1973 AMC 12 Problem 31

Question

In the following equation, each of the letters represents uniquely a different digit in base ten:

(YE)\cdot(ME)=TTT.

The sum

E+M+T+Y

equals

19

20

21

22

24

Use

TTT=111T=3\cdot37\cdot T

The prime

37

must divide one of the two-digit factors; test its two-digit multiples ending in the common digit

E

Since

TTT=111T=3\cdot37\cdot T,

the prime

37

divides one of

YE

and

ME.

A two-digit multiple of

37

is

37

or

74.

The value

74

is impossible: the other two-digit factor ending in

4

is at least

14,

and

74\cdot14\gt999.

Hence one factor is

37,

so

E=7.

The units digit of the product is the units digit of

7^2,

so

T=9.

Thus the product is

999,

and the other factor is

\frac{999}{37}=27.

The four digits are

2,

3,

7,

9,

whose sum is

21.

Therefore, the correct answer is C.

Problem 31 in other years