Problem 13 · 2009 AMC 8
Easy
Counting & Probability
last-digit
A three-digit integer contains one of each of the digits 1, 3, and 5. What is the probability that the integer is divisible by 5?
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Answer: B — 1/3.
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Hint 1 of 2
Divisibility by 5 depends ONLY on the last digit — here that means the units digit must be the 5. The other two digits don't matter, so ignore them.
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Hint 2 of 2
By symmetry, each of 1, 3, 5 is equally likely to be the units digit. So you don't even need to count all the arrangements.
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Approach: only the units digit matters — use symmetry
- A number is a multiple of 5 exactly when its last digit is 0 or 5. Our digits are 1, 3, 5, so we need the 5 sitting in the units place.
- The three digits are placed at random, and there's nothing special about any one slot — so the 5 lands in the units place with probability 1/3 (just as 1 or 3 each would).
- Why this transfers: a divisibility rule that reads only the last digit lets you collapse a whole-number question to one position — then symmetry handles the probability without listing every arrangement.
Another way — count arrangements directly:
- All orderings: 3! = 6. Those ending in 5: fix 5 last, arrange the rest: 2! = 2.
- 2/6 = 1/3.
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