2011 AMC 10B Problem 5

Attempt Problem 5 of the 2011 AMC 10B 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 2011 AMC 10B solutions, or check the answer key.

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

5.

In multiplying two positive integers aa and b,b, Ron reversed the digits of the two-digit number a.a. His erroneous product was 161.161. What is the correct value of the product of aa and b?b?
116116
161 161
204 204
214214
224224
Answer: E
Concepts:digitsprime factorization
Small hint:
Factor 161161
Big hint:
The only two-digit factor tells you the reversed value of aa
Solution:
The number 161161 is equal to 7⋅23.7\cdot 23. There are no other pairs of numbers that multiply to 161161 besides 1⋅161,1\cdot 161, so 2323 is the only two digit factor. Thus, 2323 is the number reversed, so he meant to get 32⋅732\cdot 7 which is 224.224.
Thus, the correct answer is E .
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.