💡Okay, so I have this problem with two parts. Let me try to work through each step carefully. First, the universal set is all real numbers, U = ℝ. Set A is defined as {x | x² - 3x - 18 ≥ 0}, and set B is {x | (x + 5)/(x - 14) ≤ 0}. Starting with part (1): I need to find the complement of B with respect to U, which is everything not in B, and then intersect that with set A. Let me first figure out what set A is. The inequality is x² - 3x - 18 ≥ 0. To solve this quadratic inequality, I should factor it if possible. Let me see: x² - 3x - 18. Hmm, factors of -18 that add up to -3... Maybe -6 and 3? Let's check: (x - 6)(x + 3) = x² - 3x - 18. Yes, that works. So the inequality becomes (x - 6)(x + 3) ≥ 0. To solve this, I can find the critical points where each factor is zero: x = 6 and x = -3. These divide the real line into three intervals: (-∞, -3), (-3, 6), and (6, ∞). I'll test each interval:1. For x < -3, say x = -4: (-4 - 6)(-4 + 3) = (-10)(-1) = 10 ≥ 0. So this interval is part of the solution.2. For -3 < x < 6, say x = 0: (0 - 6)(0 + 3) = (-6)(3) = -18 < 0. Not part of the solution.3. For x > 6, say x = 7: (7 - 6)(7 + 3) = (1)(10) = 10 ≥ 0. This interval is part of the solution.Also, since the inequality is ≥ 0, the endpoints x = -3 and x = 6 are included. So set A is (-∞, -3] ∪ [6, ∞).Next, set B is defined by (x + 5)/(x - 14) ≤ 0. To solve this rational inequality, I need to find where the expression is negative or zero. The critical points are where the numerator or denominator is zero: x = -5 and x = 14. These divide the real line into three intervals: (-∞, -5), (-5, 14), and (14, ∞).Testing each interval:1. For x < -5, say x = -6: (-6 + 5)/(-6 - 14) = (-1)/(-20) = 1/20 > 0. Not part of the solution.2. For -5 < x < 14, say x = 0: (0 + 5)/(0 - 14) = 5/(-14) = -5/14 < 0. This interval is part of the solution.3. For x > 14, say x = 15: (15 + 5)/(15 - 14) = 20/1 = 20 > 0. Not part of the solution.Also, check the critical points:- At x = -5: ( -5 + 5 ) / ( -5 - 14 ) = 0 / (-19) = 0, which satisfies the inequality, so x = -5 is included.- At x = 14: The denominator is zero, so the expression is undefined. So x = 14 is excluded.Therefore, set B is [-5, 14).Now, the complement of B with respect to U is everything not in B. Since B is [-5, 14), the complement is (-∞, -5) ∪ [14, ∞).Now, I need to find the intersection of this complement with set A. Set A is (-∞, -3] ∪ [6, ∞), and the complement of B is (-∞, -5) ∪ [14, ∞). So, let's find the overlap between these two sets:1. (-∞, -5) and (-∞, -3]: The overlap is (-∞, -5), since -5 is less than -3.2. [14, ∞) and [6, ∞): The overlap is [14, ∞), since 14 is greater than 6.Therefore, the intersection is (-∞, -5) ∪ [14, ∞).Moving on to part (2): Set C is {x | 2a < x < a + 1}, and it's given that B ∩ C = C. This means that C is a subset of B. So, for C to be a subset of B, every element of C must be in B. Since B is [-5, 14), C must lie entirely within [-5, 14).First, let's consider when C is non-empty. For C to be non-empty, the lower bound must be less than the upper bound: 2a < a + 1. Solving this: 2a < a + 1 ⇒ 2a - a < 1 ⇒ a < 1.If a < 1, then C is non-empty. Now, for C to be a subset of B, the entire interval (2a, a + 1) must lie within [-5, 14). So, we need:1. The lower bound of C, 2a, must be ≥ -5.2. The upper bound of C, a + 1, must be ≤ 14.So, we have two inequalities:1. 2a ≥ -5 ⇒ a ≥ -5/2.2. a + 1 ≤ 14 ⇒ a ≤ 13.But we also have the condition that a < 1 for C to be non-empty. So combining these:-5/2 ≤ a < 1.Additionally, we should check if a = 1 is allowed. If a = 1, then C becomes {x | 2 < x < 2}, which is an empty set. Since the empty set is a subset of any set, including B, a = 1 is also acceptable.So, considering both cases:- If C is non-empty (a < 1), then -5/2 ≤ a < 1.- If C is empty (a ≥ 1), it's still a subset of B.Therefore, combining both, the range of a is a ≥ -5/2.Wait, let me verify that. If a ≥ 1, C is empty, which is a subset of B. So, overall, a can be any real number greater than or equal to -5/2.Yes, that makes sense. So the range of a is a ≥ -5/2.**Final Answer**(1) boxed{(-infty, -5) cup [14, +infty)}(2) boxed{a geq -dfrac{5}{2}}

💡First, identify the dimensions of each rectangle. Each rectangle has a common base width of 2 units and lengths of 1, 8, and 27 units.Next, calculate the area of each rectangle by multiplying the width by the length:- For the first rectangle: (2 times 1 = 2)- For the second rectangle: (2 times 8 = 16)- For the third rectangle: (2 times 27 = 54)Then, sum up the areas of all three rectangles:[2 + 16 + 54 = 72]Finally, the sum of the areas of the three rectangles is 72.

