1955 AMC 12 Problem 37

Question

A three-digit number has, from left to right, the digits

h,

t,

and

u

with

h>u.

When the number with the digits reversed is subtracted from the original number, the units’ digit in the difference is

4.

The next two digits, from right to left, are:

5

and

9

9

and

5

impossible to tell

5

and

4

4

and

5

Subtract algebraically:

(100h+10t+u)

{}-(100u+10t+h)=99(h-u)

Find the digit

h-u

for which

99(h-u)

ends in

4

The difference is

99(h-u).

Since

h-u

is an integer from

1

through

9,

and the units digit is

4,

we need

9(h-u)\equiv4\pmod{10}.

This gives

h-u=6,

so the difference is

99\cdot6=594.

Moving from right to left after the units digit

4,

the next digits are

9

and

5.

Thus, the correct answer is B.

Problem 37 in other years