1953 AMC 12 Problem 24

Attempt Problem 24 of the 1953 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 1953 AMC 12 solutions, or check the answer key.

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24.

a,

b

and

c

are positive integers less than

10,

then

(10a+b)(10a+c)

equals

100a(a+1)+bc

if:

b+c=10

b=c

a+b=10

a=b

a+b+c=10

Expand both sides and cancel their common terms

The only unmatched terms are

10a(b+c)

and

100a

Expanding and canceling

100a^2+bc

from both sides leaves

10a(b+c)=100a.

Since

a

is positive, this is equivalent to

b+c=10.

Thus, the correct answer is A.

Problem 24 in other years