2019 AMC 10B Problem 14

Question

The base-ten representation for

19!

is

121,

6T5,

100,

40M,

832,

H00,

where

T,

M,

and

H

denote digits that are not given. What is

T+M+H?

3

8

12

14

17

Use divisibility by

1000

,

9

, and

11

The unknown digits are constrained by the digit-sum and alternating-sum tests

Because

19!

is divisible by

1000

, its last three digits are zero, so

H=0

.

Since

19!

is divisible by

9

, its digit sum

33+T+M

is divisible by

9

. Hence

T+M\equiv3\pmod9

, so

T+M

is either

3

or

12

.

Divisibility by

11

says the alternating digit sum

T-M-7

is divisible by

11

, so

T-M\equiv7\pmod{11}

. Checking the digit possibilities from these two congruences gives

T=4

and

M=8

. Therefore

T+M+H=4+8+0=12

.

Thus, the answer is C .

Problem 14 in other years