2011 AMC 10B Problems/Problem 5

Problem

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 $b$ ?

$\textbf{(A)}\ 116 \qquad\textbf{(B)}\ 161 \qquad\textbf{(C)}\ 204 \qquad\textbf{(D)}\ 214 \qquad\textbf{(E)}\ 224$

Solution

We have $161 = 7 \cdot 23.$ Since $a$ has two digits, the factors must be $23$ and $7,$ so $a = 32$ and $b = 7.$ Then, $ab = 7 \times 32 = \boxed{\mathrm{\textbf{(E)}\ } 224}.$

Solution 2(cross-out/guess n check)

We can cross out our answer choices. (B)161 would obviously not work, a quick look at (D)214 tells us that it only has two factor pairs which are both three digits, (A)116 is too low. This leaves (C)204 and (E)224. From here, we can simply guess and check to get (E)224.

P.S. This solution is NOT advised, but is still listed as an alternate way.

Video Solution

https://youtu.be/b3Vorx_bnpU

~savannahsolver

2011 AMC 10B (Problems • Answer Key • Resources)
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2011 AMC 10B Problems/Problem 5

Contents

Problem

Solution

Solution 2(cross-out/guess n check)

Video Solution

See Also