2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 21 Hard

Mary chose an even 4-digit number n. She wrote down all the divisors of n in increasing order from left to right: 1, 2,\cdots, \frac{n}{2}, n. At some moment Mary wrote 323 as a divisor of n. What is the smallest possible value of the next divisor written to the right of 323? (2018 AMC 10B Problem, Question#21)

  • A.

    324

  • B.

    330

  • C.

    340

  • D.

    361

  • E.

    646

Answer:C

Prime factorizing 323 gives you 17\cdot 19. The desired answer needs to be a multiple of 17 or 19 , because if it is not a multiple of 17 or 19, the LCM, or the least possible value for n, will not be more than 4 digits. Looking at the answer choices 340, is the smallest number divisible by 17 or 19. Checking, we can see that n would be 6460.

Let the next largest divisor be k. Suppose gcd (k,323)=1. Then,as 323|n,k|n, therefore, 323·k|n. However, because k>323, 323k>323\cdot 324>9999.

Therefore, gcd (k,323)>1. Note that 323=17\cdot 19. Therefore, the smallest the gcd can be is 17 and our answer is 323 +17=340.