Problem 23 · 2011 AMC 8
Hard
Counting & Probability
caseworkpermutations-with-restrictions
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?
Show answer
Answer: D — 84.
Show hints
Hint 1 of 2
Decode the constraints into a digit pool first: ‘5 is the largest digit’ means no digit exceeds 5 (pool is {0,1,2,3,4,5}) and 5 must actually appear. ‘Multiple of 5’ pins the units digit to 0 or 5.
Still stuck? Show hint 2 →
Hint 2 of 2
The two units choices (0 vs 5) behave very differently for the leading-zero rule, so split into those two cases — that's the natural fault line.
Show solution
Approach: casework on the units digit, after fixing the digit pool
- Case units = 0: the 5 must live in one of the three front slots, plus two more distinct digits from {1,2,3,4}. Pick those two: C(4,2) = 6 ways; arrange the three front digits (5 and the two picks): 3! = 6 ways. That's 6×6 = 36, and a leading 0 is impossible here since 0 is parked at the end.
- Case units = 5: the front three are distinct digits from {0,1,2,3,4}. All arrangements: 5·4·3 = 60; subtract the bad ones with 0 in front (0 _ _ with the other two from the remaining 4: 4·3 = 12). That's 60 − 12 = 48.
- Total: 36 + 48 = 84.
- Worth keeping: with several restrictions, translate each into ‘which digits are allowed’ and ‘which must appear’ before counting — then case-split on the rule (divisibility) that controls a single position.
Mark:
· log in to save