, that wasn't too bad. Now, moving on to the second circle: (x^2 + y^2 - 8x + 12 = 0). Again, I need to write this in standard form. This time, the (x) term has a coefficient, so I'll complete the square for the (x) terms.Let me rearrange the equation:(x^2 - 8x + y^2 + 12 = 0).Grouping the (x) terms:((x^2 - 8x) + y^2 + 12 = 0).To complete the square for (x^2 - 8x), I take half of (-8), which is (-4), and square it to get (16). So I add and subtract 16:((x^2 - 8x + 16 - 16) + y^2 + 12 = 0).This simplifies to:((x - 4)^2 - 16 + y^2 + 12 = 0).Combining the constants:((x - 4)^2 + y^2 - 4 = 0).Adding 4 to both sides:((x - 4)^2 + y^2 = 4).So, the center of the second circle is at ((4, 0)) and the radius is (sqrt{4} = 2).Now that I have both centers and radii, I can determine the positional relationship. I remember that the distance between the centers compared to the sum and difference of the radii tells us how the circles are positioned relative to each other.First, let's find the distance between the centers. The centers are at ((0, 3)) and ((4, 0)). Using the distance formula:(d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}).Plugging in the coordinates:(d = sqrt{(4 - 0)^2 + (0 - 3)^2} = sqrt{16 + 9} = sqrt{25} = 5).So the distance between the centers is 5 units.Now, let's compare this distance to the sum and difference of the radii. The radii are 3 and 2, so the sum is (3 + 2 = 5) and the difference is (3 - 2 = 1).Since the distance between the centers is equal to the sum of the radii ((d = 5 = 3 + 2)), this means the circles are externally tangent to each other. If the distance were greater than the sum, they would be separate; if it were less, they would overlap. If the distance were equal to the difference, one would be inside the other and tangent. Since it's equal to the sum, they touch at exactly one point externally.Wait, let me just double-check to make sure I didn't make a mistake. The centers are 5 units apart, and the sum of the radii is also 5. So yes, they should be externally tangent. I don't think they overlap because the distance isn't less than the sum, and they aren't separate because the distance isn't greater. So, yeah, external tangency makes sense.I think that's it. The positional relationship is that they are externally tangent.

💡Alright, so I've got these two combinatorics problems to solve. Let me take them one at a time.**Problem 1:** How many different ways are there to choose 4 letters from the multiset {a, b, b, c, c, c, d, d, d, d}?Okay, so first, I need to understand what's being asked here. We have a set of letters where some letters are repeated: there's one 'a', two 'b's, three 'c's, and four 'd's. We need to choose 4 letters from this set, and we want to know how many different ways we can do that.I remember that when dealing with combinations where there are repeated elements, we need to account for the multiplicities. So, it's not just a simple combination problem because some letters can be chosen multiple times, but only up to their available counts.Maybe I can approach this by considering the different cases based on how many of each letter are chosen. Since we're choosing 4 letters, the possible distributions of letters can vary. For example, we could have all four letters being the same, or three of one kind and one of another, and so on.Let me list out the possible distributions:1. All four letters are the same.2. Three letters are the same, and one is different.3. Two letters are the same, and the other two are different.4. Two letters are the same, and another two are the same (i.e., two pairs).5. All four letters are different.Now, let's analyze each case.**Case 1: All four letters are the same.**Looking at the multiset, only 'd' appears four times. So, the only possibility here is choosing four 'd's. That's 1 way.**Case 2: Three letters are the same, and one is different.**We need to choose a letter that appears at least three times, and then choose a different letter that appears at least once.From the multiset, 'c' appears three times, and 'd' appears four times. So, we can have three 'c's and one other letter, or three 'd's and one other letter.Let's break it down:- Three 'c's and one other letter: The other letter can be 'a', 'b', or 'd'. That's 3 possibilities.- Three 'd's and one other letter: The other letter can be 'a', 'b', or 'c'. That's another 3 possibilities.So, total for Case 2: 3 + 3 = 6 ways.**Case 3: Two letters are the same, and the other two are different.**Here, we need to choose one letter that appears at least twice, and then two different letters from the remaining letters.First, choose the letter to be repeated twice. The possible letters are 'b', 'c', and 'd' since they have at least two copies.