💡First, I need to find integers ( n ) between 1 and 150 such that the greatest common divisor (gcd) of 21 and ( n ) is 3.To achieve this, ( n ) must be divisible by 3 but not by 7, since 21 is ( 3 times 7 ).I'll start by determining how many multiples of 3 are there between 1 and 150. The largest multiple of 3 within this range is 150, so there are ( frac{150}{3} = 50 ) multiples of 3.Next, I need to exclude the numbers that are multiples of both 3 and 7, which are multiples of 21. The largest multiple of 21 within the range is 147, so there are ( frac{147}{21} = 7 ) multiples of 21.Finally, subtracting the multiples of 21 from the multiples of 3 gives the count of numbers where the gcd with 21 is exactly 3: ( 50 - 7 = 43 ).

💡First, I need to determine how many crayons Jacob has. Since Jacob has two fewer crayons than Dina, and Dina has 28 crayons, I can calculate Jacob's crayons by subtracting 2 from 28.Next, I will add up the number of crayons each person has: Wanda has 62 crayons, Dina has 28 crayons, and Jacob has 26 crayons. By summing these amounts, I will find the total number of crayons they have together.

💡First, I recognize that the square of 7 is 49 and the square of 8 is 64.Since 50 is between 49 and 64, it follows that the square root of 50 must be between 7 and 8.To find the product of these two consecutive integers, I multiply 7 by 8, which equals 56.Therefore, the product of the two integers between which the square root of 50 lies is 56.

💡Alright, so I have this problem here: "A fair six-sided die is rolled eight times. What is the probability that at least one of the rolls results in an odd number?" Hmm, okay, let's see. I've studied probability before, but I need to make sure I understand this correctly.First, let's break down what's being asked. We're rolling a die eight times, and we want the probability that at least one roll is odd. A six-sided die has numbers 1 through 6. The odd numbers are 1, 3, and 5, so there are three odd numbers and three even numbers (2, 4, 6). That means the probability of rolling an odd number on a single roll is 3 out of 6, which simplifies to 1/2. Similarly, the probability of rolling an even number is also 1/2.Now, the problem is asking for the probability of getting at least one odd number in eight rolls. That sounds a bit tricky because "at least one" could mean one odd, two odds, up to eight odds. Calculating all those probabilities individually and adding them up seems complicated. There must be a smarter way to approach this.Wait, I remember something about complementary probability. Instead of calculating the probability of getting at least one odd, maybe I can calculate the probability of the opposite event, which is getting no odd numbers at all in eight rolls. Then, subtracting that probability from 1 should give me the desired probability.Okay, let's try that. The probability of rolling an even number on a single roll is 1/2, as we established earlier. Since each roll is independent, the probability of rolling even numbers on all eight rolls is (1/2) multiplied by itself eight times. So, that would be (1/2)^8.Calculating (1/2)^8: 1/2 is 0.5, and 0.5^8 is 0.00390625. So, the probability of rolling all even numbers in eight rolls is 0.00390625.Now, to find the probability of getting at least one odd number, I subtract this value from 1. So, 1 - 0.00390625 equals 0.99609375.To express this as a fraction, since (1/2)^8 is 1/256, subtracting that from 1 gives 255/256. So, the probability is 255/256.Let me double-check to make sure I didn't make a mistake. The logic seems sound: calculating the complement and subtracting from 1 is a common technique in probability. Each roll is independent, so multiplying the probabilities is correct. Yes, I think this is right.Just to be thorough, let's consider what 255/256 is approximately. 256 divided by 256 is 1, and 255/256 is just slightly less than 1, which makes sense because it's highly likely to roll at least one odd number in eight tries. The decimal equivalent, 0.99609375, also aligns with this intuition.So, I feel confident that the probability of rolling at least one odd number in eight rolls of a fair six-sided die is 255/256.

