2018 AMC 10B Problem 21

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,

\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?

324

330

340

361

646

323 = 17 \cdot 19,

and

n

is an even

4

-digit multiple of

323

For a candidate next divisor

d,

n

must be a multiple of

\operatorname{lcm}(323, d);

find the smallest

d > 323

keeping that under

10000

Let

d

be the next divisor after

323.

If

\gcd(d,323)=1,

then

n

is a multiple of

323d>323^2>9999,

impossible for a four-digit number. Thus

\gcd(d,323)>1.

Since

323=17\cdot19,

this gcd is at least

17.

It also divides

d-323,

so

d-323\ge17

and hence

d\ge340.

This bound is attained: for

n=6460=2^2\cdot5\cdot17\cdot19,

both

323

and

340

are divisors. The lower bound shows there is no divisor between them. Thus the smallest possible next divisor is

340,

and C is the correct answer.

Problem 21 in other years