2009 AMC 8 Problems/Problem 16: Difference between revisions
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<math> \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24</math> | <math> \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24</math> | ||
==Solution== | |||
With the digits listed from least to greatest, the <math>3</math>-digit integers are <math>138,146,226,234</math>. <math>226</math> can be arranged in <math>\frac{3!}{2!} = 3</math> ways, and the other three can be arranged in <math>3!=6</math> ways. There are <math>3+6(3) = \boxed{\textbf{(D)}\ 21}</math> <math>3</math>-digit positive integers. | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=15|num-a=17}} | {{AMC8 box|year=2009|num-b=15|num-a=17}} | ||
Revision as of 16:19, 25 December 2012
Problem
How many 
-digit positive integers have digits whose product equals 
?
Solution
With the digits listed from least to greatest, the -digit integers are 
. 
can be arranged in 
ways, and the other three can be arranged in 
ways. There are 
-digit positive integers.
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