💡First, I recognize that the total sum of the two parts is 66, and the larger part is given as 50. This means the smaller part must be 16.Next, I set up the equation based on the given condition: a certain fraction of the larger part exceeds 0.625 of the smaller part by 10. Let's denote this fraction as ( f ).Substituting the known values into the equation:[f times 50 = 0.625 times 16 + 10]Calculating ( 0.625 times 16 ) gives 10. Therefore, the equation simplifies to:[50f = 20]Solving for ( f ) by dividing both sides by 50:[f = frac{20}{50} = 0.4]Thus, the fraction of the larger part is 0.4.

💡Alright, so I've got this problem here where Morgane has colored the integers from 1 to n, each in exactly one color. The condition given is that for any two positive integers a and b where a < b and a + b ≤ n, at least two of the integers among a, b, and a + b are of the same color. I need to prove that there exists a color that has been used for at least 2n/5 integers.Hmm, okay, let's break this down. First, I need to understand what the condition is really saying. For any two numbers a and b where a is less than b and their sum is still within the range up to n, at least two of these three numbers—either a and b, a and a+b, or b and a+b—must share the same color. So, this is a kind of constraint on how the colors can be assigned.I think this is related to something called the pigeonhole principle, where if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. Maybe I can apply a similar idea here. If the colors are spread out too thinly, the condition might not hold, so there must be a color that's used a significant number of times.Let me try to formalize this. Suppose there are k different colors used. Let x_i be the number of integers colored with the i-th color. So, the sum of all x_i from i=1 to k is n. Now, if each color is used less than 2n/5 times, then each x_i ≤ 2n/5 - 1. That would mean that the number of colors k must be at least n divided by (2n/5 - 1). Let me compute that:k ≥ n / (2n/5 - 1) = n / ((2n - 5)/5) = 5n / (2n - 5)Simplifying this, for large n, this is roughly 5/2, but for smaller n, it might be larger. Wait, but n is a positive integer, so 2n - 5 must be positive, meaning n ≥ 3. For n=3, 2n -5=1, so k ≥ 15/1=15, which is way more than n=3, which doesn't make sense. Hmm, maybe my approach is flawed.Perhaps I need to think differently. Maybe instead of counting the number of colors, I should look at the pairs (a, b) and how they relate to the colors. For each pair a < b with a + b ≤ n, at least two of a, b, a + b share the same color.Let me consider the total number of such pairs. For a given a, the maximum b is such that a + b ≤ n, so b ≤ n - a. Since a < b, b must be at least a + 1. So for each a from 1 to floor((n-1)/2), the number of b's is n - a - a = n - 2a.But this might get complicated. Maybe I can use an averaging argument. If each color is used at most 2n/5 times, then the number of monochromatic pairs (pairs where both elements are the same color) is limited. But the condition requires that for each pair a < b with a + b ≤ n, at least two elements are the same color, which might imply a certain number of monochromatic pairs.Alternatively, maybe I can construct a graph where each number from 1 to n is a vertex, and edges connect a and b if a + b is also in the set. Then the condition implies that in this graph, every edge is part of a triangle where at least two vertices are the same color. But I'm not sure if this graph approach is helpful.Wait, another idea: if I consider the numbers in terms of their binary representations or something, but that might not be directly useful. Maybe I should think about the problem