Difference between revisions of "2011 AMC 10B Problems/Problem 5"
m |
|||
Line 12: | Line 12: | ||
{{AMC10 box|year=2011|ab=B|num-b=4|num-a=6}} | {{AMC10 box|year=2011|ab=B|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Revision as of 13:11, 4 July 2013
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
Since the was the erroneous product of two integers, one that had two digits, the two factors have to be and . Reverse the digits of the two-digit number so that is . Find the product of and .
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.