2018 AMC 10B Problem 21

Attempt Problem 21 of the 2018 AMC 10B 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 10B solutions, or check the answer key.

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21.

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, \frac{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:factorleast common multipleprime factorization
Difficulty rating: 2100
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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