1962 AMC 12 Problem 22

Attempt Problem 22 of the 1962 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 1962 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

22.

The number 121b,121_b, written in the integral base b,b, is the square of an integer, for:
b=10,b=10, only
b=10b=10 and b=5,b=5, only
2≤b≤102\le b\le10
b>2b\gt2
no value of bb
Answer: D
Concepts:number baseplace valueperfect square
Small hint:
Convert 121b121_b to an expression in bb
Big hint:
Remember that the digit 22 requires b>2b\gt2
Solution:
In ordinary notation, 121b=b2+2b+1=(b+1)2. 121_b=b^2+2b+1=(b+1)^2. This is a square for every allowable base. Since digit 22 occurs, precisely the integral bases b>2b\gt2 are allowable.
Thus, the correct answer is D.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.