2018 AMC 12B Problem 19

Attempt Problem 19 of the 2018 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12B solutions, or check the answer key.

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19.

Mary chose an even 44-digit number n.n. She wrote down all the divisors of nn in increasing order from left to right: 1,2,,n2,n.1, 2, \ldots, \tfrac{n}{2}, 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?

324324

330330

340340

361361

646646

Answer: C
Concepts:divisibilitygreatest common divisorprime factorization
Difficulty rating: 2170
Solution:

Let dd be the next divisor after 323.323. If gcd(d,323)=1,\gcd(d,323)=1, then 323d323d divides n,n, forcing n323d>3232>9999,n\ge323d\gt323^2\gt9999, impossible for a 44-digit number. So dd shares a prime factor with 323=1719.323=17\cdot19.

Then d323gcd(d,323)17,d-323\ge\gcd(d,323)\ge17, so d340.d\ge340. Indeed d=340=1720d=340=17\cdot20 occurs for n=171920=6460,n=17\cdot19\cdot20=6460, which is even and 44-digit.

Thus, the correct answer is C.

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