Difference between revisions of "2011 AMC 8 Problems/Problem 23"
(→Solution) |
(Part of Last Edit; didn't mean to save. Adding Alternate Solution.) |
||
| Line 13: | Line 13: | ||
==Unofficial Alternate Solution== | ==Unofficial Alternate Solution== | ||
| − | We make four cases based off where the multiple of <math>5</math> is. | + | We make four cases based off where the multiple of <math>5</math> digit (<math>0</math> or <math>5</math>) is. The number has to end with either <math>5</math> or <math>0</math> since it's a multiple of <math>5</math>. In all but the last case, the <math>5</math> and <math>0</math> are used at the end and in another spot which separates the cases. |
| − | Case 1: The first digit can't be <math>0</math>, so it must be <math>5</math>. There are <math>4\cdot3</math> to choose the | + | Case 1: The first digit can't be <math>0</math>, so it must be <math>5</math>. There are <math>4\cdot3</math> to choose the middle two digits. After that, the last digit has to be <math>0</math>, so there are a total of <math>1\cdot4\cdot3\cdot1=12</math> numbers. |
| + | |||
| + | Case 2: The second digit can be <math>0</math> or <math>5</math>, leaving <math>2</math> choices. The first and third numbers can be chosen in <math>4\cdot3</math> ways, like last time. The last digit has to be <math>0</math> or <math>5</math>, but not the one we already used. There are a total of <math>4\cdot2\cdot3\cdot1=24</math> numbers. | ||
| + | |||
| + | Case 3: There are the same choices, but the digits <math>0</math> and <math>5</math> are at the last and second-to-last spots. So there are <math>4\cdot3\cdot2\cdot1=24</math> numbers again. | ||
| + | |||
| + | Case 4: There are <math>4\cdot3\cdot2</math> ways to choose the first three numbers. There has to be a <math>5</math> in the number because the largest digit is <math>5</math>. Coincidentally, there are <math>4\cdot3\cdot2\cdot1</math> numbers again. | ||
| + | |||
| + | There are a total of <math>12 + 24 \cdot 3=\boxed{84}</math> numbers. | ||
| + | |||
| + | [size=50]Alternate solution by Sotowa[/size] | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=22|num-a=24}} | {{AMC8 box|year=2011|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:26, 7 August 2020
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 
Unofficial Alternate Solution
We make four cases based off where the multiple of digit ( or ) is. The number has to end with either or since it's a multiple of . In all but the last case, the and are used at the end and in another spot which separates the cases.
Case 1: The first digit can't be , so it must be . There are to choose the middle two digits. After that, the last digit has to be , so there are a total of 
numbers.
Case 2: The second digit can be or , leaving 
choices. The first and third numbers can be chosen in ways, like last time. The last digit has to be or , but not the one we already used. There are a total of 
numbers.
Case 3: There are the same choices, but the digits and are at the last and second-to-last spots. So there are 
numbers again.
Case 4: There are 
ways to choose the first three numbers. There has to be a in the number because the largest digit is . Coincidentally, there are 
numbers again.
There are a total of 
numbers.
[size=50]Alternate solution by Sotowa[/size]
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
