1955 AMC 12 Problem 37

Attempt Problem 37 of the 1955 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1955 AMC 12 solutions, or check the answer key.

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37.

A three-digit number has, from left to right, the digits h,h, t,t, and uu with h>u.h>u. When the number with the digits reversed is subtracted from the original number, the units’ digit in the difference is 4.4. The next two digits, from right to left, are:

55 and 99

99 and 55

impossible to tell

55 and 44

44 and 55

Answer: B
Concepts:digit reversalmodular arithmeticsubtraction
Difficulty rating: 1570
Small Hint:

Subtract algebraically: (100h+10t+u)(100h+10t+u) (100u+10t+h)=99(hu){}-(100u+10t+h)=99(h-u)

Big Hint:

Find the digit huh-u for which 99(hu)99(h-u) ends in 44

Solution:

The difference is 99(hu).99(h-u). Since huh-u is an integer from 11 through 9,9, and the units digit is 4,4, we need 9(hu)4(mod10).9(h-u)\equiv4\pmod{10}. This gives hu=6,h-u=6, so the difference is 996=594. 99\cdot6=594. Moving from right to left after the units digit 4,4, the next digits are 99 and 5.5.

Thus, the correct answer is B.

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Problem 37 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12