💡Alright, so I have this problem here about finding the minimum possible sum of the three dimensions of a rectangular box. The volume is given as 2310 cubic inches, and all the dimensions are positive integers. The answer choices are 38, 40, 42, 44, and 46. Hmm, okay, let's try to figure this out step by step.First off, I know that the volume of a rectangular box is calculated by multiplying its length, width, and height. So, if I denote the dimensions as ( a ), ( b ), and ( c ), then the volume ( V ) is:[V = a times b times c = 2310]My goal is to find positive integers ( a ), ( b ), and ( c ) such that their product is 2310, and their sum ( a + b + c ) is as small as possible. I remember that for a given volume, the shape that minimizes the surface area (and hence, in some cases, the sum of dimensions) is a cube. But since 2310 isn't a perfect cube, the dimensions won't all be equal. However, the idea is to make the dimensions as close to each other as possible to minimize their sum.So, maybe I should start by finding the cube root of 2310 to get an approximate idea of what each dimension should be around. Let me calculate that:[sqrt[3]{2310} approx 13.2]Okay, so each dimension should be around 13 inches. But since they have to be integers, I'll need to find integers close to 13 that multiply to 2310.But before that, maybe it's better to factorize 2310 into its prime factors. That might help me distribute the factors among the three dimensions more evenly.Let's factorize 2310:2310 divided by 2 is 1155.1155 divided by 3 is 385.385 divided by 5 is 77.77 divided by 7 is 11.And 11 is a prime number.So, the prime factors of 2310 are:[2310 = 2 times 3 times 5 times 7 times 11]Now, I need to split these prime factors into three groups such that the product of the numbers in each group is as close as possible to each other. Let me list the prime factors again: 2, 3, 5, 7, 11.I have five prime factors, and I need to distribute them into three dimensions. So, some dimensions will have more factors than others.Let me try to assign the factors in a way that the resulting dimensions are as close as possible.Starting with the largest prime factor, which is 11. If I assign 11 to one dimension, then I have 2, 3, 5, and 7 left.Next, let's look at the next largest factor, which is 7. If I pair 7 with 2, that gives me 14. Alternatively, pairing 7 with 3 gives me 21, and pairing 7 with 5 gives me 35.Wait, 35 is quite large, so maybe pairing 7 with 2 or 3 is better to keep the dimensions smaller.Let me try pairing 7 with 3 to get 21. So, one dimension is 11, another is 21, and then the remaining factors are 2 and 5, which multiply to 10. So, the three dimensions would be 10, 11, and 21.Let me check the sum: 10 + 11 + 21 = 42.Is this the smallest possible sum? Let me see if I can get a smaller sum by distributing the factors differently.What if I pair 7 with 2 instead? That would give me 14. Then, the remaining factors are 3 and 5, which multiply to 15. So, the dimensions would be 11, 14, and 15.Calculating the sum: 11 + 14 + 15 = 40. Hmm, that's smaller than 42. So, 40 is a better sum.Wait, can I get even smaller? Let me try another combination.What if I pair 5 with 3? That gives me 15. Then, pairing 7 with 2 gives me 14, and then we have 11 left. So, dimensions are 11, 14, 15, same as before, sum is 40.Alternatively, what if I pair 5 with 2? That gives me 10. Then, pairing 7 with 3 gives me 21, and we have 11 left. That's the same as the first combination, sum is 42.Is there a way to get the dimensions closer to each other? Maybe 10, 14, and 16.5? Wait, no, they have to be integers.Wait, let me think differently. Maybe I can combine more factors into one dimension.For example, if I take 2, 3, and 5 together, that's 30. Then, 7 and 11 are separate. But that would give dimensions of 7, 11, and 30, which sum up to 48. That's worse.Alternatively, if I take 2 and 7 together, that's 14. Then, 3 and 5 together