2011 AMC 10B Problem 5

Below is the professionally curated solution for Problem 5 of the 2011 AMC 10B, 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.

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Concepts:digits prime factorization

Difficulty rating: 960

In multiplying two positive integers $a$ and $b,$ Ron reversed the digits of the two-digit number $a.$ His erroneous product was $161.$ What is the correct value of the product of $a$ and $b?$

$116$

$161$

$204$

$214$

$224$

Solution:

The number $161$ is equal to $7\cdot 23.$ There are no other pairs of numbers that multiply to $161$ besides $1\cdot 161,$ so $23$ is the only two digit factor. Thus, $23$ is the number reversed, so he meant to get $32\cdot 7$ which is $224.$

Thus, the correct answer is E .

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