2018 AMC 10B Problem 21

Below is the professionally curated solution for Problem 21 of the 2018 AMC 10B, 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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Concepts:factorleast common multipleprime factorization

Difficulty rating: 2100

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

Solution:

Factor 323=1719.323 = 17 \cdot 19. Since it divides the even number n,n, nn is a multiple of 21719=646.2 \cdot 17 \cdot 19 = 646. For the next divisor d,d, nn has to be a multiple of lcm(323,d).\operatorname{lcm}(323, d). Something like 324324 or 330=23511330 = 2 \cdot 3 \cdot 5 \cdot 11 shares no factor with 323,323, which pushes n323324>9999,n \ge 323 \cdot 324 > 9999, too big. But d=340=22517d = 340 = 2^2 \cdot 5 \cdot 17 gives lcm(323,340)=2251719=6460,\operatorname{lcm}(323, 340) = 2^2 \cdot 5 \cdot 17 \cdot 19 = 6460, an even 44-digit number whose divisor list jumps straight from 323323 to 340.340. So the smallest possible next divisor is 340.340. Thus, C is the correct answer.

Problem 21 in Other Years