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2011 AMC 12B Problem 4

Problem 4 of 25EasierNumber Theory

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?

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Solution

Since 161=7⋅23,161=7\cdot23, the only two-digit factor is 23.23. This must be the reversed value of a,a, so the true value of aa is 32,32, and b=7.b=7. The correct product is 32⋅7=224. 32\cdot7=224. Thus, the correct answer is E.

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Concepts: prime factorization · digits

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.