- If we choose 'b' as the repeated letter: Then we need to choose two different letters from the remaining letters: 'a', 'c', 'c', 'd', 'd', 'd', 'd'. But since we're choosing two different letters, we have 'a', 'c', and 'd' as options. So, the number of ways is C(3,2) = 3. - If we choose 'c' as the repeated letter: Similarly, we need to choose two different letters from 'a', 'b', 'b', 'd', 'd', 'd', 'd'. The distinct letters are 'a', 'b', and 'd'. So, again, C(3,2) = 3.- If we choose 'd' as the repeated letter: We need to choose two different letters from 'a', 'b', 'b', 'c', 'c', 'c'. The distinct letters are 'a', 'b', and 'c'. So, C(3,2) = 3.So, for each choice of the repeated letter, we have 3 ways, and there are 3 choices for the repeated letter. So, total for Case 3: 3 * 3 = 9 ways.Wait, but hold on. Is that correct? Because when we choose two different letters, we have to consider that some letters have multiple copies, but since we're only choosing one of each, the multiplicities beyond one don't matter here. So, yes, I think that's correct.**Case 4: Two letters are the same, and another two are the same (i.e., two pairs).**Here, we need to choose two different letters, each appearing at least twice, and take two copies of each.Looking at the multiset, the letters that appear at least twice are 'b', 'c', and 'd'.We need to choose two distinct letters from these three. The number of ways is C(3,2) = 3.So, for each pair of letters, we can form a combination of two pairs. For example, two 'b's and two 'c's, two 'b's and two 'd's, or two 'c's and two 'd's.So, total for Case 4: 3 ways.**Case 5: All four letters are different.**We need to choose four distinct letters from the available letters: 'a', 'b', 'c', 'd'. Since there are exactly four distinct letters, there's only one way to choose all four: one 'a', one 'b', one 'c', and one 'd'.So, total for Case 5: 1 way.Now, let's sum up all the cases:- Case 1: 1- Case 2: 6- Case 3: 9- Case 4: 3- Case 5: 1Total number of ways: 1 + 6 + 9 + 3 + 1 = 20.Wait, that doesn't seem right. I thought the answer was 175, but that was for a different problem. Hmm, maybe I confused the two problems.Wait, no, the first problem is about choosing 4 letters, and the second is about forming four-digit numbers. So, perhaps I need to consider permutations instead of combinations for the second problem.But for the first problem, since it's about choosing letters, and the order doesn't matter, combinations are appropriate. So, 20 ways.But wait, I think I might have missed something. Let me double-check.In Case 3, when choosing two of one letter and two different letters, I considered choosing two different letters from the remaining. But actually, since some letters have multiple copies, does that affect the count?Wait, no, because we're choosing only one of each different letter, so the multiplicities beyond one don't matter. So, I think that part is correct.Similarly, in Case 2, choosing three of one letter and one different, I considered all possibilities, and that seems correct.Case 4, two pairs, is straightforward.Case 5, all different, is just one way.So, total 20 ways.But wait, I recall that when dealing with multiset combinations, sometimes it's better to use generating functions or the stars and bars method. Maybe I should try that approach to verify.Let me try using generating functions.The generating function for each letter is:- For 'a': 1 + x (since there's only one 'a')- For 'b': 1 + x + x^2 (since there are two 'b's)- For 'c': 1 + x + x^2 + x^3 (since there are three 'c's)- For 'd': 1 + x + x^2 + x^3 + x^4 (since there are four 'd's)So, the generating function is:G(x) = (1 + x)(1 + x + x^2)(1 + x + x^2 + x^3)(1 + x + x^2 + x^3 + x^4)We need the coefficient of x^4 in this product.Let me compute this step by step.First, multiply (1 + x) and (1 + x + x^2):(1 + x)(1 + x + x^2) = 1*(1 + x + x^2) + x*(1 + x + x^2) = 1 + x + x^2 + x + x^2 + x^3 = 1 + 2x + 2x^2 + x^3Now, multiply this by (1 + x + x^2 + x^3):Let me denote the previous result as A(x) = 1 + 2x + 2x^2 + x^3So, A(x)*(1 + x + x^2 + x^3) = (1 + 2x + 2x^2 + x^3)*(1 + x + x^2 + x^3)Let's compute this:1*(1 + x + x^2 + x^3) = 1 + x + x^2 + x^32x*(1 + x + x^2 + x^3) = 2x + 2x^2 + 2x^3 + 2x^42x^2*(1 + x + x^2 + x^3) = 2x^2 + 2x^3 + 2x^4 + 2x^5x^3*(1 + x + x^2 + x^3) = x^3 + x^4 + x^5 + x^6Now, add them all together:1 + x + x^2 + x^3+ 2x + 2x^2 + 2x^3 + 2x^4+ 2x^2 + 2x^3 + 2x^4 + 2x^5+ x^3 + x^4 + x^5 + x^6Now, combine like terms:- Constant term: 1- x: 1x + 2x = 3x- x^2: 1x^2 + 2x^2 + 2x^2 = 5x^2- x^3: 1x^3 + 2x^3 + 2x^3 + 1x^3 = 6x^3- x^4: 2x^4 + 2x^4 + 1x^4 = 5x^4- x^5: 2x^5 + 1x^5 = 3x^5- x^6: 1x^6So, the result is: 1 + 3x + 5x^2 + 6x^3 + 5x^4 + 3x^5 + x^6Now, we need to multiply this by the last generating function for 'd': (1 + x + x^2 + x^3 + x^4)Let me denote the previous result as B(x) = 1 + 3x + 5x^2 + 6x^3 + 5x^4 + 3x^5 + x^6So, B(x)*(1 + x + x^2 + x^3 + x^4) = ?We need the coefficient of x^4 in this product.To find the coefficient of x^4, we can consider all pairs of exponents from B(x) and the 'd' generating function that add up to 4.So, let's list the coefficients of B(x):- x^0: 1- x^1: 3- x^2: 5- x^3: 6- x^4: 5- x^5: 3- x^6: 1And the 'd' generating function has coefficients:- x^0: 1- x^1: 1- x^2: 1- x^3: 1- x^4: 1Now, to get x^4, we need to sum the products of coefficients where the exponents add up to 4.So:- From B(x)^0 * 'd'^4: 1 * 1 = 1- From B(x)^1 * 'd'^3: 3 * 1 = 3- From B(x)^2 * 'd'^2: 5 * 1 = 5- From B(x)^3 * 'd'^1: 6 * 1 = 6- From B(x)^4 * 'd'^0: 5 * 1 = 5Adding these up: 1 + 3 + 5 + 6 + 5 = 20So, the coefficient of x^4 is 20, which matches our earlier case analysis.Therefore, the number of ways to choose 4 letters is 20.Wait, but earlier I thought the answer was 175, but that was for a different problem. This one is 20.Okay, that makes sense. So, the first problem's answer is 20.Now, moving on to the second problem.**Problem 2:** How many different four-digit numbers can be formed by choosing 4 digits from {1, 2, 2, 3, 3, 3, 4, 4, 4, 4}?Alright, so similar to the first problem, but now we're forming four-digit numbers, which means the order matters. So, this is a permutation problem with multiset.We have the digits: one '1', two '2's, three '3's, and four '4's.We need to form four-digit numbers, so we're arranging 4 digits where some digits can be repeated up to their available counts.Again, I can approach this by considering different cases based on the number of repeated digits.Possible cases:1. All four digits are the same.2. Three digits are the same, and one is different.3. Two digits are the same, and the other two are different.4. Two digits are the same, and another two are the same (i.e., two pairs).5. All four digits are different.Let's analyze each case.**Case 1: All four digits are the same.**From the multiset, only '4' appears four times. So, the only possibility is four '4's. That's 1 number: 4444.**Case 2: Three digits are the same, and one is different.**We need a digit that appears at least three times, and another digit that appears at least once.From the multiset, '3' appears three times, and '4' appears four times. So, we can have three '3's and one other digit, or three '4's and one other digit.Let's break it down:- Three '3's and one other digit: The other digit can be '1', '2', or '4'. That's 3 possibilities.- Three '4's and one other digit: The other digit can be '1', '2', or '3'. That's another 3 possibilities.So, total for Case 2: 3 + 3 = 6 combinations.But since we're forming numbers, the order matters. For each combination, we need to count the number of distinct permutations.For each case:- Three '3's and one other digit: The number of distinct permutations is 4 (positions for the different digit). So, for each of the 3 possibilities, we have 4 permutations: 3 * 4 = 12.- Similarly, three '4's and one other digit: 3 possibilities, each with 4 permutations: 3 * 4 = 12.So, total for Case 2: 12 + 12 = 24 numbers.Wait, but hold on. When we have three '3's and one '4', is that different from three '4's and one '3'? Yes, because the digits are different, so they form different numbers.So, yes, 24 is correct.**Case 3: Two digits are the same, and the other two are different.**Here, we need to choose one digit to repeat twice, and then two different digits from the remaining.First, choose the digit to be repeated twice. The possible digits are '2', '3', and '4' since they have at least two copies.- If we choose '2' as the repeated digit: Then we need to choose two different digits from the remaining: '1', '3', '3', '4', '4', '4', '4'. The distinct digits are '1', '3', and '4'. So, the number of ways to choose two different digits is C(3,2) = 3.- If we choose '3' as the repeated digit: Similarly, we need to choose two different digits from '1', '2', '2', '4', '4', '4', '4'. The distinct digits are '1', '2', and '4'. So, C(3,2) = 3.- If we choose '4' as the repeated digit: We need to choose two different digits from '1', '2', '2', '3', '3', '3'. The distinct digits are '1', '2', and '3'. So, C(3,2) = 3.So, for each choice of the repeated digit, we have 3 ways to choose the other two digits, and there are 3 choices for the repeated digit. So, total combinations: 3 * 3 = 9.But again, since we're forming numbers, we need to consider permutations.For each combination, we have two identical digits and two distinct digits. The number of distinct permutations is 4! / 2! = 12.So, for each of the 9 combinations, we have 12 permutations: 9 * 12 = 108 numbers.Wait, is that correct? Let me think.Actually, the formula for permutations of multiset is n! / (k1! * k2! * ... * km!), where ki is the number of times each element is repeated.In this case, we have two identical digits and two distinct digits, so the number of permutations is 4! / 2! = 24 / 2 = 12. Yes, that's correct.So, 9 combinations * 12 permutations each = 108.**Case 4: Two digits are the same, and another two are the same (i.e., two pairs).**Here, we need to choose two different digits, each appearing at least twice, and take two copies of each.From the multiset, the digits that appear at least twice are '2', '3', and '4'.We need to choose two distinct digits from these three. The number of ways is C(3,2) = 3.For each pair of digits, we have two pairs, so the number of distinct permutations is 4! / (2! * 2!) = 6.So, for each of the 3 combinations, we have 6 permutations: 3 * 6 = 18 numbers.**Case 5: All four digits are different.**We need to choose four distinct digits from the available digits: '1', '2', '3', '4'. Since there are exactly four distinct digits, there's only one set of digits: '1', '2', '3', '4'.But since we're forming numbers, the number of permutations is 4! = 24.So, total for Case 5: 24 numbers.Now, let's sum up all the cases:- Case 1: 1- Case 2: 24- Case 3: 108- Case 4: 18- Case 5: 24Total number of four-digit numbers: 1 + 24 + 108 + 18 + 24 = 175.Wait, that seems high, but considering the permutations, it might be correct.Let me verify using generating functions again.The generating function for each digit is:- For '1': 1 + x (since there's only one '1')- For '2': 1 + x + x^2 (since there are two '2's)- For '3': 1 + x + x^2 + x^3 (since there are three '3's)- For '4': 1 + x + x^2 + x^3 + x^4 (since there are four '4's)But since we're forming four-digit numbers, we need to consider permutations, not combinations. So, the generating function approach for combinations won't directly give us the number of permutations.Alternatively, we can think of it as arranging the digits with possible repetitions, considering the multiplicities.But I think the case analysis is more straightforward here, and it gave us 175, which seems reasonable.So, the second problem's answer is 175.**Final Answer**1. boxed{20}2. boxed{175}

💡Alright, so I have this problem where I need to find all two-digit prime numbers ( p ) such that there exists a prime number ( q ) where ( 100q + p ) is a perfect square. Then, I have to sum all those ( p ) values. Hmm, okay, let me try to break this down step by step.First, let me understand what the problem is asking. I need to find primes ( p ) (which are two-digit numbers) and primes ( q ) such that when I multiply ( q ) by 100 and add ( p ), the result is a perfect square. So, mathematically, this can be written as:[100q + p = k^2]where ( k ) is some integer. My goal is to find all such ( p ) and then sum them up.Okay, so let's think about this. Since ( 100q + p ) is a perfect square, ( k^2 ) must be a number that ends with the digits of ( p ). Because ( p ) is a two-digit number, ( k^2 ) must end with those two digits. So, the last two digits of ( k^2 ) must be equal to ( p ).This means that ( p ) must be a two-digit prime number that is also the last two digits of some perfect square. So, my first task is to figure out which two-digit primes can be the last two digits of a perfect square.I remember that the last two digits of a square number repeat in cycles, so maybe I can list all possible two-digit endings of square numbers and see which of those are primes.Let me recall that the possible last two digits of square numbers are limited. For example, numbers ending with 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, or 96 can be squares. Wait, is that right? Let me verify.Actually, the last two digits of square numbers can only be certain combinations. For example, squares modulo 100 repeat every 100 numbers, so I can list all possible squares modulo 100.Alternatively, I can think about the possible residues of squares modulo 100. Since 100 is 4*25, and using the Chinese Remainder Theorem, I can consider squares modulo 4 and modulo 25.But maybe that's getting too complicated. Let me instead list all possible squares modulo 100 by squaring numbers from 0 to 99 and noting the last two digits.Wait, that might take too long, but