# SOLUTION: consider the following infinite sequence of numbers 17, 24, 31, 38, 45, 52. what is the least term in the sequence that is a multiple of 2, 3, 4, 5, 6?

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 Question 404050: consider the following infinite sequence of numbers 17, 24, 31, 38, 45, 52. what is the least term in the sequence that is a multiple of 2, 3, 4, 5, 6? Found 2 solutions by Edwin McCravy, richard1234:Answer by Edwin McCravy(13211)   (Show Source): You can put this solution on YOUR website! ``` 17, 24, 31, 38, 45, 52 That is an arithmetic sequence with a1 = 17 and d = 7 The nth term of an arithmetic sequence is given by an = a1 + (n - 1)d an = 17 + (n - 1)(7) an = 17 + 7n - 7 an = 10 + 7n To be a multiple of 2, 3, 4, 5, 6, is to be a multiple of 2*2*3*5 = 60 So we are looking for the smallest integer k so that 10 + 7n = 60k Write the 10 and the 60 in terms of their nearest multiples of 7. 10 is nearest 7 and 60 is nearest 63 (7 + 3) + 7n = (63 - 3)k 7 + 3 + 7n = 63k - 3k Divide through by 7 1 + + n = 9k - k Isolate the fractional terms: + k = 9k - n - 1 Since the right side is an integer, the left side is also an integer. Let that integer be A. + k = A and 9k - n - 1 = A Clear of fractions: 3 + 3k = 7A Write the 7 in terms of its nearest multiple of 3, which is 6 3 + 3k = (6 + 1)A 3 + 3k = 6A + A Divide through by 3 1 + k = 2A + Isolate the fraction: 1 + k - 2A = Since the left side is an integer, the right side is also an integer. Let that integer be B. = B and 1 + k - 2A = B Clear of fractions: A = 3B Substituting in 1 + k - 2A = B 1 + k - 2(3B) = B 1 + k = 6B + B 1 + k = 7B k = 7B - 1 Substituting for A and k in 9k - n - 1 = A 9(7B - 1) - n - 1 = 3B 63B - 9 - n - 1 = 3B 63B - 10 - n = 3B 60B - 10 = n The smallest possible value of B that would produce a positive value for n is 1, thus B = 1 60(1) - 10 = n 50 = n and k = 7B - 1 = 7(1) - 1 = 6 Therefore the desired term is the 50th. That term can be gotten from either 10 + 7n = 10 + 7(50) = 10 + 350 = 360 or from 60k = 60(6) = 360. So the answer is 360, which is the 50th term. Edwin``` Answer by richard1234(6688)   (Show Source): You can put this solution on YOUR website! We want the least term that is a multiple of the LCM of 2, 3, ..., 6, or 60. Our sequence is denoted by 17 + 7k, where k is an integer. Writing everything modulo 2, 3, and 5, we obtain 1 + k ≡ 0 (mod 2) --> k ≡ 1 (mod 2) 2 + k ≡ 0 (mod 3) --> k ≡ 1 (mod 3) 2 + 2k ≡ 0 (mod 5) --> 1 + k ≡ 0 (mod 5) --> k ≡ 4 (mod 5) The first two equations imply k ≡ 1 (mod 6), so k can be 1, 7, 13, 19, 25, 31, ... The first value of k that is 4 modulo 5 is 19, so the smallest term in the sequence that satisfies is 17 + 7(19), or 150.