is 15, and 11 is separate. So, 11, 14, 15, sum is 40.Alternatively, if I take 3 and 7 together, that's 21. Then, 2 and 5 is 10, and 11 is separate. So, 10, 11, 21, sum is 42.Alternatively, if I take 2, 3, and 7 together, that's 42. Then, 5 and 11 are separate. So, 5, 11, 42, sum is 58. That's worse.Alternatively, if I take 2, 5, and 7 together, that's 70. Then, 3 and 11 are separate. So, 3, 11, 70, sum is 84. That's way worse.Alternatively, if I take 3, 5, and 7 together, that's 105. Then, 2 and 11 are separate. So, 2, 11, 105, sum is 118. That's even worse.So, it seems that pairing 2 and 7 to get 14, and 3 and 5 to get 15, with 11 left alone, gives me the smallest sum so far, which is 40.Wait, but let me check if there's another way to distribute the factors.What if I take 2 and 11 together? That's 22. Then, 3 and 5 and 7 together is 105. So, dimensions are 22 and 105, but that's only two dimensions, I need three. Wait, no, 22, 3, and 5*7=35. So, 3, 22, 35. Sum is 3 + 22 + 35 = 60. That's worse than 40.Alternatively, 2 and 5 together is 10, 3 and 7 together is 21, and 11. So, 10, 11, 21, sum is 42.Alternatively, 2 and 3 together is 6, 5 and 7 together is 35, and 11. So, 6, 11, 35, sum is 52. Worse.Alternatively, 2 and 7 together is 14, 3 and 11 together is 33, and 5. So, 5, 14, 33, sum is 52. Also worse.Alternatively, 3 and 11 together is 33, 2 and 5 together is 10, and 7. So, 7, 10, 33, sum is 50. Still worse.Alternatively, 5 and 11 together is 55, 2 and 3 together is 6, and 7. So, 6, 7, 55, sum is 68. Worse.Alternatively, 2 and 11 together is 22, 3 and 5 together is 15, and 7. So, 7, 15, 22, sum is 44. That's better than 42, but worse than 40.Wait, 44 is one of the answer choices, but I already have a sum of 40, which is lower.Is there a way to get the sum lower than 40?Let me see. Maybe if I can find three numbers closer to each other than 11, 14, 15.What's the cube root of 2310? Approximately 13.2, as I calculated earlier. So, ideally, each dimension should be around 13.Is there a way to get dimensions closer to 13?Let me see. If I can split the factors such that the dimensions are 10, 14, 15, which are 10, 14, 15. The closest to 13 is 14 and 15, but 10 is a bit away.Alternatively, if I can get dimensions like 11, 12, 17.5, but they have to be integers.Wait, let me try another approach. Maybe instead of pairing 2 and 7, I can pair 2 with 3 and 5.So, 2*3*5=30, and then 7 and 11. But that gives 7, 11, 30, sum is 48, which is worse.Alternatively, 2*3=6, 5*7=35, and 11. So, 6, 11, 35, sum is 52.Alternatively, 2*5=10, 3*7=21, and 11. So, 10, 11, 21, sum is 42.Alternatively, 3*5=15, 2*7=14, and 11. So, 11, 14, 15, sum is 40.I think that's the best I can do. Let me check if there's another combination.What if I take 2*11=22, 3*5=15, and 7. So, 7, 15, 22, sum is 44.Alternatively, 2*7=14, 3*5=15, and 11. So, 11, 14, 15, sum is 40.Alternatively, 3*7=21, 2*5=10, and 11. So, 10, 11, 21, sum is 42.Alternatively, 5*7=35, 2*3=6, and 11. So, 6, 11, 35, sum is 52.Alternatively, 2*3*7=42, 5, and 11. So, 5, 11, 42, sum is 58.Alternatively, 2*5*7=70, 3, and 11. So, 3, 11, 70, sum is 84.Alternatively, 3*5*7=105, 2, and 11. So, 2, 11, 105, sum is 118.Alternatively, 2*3*5=30, 7, and 11. So, 7, 11, 30, sum is 48.Alternatively, 2*3*11=66, 5, and 7. So, 5, 7, 66, sum is 78.Alternatively, 2*5*11=110, 3, and 7. So, 3, 7, 110, sum is 120.Alternatively, 3*5*11=165, 2, and 7. So, 2, 7, 165, sum is 174.Alternatively, 2*7*11=154, 3, and 5. So, 3, 5, 154, sum is 162.Alternatively, 3*7*11=231, 2, and 5. So, 2, 5, 231, sum is 238.Alternatively, 5*7*11=385, 2, and 3. So, 2, 3, 385, sum is 390.So, from all these combinations, the smallest sum I can get is 40, which comes from the dimensions 11, 14, and 15.Wait, but let me double-check if there's a way to get the dimensions closer to 13.2, maybe 12, 13, and something else.Is 12 a factor of 2310? Let's see. 2310 divided by 12 is 192.5, which is not an integer, so 12 isn't a factor.How about 13? 