Difference between revisions of "2011 AMC 12B Problems/Problem 4"
m (→Solution) |
|||
| Line 15: | Line 15: | ||
<cmath> a*b = 32*7 = </cmath> | <cmath> a*b = 32*7 = </cmath> | ||
| − | <cmath> = \boxed{224\ \((E)} </cmath> | + | <cmath> = \boxed{224\ \(\textbf{(E)}} </cmath> |
== See also == | == See also == | ||
{{AMC12 box|year=2011|before=Problem 3|num-a=5|ab=B}} | {{AMC12 box|year=2011|before=Problem 3|num-a=5|ab=B}} | ||
Revision as of 13:19, 6 March 2011
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)}}\] (Error compiling LaTeX. Unknown error_msg)
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 | |