in terms of arithmetic progressions or something similar.Let me try to think about specific cases. Suppose n=5. Then the numbers are 1,2,3,4,5. Let's see what pairs (a,b) satisfy a < b and a + b ≤5.For a=1, b can be 2,3,4 (since 1+2=3, 1+3=4, 1+4=5).For a=2, b can be 3 (since 2+3=5).a=3, b would have to be 4, but 3+4=7 >5, so no.Similarly, a=4, b=5 would be 9>5, so no.So total pairs: (1,2), (1,3), (1,4), (2,3).Now, for each of these pairs, at least two of a, b, a+b must be the same color.Let's see what this implies.Take pair (1,2): 1,2,3 must have at least two same colors.Pair (1,3): 1,3,4 must have at least two same colors.Pair (1,4): 1,4,5 must have at least two same colors.Pair (2,3): 2,3,5 must have at least two same colors.So, for n=5, let's try to color the numbers such that this condition holds and see if any color is used at least 2n/5=2*5/5=2 times.Wait, 2n/5=2, so we need at least one color used twice. But n=5, so if we have 5 colors, each used once, but that would violate the condition because for each pair, we need at least two same colors. So, in reality, we can't have all colors unique.Let me try to color them with two colors, say red and blue.Suppose 1 is red. Then, for pair (1,2), either 1 and 2 are red, or 1 and 3 are red, or 2 and 3 are red.If 1 is red, let's say 2 is blue. Then, for pair (1,2), since 1 is red and 2 is blue, 3 must be red or blue. If 3 is red, then in pair (1,3), 1 and 3 are red, which is fine. Then, pair (1,4): 1 is red, so either 4 is red or 5 is red. If 4 is red, then pair (2,3): 2 is blue, 3 is red, so 5 must be blue or red. If 5 is blue, then pair (2,3): 2 and 5 are blue, which is fine. So, let's see the coloring:1: red2: blue3: red4: red5: blueNow, check all pairs:(1,2): 1(red), 2(blue), 3(red) → 1 and 3 are red, good.(1,3): 1(red), 3(red), 4(red) → all red, good.(1,4): 1(red), 4(red), 5(blue) → 1 and 4 are red, good.(2,3): 2(blue), 3(red), 5(blue) → 2 and 5 are blue, good.So, this coloring works. Now, how many times is each color used?Red: 1,3,4 → 3 times.Blue: 2,5 → 2 times.So, red is used 3 times, which is more than 2n/5=2. So, in this case, the conclusion holds.But this is just for n=5. I need to generalize this.Maybe I can use induction. Suppose the statement holds for n=k, then show it holds for n=k+1. But I'm not sure if that's the best approach.Alternatively, maybe I can use double counting or some combinatorial argument.Let me think about the total number of pairs (a,b) with a < b and a + b ≤ n. Let's denote this number as P.For each a from 1 to floor(n/2), the number of b's is n - a - a = n - 2a.So, P = sum_{a=1}^{floor(n/2)} (n - 2a).Let me compute this sum.If n is even, say n=2m, then P = sum_{a=1}^{m} (2m - 2a) = 2 sum_{a=1}^{m} (m - a) = 2 sum_{k=0}^{m-1} k = 2*(m(m-1)/2) = m(m-1).If n is odd, say n=2m+1, then P = sum_{a=1}^{m} (2m+1 - 2a) = sum_{a=1}^{m} (2(m - a) +1) = 2 sum_{a=1}^{m} (m - a) + sum_{a=1}^{m} 1 = 2*(m(m-1)/2) + m = m(m-1) + m = m^2.So, in general, P is roughly n^2/4.Now, each pair (a,b) contributes to the condition that at least two of a, b, a+b are the same color. So, for each such pair, we have a monochromatic pair among a, b, a+b.Now, suppose that each color is used at most c times. Then, the number of monochromatic pairs for each color is at most C(c,2). So, the total number of monochromatic pairs across all colors is at most k*C(c,2), where k is the number of colors.But we know that the total number of required monochromatic pairs is at least P, because each pair (a,b) must have at least one monochromatic pair among a, b, a+b.Wait, but actually, for each pair (a,b), at least one of the three pairs (a,b), (a,a+b), (b,a+b) must be monochromatic. So, the total number