2310 divided by 13 is approximately 177.69, which isn't an integer.14? 2310 divided by 14 is 165, which is an integer. So, 14 is a factor.Wait, so if I take 14 as one dimension, then the other two dimensions would be factors of 165 (since 2310 / 14 = 165).Now, factorizing 165: 165 = 5 × 33, and 33 = 3 × 11. So, 165 = 3 × 5 × 11.So, the other two dimensions could be 3 and 55 (since 3 × 55 = 165), or 5 and 33, or 11 and 15.So, if I take 14, 11, and 15, that's the same as before, sum is 40.Alternatively, 14, 5, and 33, sum is 52.Alternatively, 14, 3, and 55, sum is 72.So, the best sum here is still 40.Alternatively, if I take 15 as one dimension, then 2310 / 15 = 154. So, 154 factors into 2 × 77, and 77 is 7 × 11. So, the other two dimensions could be 2 and 77, or 7 and 11.So, dimensions would be 15, 2, 77, sum is 94, or 15, 7, 11, sum is 33. Wait, 15 + 7 + 11 is 33? That can't be right because 15 × 7 × 11 is 1155, not 2310. Wait, no, 2310 / 15 = 154, which is 14 × 11, not 7 × 11.Wait, I think I made a mistake here. 154 is 14 × 11, so the dimensions would be 15, 14, and 11, which is the same as before, sum is 40.So, yeah, that's consistent.Alternatively, if I take 10 as one dimension, 2310 / 10 = 231. 231 factors into 3 × 77, which is 3 × 7 × 11. So, dimensions could be 10, 3, 77, sum is 90, or 10, 7, 33, sum is 50, or 10, 11, 21, sum is 42.So, again, the smallest sum here is 42.Alternatively, if I take 21 as one dimension, 2310 / 21 = 110. 110 factors into 10 × 11, so dimensions would be 21, 10, 11, sum is 42.Alternatively, 21, 2, 55, sum is 78.Alternatively, 21, 5, 22, sum is 48.So, again, the smallest sum is 42.Alternatively, if I take 30 as one dimension, 2310 / 30 = 77. 77 is 7 × 11, so dimensions would be 30, 7, 11, sum is 48.Alternatively, 30, 1, 77, sum is 108.So, sum is 48.Alternatively, taking 33 as one dimension, 2310 / 33 = 70. 70 factors into 2 × 5 × 7, so dimensions could be 33, 2, 35, sum is 70, or 33, 5, 14, sum is 52, or 33, 7, 10, sum is 50.So, the smallest sum here is 50.Alternatively, taking 35 as one dimension, 2310 / 35 = 66. 66 factors into 2 × 3 × 11, so dimensions could be 35, 2, 33, sum is 60, or 35, 3, 22, sum is 60, or 35, 6, 11, sum is 52.So, the smallest sum here is 52.Alternatively, taking 55 as one dimension, 2310 / 55 = 42. 42 factors into 2 × 3 × 7, so dimensions could be 55, 2, 21, sum is 78, or 55, 3, 14, sum is 72, or 55, 6, 7, sum is 68.So, the smallest sum here is 68.Alternatively, taking 70 as one dimension, 2310 / 70 = 33. 33 factors into 3 × 11, so dimensions could be 70, 3, 11, sum is 84.Alternatively, 70, 1, 33, sum is 104.So, sum is 84.Alternatively, taking 77 as one dimension, 2310 / 77 = 30. 30 factors into 2 × 3 × 5, so dimensions could be 77, 2, 15, sum is 94, or 77, 3, 10, sum is 80, or 77, 5, 6, sum is 88.So, the smallest sum here is 80.Alternatively, taking 105 as one dimension, 2310 / 105 = 22. 22 factors into 2 × 11, so dimensions could be 105, 2, 11, sum is 118.Alternatively, 105, 1, 22, sum is 128.So, sum is 118.Alternatively, taking 110 as one dimension, 2310 / 110 = 21. 21 factors into 3 × 7, so dimensions could be 110, 3, 7, sum is 120.Alternatively, 110, 1, 21, sum is 132.So, sum is 120.Alternatively, taking 154 as one dimension, 2310 / 154 = 15. 15 factors into 3 × 5, so dimensions could be 154, 3, 5, sum is 162.Alternatively, 154, 1, 15, sum is 170.So, sum is 162.Alternatively, taking 165 as one dimension, 2310 / 165 = 14. 14 factors into 2 × 7, so dimensions could be 165, 2, 7, sum is 174.Alternatively, 165, 1, 14, sum is 180.So, sum is 174.Alternatively, taking 231 as one dimension, 2310 / 231 = 10. 10 factors into 2 × 5, so dimensions could be 231, 2, 5, sum is 238.Alternatively, 231, 1, 10, sum is 242.So, sum is 238.Alternatively, taking 385 as one dimension, 2310 / 385 = 6. 