1971 AMC 12 Problem 25

Attempt Problem 25 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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25.

A teenage boy wrote his own age after his father’s. From this new four-place number, he subtracted the absolute value of the difference of their ages to get

4{,}289.

The sum of their ages was:

48

52

56

59

64

Answer: D

Concepts:ages digits Diophantine Equation modular arithmetic

Difficulty rating: 1880

Small hint:

Let the father’s age be

f

and the boy’s age be

b

Big hint:

The concatenated number is

100f+b

, so use

99f+2b=4289

Solution:

Let the father and boy be

f

and

b

years old. Since

f\gt b,

\begin{aligned} 100f+b-(f-b)&=4289,\\ 99f+2b&=4289. \end{aligned}

The boy is a teenager, so

13\le b\le19.

Reducing modulo

9

gives

2b\equiv5\pmod9,

hence

b=16.

Then

f=\frac{4289-32}{99}=43,

and

f+b=59.

Therefore, the correct answer is D.