of monochromatic pairs needed is at least P.But each monochromatic pair can cover multiple such conditions. For example, if a and b are the same color, then this covers the pair (a,b). Similarly, if a and a+b are the same color, this covers the pair (a,b), and similarly for b and a+b.So, the total number of monochromatic pairs needed is at least P, but each monochromatic pair can cover multiple conditions.Wait, actually, each monochromatic pair (x,y) can cover all pairs (a,b) where either a=x and b=y, or a=x and a+b=y, or a=y and a+b=x. But this seems complicated.Maybe instead of counting the number of monochromatic pairs, I should think about the number of times a color appears in the numbers a, b, a+b.Alternatively, perhaps I can use the concept of additive energy or something similar from additive combinatorics, but I'm not sure.Wait, another idea: consider the set S of numbers colored with a particular color, say color C. The condition implies that S cannot contain both a and b without also containing a+b, unless a+b is colored with C as well. But actually, the condition is weaker: it just requires that at least two of a, b, a+b are in C.So, if S is the set of numbers colored with C, then for any a < b with a + b ≤ n, either a ∈ S, b ∈ S, or a + b ∈ S, but not necessarily all three.This is similar to a sumset condition. Maybe I can use some sumset estimates.Alternatively, perhaps I can use the concept of minimal overlapping. If a color is used too sparingly, then there might be too many pairs (a,b) where neither a nor b nor a+b is in that color, violating the condition.Wait, let's think about it in terms of graph theory. Imagine a graph where each vertex is a number from 1 to n, and there is an edge between a and b if a + b is also in the set. Then, the condition is that for every edge (a,b), at least two of a, b, a+b are the same color.But I'm not sure if this helps directly.Maybe I can use the probabilistic method. Suppose that each color is used less than 2n/5 times, then show that the probability of satisfying the condition is less than 1, leading to a contradiction.But perhaps a simpler approach is to use the pigeonhole principle more directly.Suppose that each color is used at most 2n/5 - 1 times. Then, the number of colors k must satisfy k ≥ n / (2n/5 - 1). As I tried earlier, but for n=5, this gives k ≥ 5 / (2 -1)=5, which is possible, but in reality, we saw that with n=5, we can have two colors, one used 3 times and the other 2 times, so the assumption that each color is used less than 2n/5=2 times is violated because one color is used 3 times.Wait, so maybe the key is that if each color is used less than 2n/5 times, then the number of colors is too large, leading to some contradiction.Alternatively, perhaps I can consider the complement: if no color is used 2n/5 times, then the number of colors is at least 5/2, but since the number of colors must be an integer, it's at least 3, but I'm not sure.Wait, let's think about the total number of pairs (a,b) with a < b and a + b ≤ n. As I computed earlier, this is roughly n^2/4. Each such pair requires that at least two of a, b, a+b are the same color. So, for each pair, we have a monochromatic pair among a, b, a+b.Now, suppose that each color is used at most c times. Then, the number of monochromatic pairs for each color is at most C(c,2). So, the total number of monochromatic pairs across all colors is at most k*C(c,2).But we need this total to be at least P, the number of pairs (a,b). So, k*C(c,2) ≥ P.If we assume that c < 2n/5, then C(c,2) < C(2n/5,2) = (2n/5)(2n/5 -1)/2 ≈ (4n^2)/50 = 2n^2/25.So, k*C(c,2) < k*(2n^2/25).But we need k*(2n^2/25) ≥ P ≈ n^2/4.Thus, k > (n^2/4) / (2n^2/25) = (25)/(8) ≈ 3.125.So, k > 3.125, meaning k ≥ 4.But if k ≥ 4, then the average number of times a color is used is n/k ≤ n/4.But we need to show that at least one color is used at least 2n/5 times, which is larger than n/4 for n ≥ 5.Wait, 2n/5 is greater than n/4 when n > 0, since 2/5 > 1/4.So, if the average is n/k ≤ n/4, but we need to show that one color is used at least 2n/5 times, which is higher than the average.This suggests that if the number of colors k is too large, the average usage per color is too low, contradicting the condition that for each pair (a,b), at least two of a, b, a+b are the same color.Therefore, there must be a color used at least 2n/5 times.Wait, but I need to make this more precise.Let me formalize this.Assume for contradiction that every color is used less than 2n/5 times. Then, the number of colors k satisfies k > n / (2n/5 -1) ≈ 5/2.But since k must be an integer, k ≥ 3.Now, the total number of monochromatic pairs is at most k*C(c,2), where c < 2n/5.So, total monochromatic pairs < k*(2n/5 choose 2) = k*( (2n/5)(2n/5 -1)/2 ) ≈ k*(2n^2/25).But we need total monochromatic pairs ≥ P ≈ n^2/4.So, k*(2n^2/25) ≥ n^2/4 ⇒ k ≥ (n^2/4) / (2n^2/25) = 25/8 ≈ 3.125.Thus, k ≥ 4.But if k ≥ 4, then the average number of times a color is used is n/k ≤ n/4.But 2n/5 > n/4 for n > 0, so this contradicts the assumption that every color is used less than 2n/5 times, because the average is n/4, and if all colors are below 2n/5, which is higher than n/4, then the total number of colors would have to be less than n/(n/4)=4, but we just showed that k ≥4, leading to a contradiction.Therefore, our assumption that every color is used less than 2n/5 times must be false. Hence, there exists a color used at least 2n/5 times.I think this makes sense. By assuming that no color is used enough times, we end up needing too many colors, which contradicts the total number of pairs that need to be covered by monochromatic pairs. Therefore, at least one color must be used sufficiently often.

💡Okay, so I have this problem about filling a tank with four valves, A, B, C, and D. The goal is to find out how long it takes to fill the tank when only valves A, B, and C are open. Hmm, let's see. First, I need to understand the given information. When all four valves are open, the tank fills in 1.2 hours. That means the combined rate of all four valves is such that they can fill 1 tank in 1.2 hours. Similarly, when only A, B, and D are open, it takes 2 hours, so their combined rate is slower. And when A, C, and D are open, it takes 1.5 hours. Also, it's given that the rate of valve D is half the rate of valve C. I think I should start by defining the rates of each valve. Let me denote the rates as a, b, c, and d for valves A, B, C, and D respectively. Since these are rates, they would be in tanks per hour. So, if I can find these rates, I can then find the combined rate of A, B, and C.Given that all four valves together fill the tank in 1.2 hours, their combined rate is 1 tank per 1.2 hours, which is 1/1.2 tanks per hour. So, a + b + c + d = 1/1.2.Similarly, when A, B, and D are open, their combined rate is 1/2 tanks per hour. So, a + b + d = 1/2.And when A, C, and D are open, their combined rate is 1/1.5 tanks per hour, which is 2/3. So, a + c + d = 2/3.Also, it's given that d = (1/2)c. So, the rate of D is half that of C. That should help in solving the equations.So, let me write down the equations:1. a + b + c + d = 1/1.2 = 5/62. a + b + d = 1/23. a + c + d = 2/34. d = (1/2)cNow, I can substitute equation 4 into equations 2 and 3.Starting with equation 2: a + b + d = 1/2. Since d = (1/2)c, substitute that in:a + b + (1/2)c = 1/2. Let's call this