6 factors into 2 × 3, so dimensions could be 385, 2, 3, sum is 390.Alternatively, 385, 1, 6, sum is 392.So, sum is 390.Alternatively, taking 462 as one dimension, 2310 / 462 = 5. So, dimensions could be 462, 5, 1, sum is 468.Alternatively, 462, 1, 5, sum is 468.So, sum is 468.Alternatively, taking 770 as one dimension, 2310 / 770 = 3. So, dimensions could be 770, 3, 1, sum is 774.Alternatively, 770, 1, 3, sum is 774.So, sum is 774.Alternatively, taking 1155 as one dimension, 2310 / 1155 = 2. So, dimensions could be 1155, 2, 1, sum is 1158.Alternatively, 1155, 1, 2, sum is 1158.So, sum is 1158.Alternatively, taking 2310 as one dimension, 2310 / 2310 = 1. So, dimensions could be 2310, 1, 1, sum is 2312.Alternatively, 2310, 1, 1, sum is 2312.So, sum is 2312.Okay, so after going through all these combinations, the smallest sum I can get is 40, which comes from the dimensions 11, 14, and 15. Wait, but let me just make sure there isn't a combination I missed. Maybe if I take 2, 3, 5, 7, 11 and try to distribute them differently.For example, if I take 2 × 3 = 6, 5 × 7 = 35, and 11. So, dimensions are 6, 11, 35, sum is 52.Alternatively, 2 × 5 = 10, 3 × 7 = 21, and 11. So, 10, 11, 21, sum is 42.Alternatively, 2 × 7 = 14, 3 × 5 = 15, and 11. So, 11, 14, 15, sum is 40.Alternatively, 3 × 5 = 15, 2 × 7 = 14, and 11. Same as above.Alternatively, 3 × 7 = 21, 2 × 5 = 10, and 11. So, 10, 11, 21, sum is 42.Alternatively, 5 × 7 = 35, 2 × 3 = 6, and 11. So, 6, 11, 35, sum is 52.Alternatively, 2 × 3 × 5 = 30, 7, and 11. So, 7, 11, 30, sum is 48.Alternatively, 2 × 3 × 7 = 42, 5, and 11. So, 5, 11, 42, sum is 58.Alternatively, 2 × 5 × 7 = 70, 3, and 11. So, 3, 11, 70, sum is 84.Alternatively, 3 × 5 × 7 = 105, 2, and 11. So, 2, 11, 105, sum is 118.Alternatively, 2 × 3 × 5 × 7 = 210, 11, and 1. So, 1, 11, 210, sum is 222.Alternatively, 2 × 3 × 5 × 11 = 330, 7, and 1. So, 1, 7, 330, sum is 338.Alternatively, 2 × 3 × 7 × 11 = 462, 5, and 1. So, 1, 5, 462, sum is 468.Alternatively, 2 × 5 × 7 × 11 = 770, 3, and 1. So, 1, 3, 770, sum is 774.Alternatively, 3 × 5 × 7 × 11 = 1155, 2, and 1. So, 1, 2, 1155, sum is 1158.Alternatively, 2 × 3 × 5 × 7 × 11 = 2310, 1, and 1. So, 1, 1, 2310, sum is 2312.So, yeah, all these combinations either give me a sum higher than 40 or the same as 40.Therefore, I think the minimum possible sum of the three dimensions is 40, which corresponds to the dimensions 11, 14, and 15.Wait, but let me just confirm once more. Is there a way to get the dimensions closer to 13.2 without making the sum larger?For example, if I can get dimensions like 12, 13, and something else. But 12 isn't a factor of 2310, as I saw earlier. 2310 divided by 12 is 192.5, which isn't an integer. Similarly, 13 isn't a factor either.What about 14? 14 is a factor, as we've seen. So, 14, 15, and 11. That's the closest I can get to 13.2 without going below 11.Alternatively, if I take 10, 14, and 15, which sum to 42, but that's worse than 40.Wait, is there a way to get dimensions like 12, 14, and something else? Let's see. 12 × 14 = 168. 2310 / 168 ≈ 13.75, which isn't an integer. So, no.Alternatively, 12 × 15 = 180. 2310 / 180 ≈ 12.83, not an integer.Alternatively, 13 × 14 = 182. 2310 / 182 ≈ 12.69, not an integer.Alternatively, 13 × 15 = 195. 2310 / 195 ≈ 11.85, not an integer.So, it seems that 11, 14, and 15 are the closest I can get to 13.2 without having non-integer dimensions.Therefore, I think the minimum possible sum is 40.

💡First, I need to identify the four consecutive numbers. Let's denote the smallest number as ( x ). Therefore, the four consecutive numbers are ( x ), ( x + 1 ), ( x + 2 ), and ( x + 3 ).Next, I'll set up an equation based on the given information that the sum of these four numbers is 102:[x + (x + 1) + (x + 2) + (x + 3) = 102]Simplifying the equation by combining like terms:[4x + 6 = 102]Now, I'll solve for ( x ):[4x = 102 - 6][4x = 96][x = frac{96}{4}][x = 24]The smallest number is 24. To find the greatest number, I'll add 3 to ( x ):[24 + 3 = 27]Therefore, the greatest number among the four consecutive numbers is 27.