Difference between revisions of "2011 AMC 12B Problems/Problem 4"
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<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 14:19, 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
\[= \boxed{224\ \(\textbf{(E)}}\] (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)
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 |