2011 AMC 8 Problem 23
Attempt Problem 23 of the 2011 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
23.
How many -digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of and is the largest digit?
Answer: D
Small hint:
Handle the two possible units digits as separate cases.
Big hint:
A multiple of must end in or .
Solution:
For a number to be divisible by the units digit must be either or
If the units digit is one of the other three digits must be The remaining two digits must be chosen from There are ways to choose the pair, and there are ways to arrange the three digits for a total of numbers.
If the units digit is there are ways to choose the thousands digit. There are ways to choose the other digits. This leaves a total of numbers for this case.
Combining both cases, we get the total number of such integers is
Thus, D is the correct answer.
Problem 23 in other years
- 1985 AMC 8
- 1986 AMC 8
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