2011 AMC 8 Problems/Problem 23
Problem
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 1
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 .
Solution 2
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.
Video Solution
https://youtu.be/OOdK-nOzaII?t=48
https://youtu.be/QWJ-zWaLtzY ~David
https://youtu.be/4Teb7HbmARc ~savannahsolver
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 |
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