2019 AMC 12B Problem 14

Question

Let

S

be the set of all positive integer divisors of

100{,}000.

How many numbers are the product of two distinct elements of

S?

98

100

117

119

121

Divisors are

2^a5^b

with

0\le a,b\le5;

products are

2^x5^y

with

0\le x,y\le10

All

11\times11

products arise; discard those obtainable only as a divisor times itself

Since

100{,}000=2^5\cdot5^5,

every divisor is

2^a5^b

with

0\le a,b\le5.

A product of two divisors is

2^x5^y

with

0\le x,y\le10,

and every such pair

(x,y)

is attainable, giving

11\cdot11=121

values.

We need two distinct divisors. A value

2^x5^y

is forced to be a divisor times itself only when both

x

and

y

have a unique split, which happens exactly when

x,y\in\{0,10\}.

Those

4

corner values (

1,\ 2^{10},\ 5^{10},\ 2^{10}5^{10}

) cannot use two distinct divisors.

The count is

121-4=117.

Thus, C is the correct answer.

Problem 14 in other years