Difference between revisions of "2011 AMC 8 Problems/Problem 23"
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==Solution== | ==Solution== | ||
| + | We can separate this into two cases. If an integer is a multiple of <math>5,</math> the last digit must be either <math>0</math> or <math>5.</math> | ||
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| + | Case 1: The last digit is <math>5.</math> The leading digit can be <math>1,2,3,</math> or <math>4.</math> Because the second digit can be <math>0</math> but not the leading digit, there are also <math>4</math> choices. The third digit cannot be the leading digit or the second digit, so there are <math>3</math> choices. The number of integers is this case is <math>4\cdot4\cdot3\cdot1=48.</math> | ||
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| + | Case 2: The last digit is <math>0.</math> Because <math>5</math> is the largest digit, one of the remaining three digits must be <math>5.</math> There are <math>3</math> ways to choose which digit should be <math>5.</math> The remaining digits can be <math>1,2,3,</math> or <math>4,</math> but since they have to be different there are <math>4\cdot3</math> ways to choose. The number of integers in this case is <math>1\cdot3\cdot4\cdot3=36.</math> | ||
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| + | Therefore the answer is <math>48+36=\boxed{\textbf{(D)}\ 84}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=22|num-a=24}} | {{AMC8 box|year=2011|num-b=22|num-a=24}} | ||
Revision as of 19:30, 25 November 2011
How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?
Solution
We can separate this into two cases. If an integer is a multiple of 
the last digit must be either 
or 
Case 1: The last digit is The leading digit can be 
or 
Because the second digit can be but not the leading digit, there are also 
choices. The third digit cannot be the leading digit or the second digit, so there are 
choices. The number of integers is this case is 
Case 2: The last digit is 
Because 
is the largest digit, one of the remaining three digits must be There are ways to choose which digit should be The remaining digits can be or 
but since they have to be different there are 
ways to choose. The number of integers in this case is 
Therefore the answer is 
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 | ||