equation 2a.Similarly, equation 3: a + c + d = 2/3. Substitute d = (1/2)c:a + c + (1/2)c = 2/3. Simplify that:a + (3/2)c = 2/3. Let's call this equation 3a.Now, let's look at equation 1: a + b + c + d = 5/6. Again, substitute d = (1/2)c:a + b + c + (1/2)c = 5/6. Simplify:a + b + (3/2)c = 5/6. Let's call this equation 1a.Now, we have:1a. a + b + (3/2)c = 5/62a. a + b + (1/2)c = 1/23a. a + (3/2)c = 2/3Hmm, maybe I can subtract equation 2a from equation 1a to eliminate a and b.So, subtracting 2a from 1a:(a + b + (3/2)c) - (a + b + (1/2)c) = 5/6 - 1/2Simplify:(0a + 0b + (3/2 - 1/2)c) = 5/6 - 3/6Which is:(1)c = 2/6 = 1/3So, c = 1/3.Now that I have c, I can find d since d = (1/2)c = (1/2)(1/3) = 1/6.Now, let's plug c = 1/3 into equation 3a:a + (3/2)(1/3) = 2/3Simplify:a + (1/2) = 2/3So, a = 2/3 - 1/2 = (4/6 - 3/6) = 1/6.Now, a = 1/6.Now, let's go back to equation 2a: a + b + (1/2)c = 1/2We know a = 1/6, c = 1/3, so:1/6 + b + (1/2)(1/3) = 1/2Simplify:1/6 + b + 1/6 = 1/2Combine like terms:(1/6 + 1/6) + b = 1/22/6 + b = 1/21/3 + b = 1/2So, b = 1/2 - 1/3 = (3/6 - 2/6) = 1/6.So, b = 1/6.Now, we have all the rates:a = 1/6, b = 1/6, c = 1/3, d = 1/6.So, the combined rate of A, B, and C is a + b + c = 1/6 + 1/6 + 1/3.Let's compute that:1/6 + 1/6 = 2/6 = 1/31/3 + 1/3 = 2/3.So, the combined rate is 2/3 tanks per hour.Therefore, the time to fill 1 tank is 1 / (2/3) = 3/2 hours, which is 1.5 hours.So, the answer is 1.5 hours, which is option C.**Final Answer**boxed{C}

💡Alright, so I have these two problems to prove: a) That a number made up of twelve 1s is divisible by 13.b) That a number made up of sixteen 1s is divisible by 17.Hmm, okay. Let me start with part a). I remember that numbers made up of all 1s are called repunits. Maybe there's something about repunits and their divisibility properties that I can use here.First, let me write down what the number looks like. For part a), it's 111111111111, which is twelve 1s. I need to show that this number is divisible by 13. I recall that 13 is a prime number, and maybe I can use some properties related to primes. There's something called Fermat's Little Theorem which says that if p is a prime number, then for any integer a not divisible by p, a^(p-1) is congruent to 1 modulo p. Maybe that can help here.Let me try to apply Fermat's Little Theorem. Here, p is 13, so p-1 is 12. If I take a as 10, since we're dealing with base 10 numbers, then 10^12 should be congruent to 1 modulo 13. Let me check that.Calculating 10^12 is a big number, but maybe I can find a pattern or simplify it. I know that 10^1 mod 13 is 10, 10^2 mod 13 is 9 (since 100 divided by 13 is 7 with a remainder of 9), 10^3 mod 13 is 12 (since 10*9=90, 90 mod 13 is 12), 10^4 mod 13 is 3 (since 10*12=120, 120 mod 13 is 3), 10^5 mod 13 is 4 (10*3=30, 30 mod 13 is 4), 10^6 mod 13 is 1 (10*4=40, 40 mod 13 is 1). Oh, interesting! So 10^6 mod 13 is 1. That means 10^6 ≡ 1 mod 13. Therefore, 10^12 is (10^6)^2, which would be 1^2 ≡ 1 mod 13. So, 10^12 ≡ 1 mod 13. That means 10^12 - 1 is divisible by 13. Now, how does that relate to the repunit number with twelve 1s? Well, the repunit number can be written as (10^12 - 1)/9. Let me verify that:(10^12 - 1)/9 = (999999999999)/9 = 111111111111. Yep, that's the number we're talking about.Since 10^12 - 1 is divisible by 13, and 9 and 13 are coprime (they have no common factors other than 1), then (10^12 - 1)/9 must also be divisible by 13. Therefore, the repunit number with twelve 1s is divisible by 13. Okay, that makes sense.Now, moving on to part b). This time, the number is sixteen 1s, so it's 1111111111111111. I need to show that this number is divisible by 17.Again, 17 is a prime number, so maybe Fermat's Little Theorem can help here too. Let's see. For p = 17, p-1 is 16. So, according to Fermat's Little Theorem, 10^16 ≡ 1 mod 17.Let me check that. Calculating 10^16 mod 17 directly would be tedious, but maybe I can find a pattern or simplify it. Let's compute powers of 10 modulo 17:10^1 mod 17 = 1010^2 mod 17 = 100 mod 17. 17*5=85, so 100 - 85 = 15. So, 10^2 ≡ 15 mod 17.10^3 mod 17 = 10*15 = 150 mod 17. 17*8=136, so 150 - 136 = 14. So, 10^3 ≡ 14 mod 17.10^4 mod 17 = 10*14 = 140 mod 17. 17*8=136, so 140 - 136 = 4. So, 10^4 ≡ 4 mod 17.10^5 mod 17 = 10*4 = 40 mod 17. 17*2=34, so 40 - 34 = 6. So, 10^5 ≡ 6 mod 17.10^6 mod 17 = 10*6 = 60 mod 17. 17*3=51, so 60 - 51 = 9. So, 10^6 ≡ 9 mod 17.10^7 mod 17 = 10*9 = 90 mod 17. 17*5=85, so 90 - 85 = 5. So, 10^7 ≡ 5 mod 17.10^8 mod 17 = 10*5 = 50 mod 17. 17*2=34, so 50 - 34 = 16. So, 10^8 ≡ 16 mod 17.10^9 mod 17 = 10*16 = 160 mod 17. 17*9=153, so 160 - 153 = 7. So, 10^9 ≡ 7 mod 17.10^10 mod 17 = 10*7 = 70 mod 17. 17*4=68, so 70 - 68 = 2. So, 10^10 ≡ 2 mod 17.10^11 mod 17 = 10*2 = 20 mod 17 = 3. So, 10^11 ≡ 3 mod 17.10^12 mod 17 = 10*3 = 30 mod 17 = 13. So, 10^12 ≡ 13 mod 17.10^13 mod 17 = 10*13 = 130 mod 17. 17*7=119, so 130 - 119 = 11. So, 10^13 ≡ 11 mod 17.10^14 mod 17 = 10*11 = 110 mod 17. 17*6=102, so 110 - 102 = 8. So, 10^14 ≡ 8 mod 17.10^15 mod 17 = 10*8 = 80 mod 17. 17*4=68, so 80 - 68 = 12. So, 10^15 ≡ 12 mod 17.10^16 mod 17 = 10*12 = 120 mod 17. 17*7=119, so 120 - 119 = 1. So, 10^16 ≡ 1 mod 17.Okay, so after all that, 10^16 ≡ 1 mod 17. That means 10^16 - 1 is divisible by 17. Now, similar to part a), the repunit number with sixteen 1s can be written as (10^16 - 1)/9. Let me verify that:(10^16 - 1)/9 = (9999999999999999)/9 = 1111111111111111. Yep, that's the number we're dealing with.Since 10^16 - 1 is divisible by 17, and 9 and 17 are coprime, then (10^16 - 1)/9 must also be divisible by 17. Therefore, the repunit number with sixteen 1s is divisible by 17.Wait, let me double-check that reasoning. So, if 10^k ≡ 1 mod p, then 10^k - 1 is divisible by p. And since the repunit is (10^k - 1)/9, and 9 and p are coprime (since p is 13 or 17, which are primes not dividing 9), then the repunit is divisible by p. That seems correct.I think I've got it. Both parts a) and b) can be proven using Fermat's Little Theorem by showing that 10 raised to the power of one less than the prime (13 and 17) is congruent to 1 modulo that prime, hence making the repunit number divisible by the prime.Just to make sure, maybe I can test with smaller numbers to see if the pattern holds. For example, for p=3, which is a prime, the repunit with 2 ones (since p-1=2) is 11. 11 divided by 3 is not an integer, but wait, 10^2 ≡ 1 mod 3, so 10^2 -1=99, which is divisible by 3, and 99/9=11, which is not divisible by 3. Hmm, that seems contradictory.Wait, no, actually, 11 is not divisible by 3, but 99 is. So, maybe the repunit itself isn't necessarily divisible by p, but (10^k -1)/9 is. In the case of p=3, (10^2 -1)/9=11, which is not divisible by 3, but 10^2 -1=99 is divisible by 3. So, perhaps the condition is that k must be such that 10^k ≡ 1 mod p, and then (10^k -1)/9 is divisible by p only if 9 and p are coprime, which they are since p is 13 or 17.Wait, in the case of p=3, 9 and 3 are not coprime, which is why (10^2 -1)/9=11 is not divisible by 3. So, in our original problems, since 13 and 17 are coprime with 9, the repunit numbers are divisible by them. That makes sense.Okay, I think my reasoning holds. So, both parts a) and b) are proven using Fermat's Little Theorem by showing that 10 raised to the power of one less than the prime (13 and 17) is congruent to 1 modulo that prime, hence making the repunit number divisible by the prime.