Difference between revisions of "2011 AMC 10B Problems/Problem 5"
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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>? |
Revision as of 16:11, 4 June 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
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 |