1971 AMC 12 Problem 11

Attempt Problem 11 of the 1971 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 1971 AMC 12 solutions, or check the answer key.

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11.

The numeral

47

in base

a

represents the same number as

74

in base

b.

Assuming both bases are positive integers, the least possible value of

a+b,

written as a Roman numeral, is:

\mathrm{XIII}

\mathrm{XV}

\mathrm{XXI}

\mathrm{XXIV}

\mathrm{XVI}

Translate the numerals into

4a+7=7b+4

Both bases exceed

7

; solve

4a-7b=-3

for the least valid pair

The equality is

4a+7=7b+4,

4a-7b=-3.

Since both digits must be valid,

a,b\gt7.

Reducing modulo

7

gives

a\equiv1\pmod7.

The first valid value is

a=15,

which gives

b=9.

Thus

a+b=24,

written

\mathrm{XXIV}.

Therefore, the correct answer is D.

Problem 11 in other years