Difference between revisions of "2011 AMC 12B Problems/Problem 4"

m (Solution)
(See also)
Line 18: Line 18:
  
 
== See also ==
 
== See also ==
{{AMC12 box|year=2011|before=Problem 3|num-a=5|ab=B}}
+
{{AMC12 box|year=2011|num-b=3|num-a=5|ab=B}}

Revision as of 17:38, 29 May 2011

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

Taking the prime factorization of $161$ reveals that it is equal to $23*7.$ Therefore, the only ways to represent $161$ as a product of two positive integers is $161*1$ and $23*7.$ Because neither $161$ nor $1$ is a two-digit number, we know that $a$ and $b$ are $23$ and $7.$ Because $23$ is a two-digit number, we know that a, with its two digits reversed, gives $23.$ Therefore, $a = 32$ and $b = 7.$ Multiplying our two correct values of $a$ and $b$ yields

\[a*b = 32*7 =\]

\[= \boxed{224\  \(\textbf{(E)}}\] (Error compiling LaTeX. Unknown error_msg)

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions