Difference between revisions of "2011 AMC 10B Problems/Problem 5"
m |
|||
| (2 intermediate revisions by the same user not shown) | |||
| Line 7: | Line 7: | ||
== Solution == | == Solution == | ||
| − | We have <math>161 = 7 \cdot 23.</math> Since <math>a</math> has two digits, the factors must be <math>23</math> and <math>7,</math> so <math>a = 32</math> and <math>b = 7.</math> Then, <math>ab = 7 \times 32 = \boxed{\mathrm{\textbf{(E) }\ } 224}.</math> | + | We have <math>161 = 7 \cdot 23.</math> Since <math>a</math> has two digits, the factors must be <math>23</math> and <math>7,</math> so <math>a = 32</math> and <math>b = 7.</math> Then, <math>ab = 7 \times 32 = \boxed{\mathrm{\textbf{(E)}\ } 224}.</math> |
==Video Solution== | ==Video Solution== | ||
Latest revision as of 12:07, 24 January 2024
Contents
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
We have 
Since has two digits, the factors must be 
and 
so 
and 
Then, 
Video Solution
~savannahsolver
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. 
