2018 AMC 12B Problem 19

Question

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,

\ldots,

\tfrac{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?

324

330

340

361

646

Factor

323=17\cdot19

If the next divisor

d

were coprime to

323,

then

n\ge323d\gt9999;

so

d

shares a prime with

323

Let

d

be the next divisor after

323.

If

\gcd(d,323)=1,

then

323d

divides

n,

forcing

n\ge323d\gt323^2\gt9999,

impossible for a

4

-digit number. So

d

shares a prime factor with

323=17\cdot19.

Then

d-323\ge\gcd(d,323)\ge17,

so

d\ge340.

Indeed

d=340=17\cdot20

occurs for

n=17\cdot19\cdot20=6460,

which is even and

4

-digit.

Thus, the correct answer is C.

Problem 19 in other years