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2018 AMC 10B Problem 21

Problem 21 of 25HarderNumber Theory

Mary chose an even 44-digit number n.n. She wrote down all the divisors of nn in increasing order from left to right: 1,1, 2,2, ,\ldots, n2,\frac{n}{2}, n.n. At some moment Mary wrote 323323 as a divisor of n.n. What is the smallest possible value of the next divisor written to the right of 323?323?

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Solution

Let dd be the next divisor after 323.323. If gcd(d,323)=1,\gcd(d,323)=1, then nn is a multiple of 323d>3232>9999,323d>323^2>9999, impossible for a four-digit number. Thus gcd(d,323)>1.\gcd(d,323)>1. Since 323=1719,323=17\cdot19, this gcd is at least 17.17. It also divides d323,d-323, so d32317d-323\ge17 and hence d340.d\ge340. This bound is attained: for n=6460=2251719,n=6460=2^2\cdot5\cdot17\cdot19, both 323323 and 340340 are divisors. The lower bound shows there is no divisor between them. Thus the smallest possible next divisor is 340,340, and C is the correct answer.

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Concepts: factor · least common multiple · prime factorization

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.