perhaps I can find a pattern or a list of possible last two digits of squares.I recall that squares modulo 100 can only end with certain two-digit combinations. For example, numbers ending with 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, or 96 are possible. Let me confirm this.Wait, actually, I think the possible last two digits of squares are more limited. Let me check:- 0^2 = 00- 1^2 = 01- 2^2 = 04- 3^2 = 09- 4^2 = 16- 5^2 = 25- 6^2 = 36- 7^2 = 49- 8^2 = 64- 9^2 = 81- 10^2 = 100 (ends with 00)- 11^2 = 121 (ends with 21)- 12^2 = 144 (ends with 44)- 13^2 = 169 (ends with 69)- 14^2 = 196 (ends with 96)- 15^2 = 225 (ends with 25)- 16^2 = 256 (ends with 56)- 17^2 = 289 (ends with 89)- 18^2 = 324 (ends with 24)- 19^2 = 361 (ends with 61)- 20^2 = 400 (ends with 00)- 21^2 = 441 (ends with 41)- 22^2 = 484 (ends with 84)- 23^2 = 529 (ends with 29)- 24^2 = 576 (ends with 76)- 25^2 = 625 (ends with 25)- 26^2 = 676 (ends with 76)- 27^2 = 729 (ends with 29)- 28^2 = 784 (ends with 84)- 29^2 = 841 (ends with 41)- 30^2 = 900 (ends with 00)- 31^2 = 961 (ends with 61)- 32^2 = 1024 (ends with 24)- 33^2 = 1089 (ends with 89)- 34^2 = 1156 (ends with 56)- 35^2 = 1225 (ends with 25)- 36^2 = 1296 (ends with 96)- 37^2 = 1369 (ends with 69)- 38^2 = 1444 (ends with 44)- 39^2 = 1521 (ends with 21)- 40^2 = 1600 (ends with 00)- 41^2 = 1681 (ends with 81)- 42^2 = 1764 (ends with 64)- 43^2 = 1849 (ends with 49)- 44^2 = 1936 (ends with 36)- 45^2 = 2025 (ends with 25)- 46^2 = 2116 (ends with 16)- 47^2 = 2209 (ends with 09)- 48^2 = 2304 (ends with 04)- 49^2 = 2401 (ends with 01)- 50^2 = 2500 (ends with 00)Hmm, I notice that after 50, the last two digits start repeating in reverse because of the way squaring works. For example, 51^2 = 2601 (ends with 01), which is the same as 49^2. Similarly, 52^2 = 2704 (ends with 04), same as 48^2, and so on.So, compiling all the unique last two digits from 0^2 to 50^2, we have:00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, 96.Now, among these, which are two-digit primes? Let me list the two-digit primes first:11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.Now, let's see which of these primes are in the list of possible last two digits of squares:- 11: Not in the list.- 13: Not in the list.- 17: Not in the list.- 19: Yes, 19 is in the list (from 14^2 = 196).- 23: Not in the list.- 29: Yes, 29 is in the list (from 23^2 = 529 and 27^2 = 729).- 31: Not in the list.- 37: Not in the list.- 41: Yes, 41 is in the list (from 21^2 = 441 and 29^2 = 841).- 43: Not in the list.- 47: Not in the list.- 53: Not in the list.- 59: Not in the list.- 61: Yes, 61 is in the list (from 19^2 = 361 and 31^2 = 961).- 67: Not in the list.- 71: Not in the list.- 73: Not in the list.- 79: Not in the list.- 83: Not in the list.- 89: Yes, 89 is in the list (from 17^2 = 289 and 33^2 = 1089).- 97: Not in the list.So, the two-digit primes that can be the last two digits of a square are: 19, 29, 41, 61, 89.Wait, but earlier I thought 19 was in the list because 14^2 ends with 96, but 14^2 is 196, which ends with 96, not 19. Wait, that's a mistake. Let me correct that.Looking back, 19 is not in the list of last two digits of squares. Let me check again:From the list of squares modulo 100, I have 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, 96.So, 19 is not among these. That was a mistake earlier. So, the correct list of two-digit primes that are last two digits of squares is: 29, 41, 61, 89.Wait, let me double-check:- 29: Yes, 23^2 = 529 ends with 29.- 41: Yes, 21^2 = 441 ends with 41.- 61: Yes, 19^2 = 361 ends with 61.- 89: Yes, 17^2 = 289 ends with 89.So, 19 is not in the list because no square ends with 19. My mistake earlier was thinking that 14^2 ends with 19, but it actually ends with 96.So, the correct list is: 29, 41, 61, 89.Now, these are the possible candidates for ( p ). But I need to ensure that for each ( p ), there exists a prime ( q ) such that ( 100q + p ) is a perfect square.So, for each ( p ) in {29, 41, 61, 89}, I need to find a prime ( q ) such that ( 100q + p = k^2 ) for some integer ( k ).Let me handle each ( p ) one by one.1. ( p = 29 ):We have ( 100q + 29 = k^2 ).So, ( k^2 equiv 29 mod 100 ).From earlier, we know that squares ending with 29 occur when ( k equiv 23 ) or ( k equiv 27 mod 50 ) because 23^2 = 529 and 27^2 = 729 both end with 29.So, let's express ( k ) as ( 50m + 23 ) or ( 50m + 27 ) for some integer ( m ).Let's take ( k = 50m + 23 ):Then, ( k^2 = (50m + 23)^2 = 2500m^2 + 2300m + 529 ).So, ( 100q + 29 = 2500m^2 + 2300m + 529 ).Subtracting 29 from both sides:( 100q = 2500m^2 + 2300m + 500 ).Divide both sides by 100:( q = 25m^2 + 23m + 5 ).Now, ( q ) must be a prime number. Let's try small values of ( m ):- ( m = 0 ): ( q = 0 + 0 + 5 = 5 ). 5 is prime. So, ( q = 5 ). Then, ( k = 50*0 + 23 = 23 ). Check: ( 100*5 + 29 = 529 = 23^2 ). Correct.- ( m = 1 ): ( q = 25 + 23 + 5 = 53 ). 53 is prime. ( k = 50*1 + 23 = 73 ). Check: ( 100*53 + 29 = 5329 = 73^2 ). Correct.- ( m = 2 ): ( q = 25*4 + 23*2 + 5 = 100 + 46 + 5 = 151 ). 151 is prime. ( k = 50*2 + 23 = 123 ). Check: ( 100*151 + 29 = 15129 = 123^2 ). Correct.- ( m = 3 ): ( q = 25*9 + 23*3 + 5 = 225 + 69 + 5 = 299 ). 299 is not prime (divisible by 13: 13*23=299). So, discard.- ( m = 4 ): ( q = 25*16 + 23*4 + 5 = 400 + 92 + 5 = 497 ). 497 is not prime (divisible by 7: 7*71=497). Discard.- ( m = 5 ): ( q = 25*25 + 23*5 + 5 = 625 + 115 + 5 = 745 ). 745 is not prime (divisible by 5: 5*149=745). Discard.So, for ( p = 29 ), we have valid primes ( q = 5, 53, 151 ), etc. So, ( p = 29 ) is valid.2. ( p = 41 ):We have ( 100q + 41 = k^2 ).So, ( k^2 equiv 41 mod 100 ).From earlier, squares ending with 41 occur when ( k equiv 21 ) or ( k equiv 29 mod 50 ) because 21^2 = 441 and 29^2 = 841 both end with 41.Express ( k ) as ( 50m + 21 ) or ( 50m + 29 ).First, ( k = 50m + 21 ):( k^2 = (50m + 21)^2 = 2500m^2 + 2100m + 441 ).So, ( 100q + 41 = 2500m^2 + 2100m + 441 ).Subtract 41:( 100q = 2500m^2 + 2100m + 400 ).Divide by 100:( q = 25m^2 + 21m + 4 ).Check if ( q ) is prime:- ( m = 0 ): ( q = 0 + 0 + 4 = 4 ). Not prime.- ( m = 1 ): ( q = 25 + 21 + 4 = 50 ). Not prime.- ( m = 2 ): ( q = 100 + 42 + 4 = 146 ). Not prime.- ( m = 3 ): ( q = 225 + 63 + 4 = 292 ). Not prime.- ( m = 4 ): ( q = 400 + 84 + 4 = 488 ). Not prime.- ( m = 5 ): ( q = 625 + 105 + 4 = 734 ). Not prime.Hmm, not promising. Let's try the other form, ( k = 50m + 29 ):( k^2 = (50m + 29)^2 = 2500m^2 + 2900m + 841 ).So, ( 100q + 41 = 2500m^2 + 2900m + 841 ).Subtract 41:( 100q = 2500m^2 + 2900m + 800 ).Divide by 100:( q = 25m^2 + 29m + 8 ).Check for primes:- ( m = 0 ): ( q = 0 + 0 + 8 = 8 ). Not prime.- ( m = 1 ): ( q = 25 + 29 + 8 = 62 ). Not prime.- ( m = 2 ): ( q = 100 + 58 + 8 = 166 ). Not prime.- ( m = 3 ): ( q = 225 + 87 + 8 = 320 ). Not prime.- ( m = 4 ): ( q = 400 + 116 + 8 = 524 ). Not prime.- ( m = 5 ): ( q = 625 + 145 + 8 = 778 ). Not prime.Hmm, again, no primes. Maybe I need to check higher ( m ), but it's getting large. Alternatively, perhaps ( p = 41 ) doesn't have a corresponding prime ( q ). Let me check for small ( k ):Let's see, ( k^2 = 100q + 41 ). So, ( k^2 ) must be greater than 41, so ( k geq 7 ).Check ( k = 21 ): ( 21^2 = 441 ). So, ( 100q + 41 = 441 ) → ( 100q = 400 ) → ( q = 4 ). Not prime.( k = 29 ): ( 29^2 = 841 ). ( 100q + 41 = 841 ) → ( 100q = 800 ) → ( q = 8 ). Not prime.( k = 71 ): ( 71^2 = 5041 ). ( 100q + 41 = 5041 ) → ( 100q = 5000 ) → ( q = 50 ). Not prime.( k = 79 ): ( 79^2 = 6241 ). ( 100q + 41 = 6241 ) → ( 100q = 6200 ) → ( q = 62 ). Not prime.( k = 121 ): ( 121^2 = 14641 ). ( 100q + 41 = 14641 ) → ( 100q = 14600 ) → ( q = 146 ). Not prime.Hmm, seems like for ( p = 41 ), ( q ) is always even and greater than 2, hence not prime. So, maybe ( p = 41 ) doesn't work. Let me check if there's a ( k ) where ( q ) is prime.Wait, maybe I missed something. Let me try ( k = 101 ): ( 101^2 = 10201 ). ( 100q + 41 = 10201 ) → ( 100q = 10160 ) → ( q = 101.6 ). Not integer.( k = 109 ): ( 109^2 = 11881 ). ( 100q + 41 = 11881 ) → ( 100q = 11840 ) → ( q = 118.4 ). Not integer.( k = 111 ): ( 111^2 = 12321 ). ( 100q + 41 = 12321 ) → ( 100q = 12280 ) → ( q = 122.8 ). Not integer.Hmm, seems like ( q ) is always even or non-integer. So, perhaps ( p = 41 ) doesn't have a corresponding prime ( q ). Therefore, ( p = 41 ) might not be valid.3. ( p = 61 ):We have ( 100q + 61 = k^2 ).So, ( k^2 equiv 61 mod 100 ).From earlier, squares ending with 61 occur when ( k equiv 19 ) or ( k equiv 31 mod 50 ) because 19^2 = 361 and 31^2 = 961 both end with 61.Express ( k ) as ( 50m + 19 ) or ( 50m + 31 ).First, ( k = 50m + 19 ):( k^2 = (50m + 19)^2 = 2500m^2 + 1900m + 361 ).So, ( 100q + 61 = 2500m^2 + 1900m + 361 ).Subtract 61:( 100q = 2500m^2 + 1900m + 300 ).Divide by 100:( q = 25m^2 + 19m + 3 ).Check if ( q ) is prime:- ( m = 0 ): ( q = 0 + 0 + 3 = 3 ). Prime. ( k = 50*0 + 19 = 19 ). Check: ( 100*3 + 61 = 361 = 19^2 ). Correct.- ( m = 1 ): ( q = 25 + 19 + 3 = 47 ). Prime. ( k = 50*1 + 19 = 69 ). Check: ( 100*47 + 61 = 4761 = 69^2 ). Correct.- ( m = 2 ): ( q = 100 + 38 + 3 = 141 ). Not prime (divisible by 3: 141/3=47). Discard.- ( m = 3 ): ( q = 225 + 57 + 3 = 285 ). Not prime. Discard.- ( m = 4 ): ( q = 400 + 76 + 3 = 479 ). 479 is prime. ( k = 50*4 + 19 = 219 ). Check: ( 100*479 + 61 = 47961 = 219^2 ). Correct.