Difference between revisions of "2011 AMC 10B Problems/Problem 5"
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| − | Since the <math>161</math> was the erroneous product of two integers, one that had two digits, the two factors have to be <math>7</math> and <math>23</math>. Reverse the digits of the two-digit number so that <math>a</math> is <math>32</math>. Find the product of <math>a</math> and <math>b</math>. <math>\longrightarrow 7 \times 32 = \boxed{(E) 224}</math> | + | Since the <math>161</math> was the erroneous product of two integers, one that had two digits, the two factors have to be <math>7</math> and <math>23</math>. Reverse the digits of the two-digit number so that <math>a</math> is <math>32</math>. Find the product of <math>a</math> and <math>b</math>. <math>\longrightarrow 7 \times 32 = \boxed{\mathrm{(E) \ } 224}</math> |
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| + | == See Also== | ||
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| + | {{AMC10 box|year=2011|ab=B|num-b=4|num-a=6}} | ||
Revision as of 15:43, 4 June 2011
Problem 5
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 | ||
