Difference between revisions of "2011 AMC 12B Problems/Problem 4"
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In multiplying two positive integers <math>a</math> and <math>b</math>, Ron reversed the digits of the two-digit number <math>a</math>. His erroneous product was <math>161.</math> What is the correct value of the product of <math>a</math> and <math>b</math>? | In multiplying two positive integers <math>a</math> and <math>b</math>, Ron reversed the digits of the two-digit number <math>a</math>. His erroneous product was <math>161.</math> What is the correct value of the product of <math>a</math> and <math>b</math>? | ||
| − | <math> | + | <math>\textbf{(A)}\ 116 \qquad |
| − | \textbf{(A)}\ 116 \qquad | ||
\textbf{(B)}\ 161 \qquad | \textbf{(B)}\ 161 \qquad | ||
\textbf{(C)}\ 204 \qquad | \textbf{(C)}\ 204 \qquad | ||
\textbf{(D)}\ 214 \qquad | \textbf{(D)}\ 214 \qquad | ||
\textbf{(E)}\ 224 </math> | \textbf{(E)}\ 224 </math> | ||
| − | |||
== Solution == | == Solution == | ||
Latest revision as of 13:13, 19 January 2021
Problem
In multiplying two positive integers 
and 
, Ron reversed the digits of the two-digit number . His erroneous product was 
What is the correct value of the product of and ?
Solution
Taking the prime factorization of 
reveals that it is equal to 
Therefore, the only ways to represent as a product of two positive integers is 
and Because neither nor 
is a two-digit number, we know that and are 
and 
Because is a two-digit number, we know that a, with its two digits reversed, gives 
Therefore, 
and 
Multiplying our two correct values of and yields
![\[a*b = 32*7 =\]](http://latex.artofproblemsolving.com/5/b/1/5b15af41adbaac2f450d801429ad90374b7824b7.png)
![\[= \boxed{224 \textbf{(E)}}\]](http://latex.artofproblemsolving.com/2/5/6/2565fe98537a4d80d87df8cc8475c4e2d9b00ab6.png)
See also
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