- ( m = 5 ): ( q = 625 + 95 + 3 = 723 ). Not prime. Discard.So, for ( p = 61 ), we have valid primes ( q = 3, 47, 479 ), etc. So, ( p = 61 ) is valid.Now, let's try the other form, ( k = 50m + 31 ):( k^2 = (50m + 31)^2 = 2500m^2 + 3100m + 961 ).So, ( 100q + 61 = 2500m^2 + 3100m + 961 ).Subtract 61:( 100q = 2500m^2 + 3100m + 900 ).Divide by 100:( q = 25m^2 + 31m + 9 ).Check for primes:- ( m = 0 ): ( q = 0 + 0 + 9 = 9 ). Not prime.- ( m = 1 ): ( q = 25 + 31 + 9 = 65 ). Not prime.- ( m = 2 ): ( q = 100 + 62 + 9 = 171 ). Not prime.- ( m = 3 ): ( q = 225 + 93 + 9 = 327 ). Not prime.- ( m = 4 ): ( q = 400 + 124 + 9 = 533 ). Not prime.- ( m = 5 ): ( q = 625 + 155 + 9 = 789 ). Not prime.So, no primes here. But since we already found valid ( q ) for ( k = 50m + 19 ), ( p = 61 ) is valid.4. ( p = 89 ):We have ( 100q + 89 = k^2 ).So, ( k^2 equiv 89 mod 100 ).From earlier, squares ending with 89 occur when ( k equiv 17 ) or ( k equiv 33 mod 50 ) because 17^2 = 289 and 33^2 = 1089 both end with 89.Express ( k ) as ( 50m + 17 ) or ( 50m + 33 ).First, ( k = 50m + 17 ):( k^2 = (50m + 17)^2 = 2500m^2 + 1700m + 289 ).So, ( 100q + 89 = 2500m^2 + 1700m + 289 ).Subtract 89:( 100q = 2500m^2 + 1700m + 200 ).Divide by 100:( q = 25m^2 + 17m + 2 ).Check if ( q ) is prime:- ( m = 0 ): ( q = 0 + 0 + 2 = 2 ). Prime. ( k = 50*0 + 17 = 17 ). Check: ( 100*2 + 89 = 289 = 17^2 ). Correct.- ( m = 1 ): ( q = 25 + 17 + 2 = 44 ). Not prime.- ( m = 2 ): ( q = 100 + 34 + 2 = 136 ). Not prime.- ( m = 3 ): ( q = 225 + 51 + 2 = 278 ). Not prime.- ( m = 4 ): ( q = 400 + 68 + 2 = 470 ). Not prime.- ( m = 5 ): ( q = 625 + 85 + 2 = 712 ). Not prime.Now, let's try the other form, ( k = 50m + 33 ):( k^2 = (50m + 33)^2 = 2500m^2 + 3300m + 1089 ).So, ( 100q + 89 = 2500m^2 + 3300m + 1089 ).Subtract 89:( 100q = 2500m^2 + 3300m + 1000 ).Divide by 100:( q = 25m^2 + 33m + 10 ).Check for primes:- ( m = 0 ): ( q = 0 + 0 + 10 = 10 ). Not prime.- ( m = 1 ): ( q = 25 + 33 + 10 = 68 ). Not prime.- ( m = 2 ): ( q = 100 + 66 + 10 = 176 ). Not prime.- ( m = 3 ): ( q = 225 + 99 + 10 = 334 ). Not prime.- ( m = 4 ): ( q = 400 + 132 + 10 = 542 ). Not prime.- ( m = 5 ): ( q = 625 + 165 + 10 = 800 ). Not prime.So, only ( m = 0 ) gives a prime ( q = 2 ). Therefore, ( p = 89 ) is valid.Wait, but let me check higher ( m ) for ( k = 50m + 17 ):- ( m = 6 ): ( q = 25*36 + 17*6 + 2 = 900 + 102 + 2 = 1004 ). Not prime.- ( m = 7 ): ( q = 25*49 + 17*7 + 2 = 1225 + 119 + 2 = 1346 ). Not prime.- ( m = 8 ): ( q = 25*64 + 17*8 + 2 = 1600 + 136 + 2 = 1738 ). Not prime.So, only ( m = 0 ) gives a prime ( q = 2 ). Therefore, ( p = 89 ) is valid.Wait, but let me check ( k = 17 ): ( 17^2 = 289 ). So, ( 100q + 89 = 289 ) → ( 100q = 200 ) → ( q = 2 ). Correct, and 2 is prime.Similarly, ( k = 33 ): ( 33^2 = 1089 ). So, ( 100q + 89 = 1089 ) → ( 100q = 1000 ) → ( q = 10 ). Not prime.So, only ( k = 17 ) gives a prime ( q = 2 ). Therefore, ( p = 89 ) is valid.Now, summarizing:- ( p = 29 ): Valid, with ( q = 5, 53, 151 ), etc.- ( p = 41 ): Seems invalid as ( q ) is not prime.- ( p = 61 ): Valid, with ( q = 3, 47, 479 ), etc.- ( p = 89 ): Valid, with ( q = 2 ).Wait, earlier I thought ( p = 41 ) might not be valid because ( q ) was always even or non-integer. Let me double-check ( p = 41 ):We have ( 100q + 41 = k^2 ). So, ( k^2 ) must be 41 more than a multiple of 100. Let's see if there's a ( k ) such that ( k^2 - 41 ) is divisible by 100 and ( q = (k^2 - 41)/100 ) is prime.Let me try ( k = 21 ): ( 21^2 = 441 ). ( 441 - 41 = 400 ). ( q = 4 ). Not prime.( k = 29 ): ( 29^2 = 841 ). ( 841 - 41 = 800 ). ( q = 8 ). Not prime.( k = 71 ): ( 71^2 = 5041 ). ( 5041 - 41 = 5000 ). ( q = 50 ). Not prime.( k = 79 ): ( 79^2 = 6241 ). ( 6241 - 41 = 6200 ). ( q = 62 ). Not prime.( k = 121 ): ( 121^2 = 14641 ). ( 14641 - 41 = 14600 ). ( q = 146 ). Not prime.( k = 129 ): ( 129^2 = 16641 ). ( 16641 - 41 = 16600 ). ( q = 166 ). Not prime.( k = 171 ): ( 171^2 = 29241 ). ( 29241 - 41 = 29200 ). ( q = 292 ). Not prime.( k = 179 ): ( 179^2 = 32041 ). ( 32041 - 41 = 32000 ). ( q = 320 ). Not prime.Hmm, seems like ( q ) is always even and greater than 2, hence not prime. Therefore, ( p = 41 ) does not have a corresponding prime ( q ). So, ( p = 41 ) is invalid.So, the valid ( p ) values are 29, 61, and 89.Now, let's sum them up:29 + 61 + 89 = 179.Wait, but earlier I thought ( p = 89 ) is valid because ( q = 2 ) is prime. Yes, that's correct.So, the sum is 29 + 61 + 89 = 179.But wait, let me double-check if there are any other ( p ) values I might have missed. Earlier, I thought ( p = 19 ) was a candidate, but realized it's not because no square ends with 19. So, only 29, 61, and 89 are valid.Therefore, the final answer is 179.

💡Okay, so I have this sequence problem here. It says that in the sequence {a_n}, a₁ is equal to 1, and each subsequent term is defined by a_{n+1} = a_n + 2^{n-1}. I need to find an expression for a_n. Hmm, let me try to figure this out step by step.First, let me write down what I know. The first term, a₁, is 1. Then, each next term is the previous term plus 2 raised to the power of (n-1). So, for example, a₂ would be a₁ + 2^{1-1} which is 1 + 2⁰ = 1 + 1 = 2. Similarly, a₃ would be a₂ + 2^{2-1} = 2 + 2¹ = 2 + 2 = 4. Let me compute a few more terms to see if I can spot a pattern.a₁ = 1 a₂ = a₁ + 2^{1-1} = 1 + 1 = 2 a₃ = a₂ + 2^{2-1} = 2 + 2 = 4 a₄ = a₃ + 2^{3-1} = 4 + 4 = 8 a₅ = a₄ + 2^{4-1} = 8 + 8 = 16 a₆ = a₅ + 2^{5-1} = 16 + 16 = 32 Wait a minute, these numbers look familiar. 1, 2, 4, 8, 16, 32... These are powers of 2. Specifically, each term seems to be 2 raised to the power of (n-1). Let me check:For n=1: 2^{1-1} = 2⁰ = 1, which matches a₁. For n=2: 2^{2-1} = 2¹ = 2, which matches a₂. For n=3: 2^{3-1} = 2² = 4, which matches a₃. And so on. So it seems like a_n = 2^{n-1} for all n.But wait, I should verify this more formally, not just by computing the first few terms. Maybe I can use mathematical induction to prove that a_n = 2^{n-1}.Let's start with the base case. For n=1, a₁ = 1, and 2^{1-1} = 1. So the base case holds.Now, assume that for some integer k ≥ 1, a_k = 2^{k-1}. This is our induction hypothesis. We need to show that a_{k+1} = 2^{(k+1)-1} = 2^k.From the given recurrence relation, a_{k+1} = a_k + 2^{k-1}. By the induction hypothesis, a_k = 2^{k-1}, so substituting that in, we get:a_{k+1} = 2^{k-1} + 2^{k-1} = 2 * 2^{k-1} = 2^k.Which is exactly what we wanted to show. Therefore, by induction, the formula a_n = 2^{n-1} holds for all positive integers n.Alternatively, I can approach this problem by recognizing that the given recurrence relation is a linear recurrence. The general solution can be found by summing the terms.Given a_{n+1} = a_n + 2^{n-1}, we can write this as a_{n+1} - a_n = 2^{n-1}. This is a first-order linear recurrence, and the solution can be found by summing both sides from n=1 to n=k.Let's write out the terms:a₂ - a₁ = 2^{1-1} = 1 a₃ - a₂ = 2^{2-1} = 2 a₄ - a₃ = 2^{3-1} = 4 ... a_{k} - a_{k-1} = 2^{(k-1)-1} = 2^{k-2}If we add all these equations together, the left side telescopes:(a₂ - a₁) + (a₃ - a₂) + (a₄ - a₃) + ... + (a_{k} - a_{k-1}) = a_{k} - a₁The right side is the sum of a geometric series:1 + 2 + 4 + ... + 2^{k-2}The sum of a geometric series with first term 1 and common ratio 2 for (k-1) terms is 2^{k-1} - 1. Therefore:a_{k} - a₁ = 2^{k-1} - 1 Since a₁ = 1, we have:a_{k} = 2^{k-1} - 1 + 1 = 2^{k-1}So again, we arrive at a_n = 2^{n-1}.Just to make sure I haven't missed anything, let me consider another approach. Maybe using generating functions or characteristic equations, but since this is a simple linear recurrence, the telescoping method seems sufficient and straightforward.Alternatively, thinking about the sequence, each term is double the previous term minus something, but in this case, it's actually adding a term that's increasing exponentially. Wait, no, actually, each term is the previous term plus 2^{n-1}, which itself is growing exponentially. So the sequence is growing exponentially as well, which aligns with the solution we found.Let me test the formula with n=5. According to the formula, a₅ should be 2^{5-1} = 16. Earlier, I computed a₅ as 16, which matches. Similarly, a₆ should be 32, which also matches my earlier computation. So, this seems consistent.Another way to think about it is that each term adds a power of 2, so starting from 1, each addition is the next power of 2, effectively building up the binary representation of the number. So, 1 is 1, adding 1 gives 2, adding 2 gives 4, adding 4 gives 8, and so on. Each time, we're doubling the previous term, which is another way of seeing that a_n = 2^{n-1}.Wait, is that accurate? Let me see. If a_{n+1} = a_n + 2^{n-1}, and a_n = 2^{n-1}, then a_{n+1} = 2^{n-1} + 2^{n-1} = 2*2^{n-1} = 2^n, which is indeed 2^{(n+1)-1}. So yes, each term is double the previous term, which is another way of confirming the formula.So, after considering multiple approaches—computing initial terms, mathematical induction, telescoping sums, and even a binary perspective—it seems clear that the expression for a_n is 2^{n-1}.I think I've covered all the bases here. I can't think of any edge cases or exceptions because the recurrence starts at n=1 and the formula holds for all subsequent terms. Therefore, I'm confident that the expression for a_n is 2^{n-1}.**Final Answer**The expression for ( a_n ) is boxed{2^{n-1}}.

💡First, let's denote the number of cookies Kenny has as ( K ).Chris has half as many cookies as Kenny, so Chris has ( frac{K}{2} ) cookies.Glenn has 24 cookies.The total number of cookies is 33, so we can set up the equation:[frac{K}{2} + K + 24 = 33]Combining like terms:[frac{3K}{2} + 24 = 33]Subtracting 24 from both sides:[frac{3K}{2} = 9]Multiplying both sides by 2:[3K = 18]Dividing both sides by 3:[K = 6]Kenny has 6 cookies. Therefore, the ratio of Glenn's cookies to Kenny's cookies is:[frac{24}{6} = 4]So, the ratio is ( 4:1 ).