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2021 Math Kangaroo

Problem 8

Problem 8 · 2021 Math Kangaroo Medium
Counting & Probability careful-counting

How many 3-digit numbers formed using only the digits 1, 3 and 5 are divisible by 5? You may use digits more than once.

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Answer: C — 9
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Hint 1 of 2
For divisibility by 5 the last digit is forced.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the free choices for the remaining digits.
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Approach: fix the last digit, then count the rest
  1. A number ending in 5 is divisible by 5; among 1, 3, 5 only 5 works for the last digit.
  2. The first two digits are each free to be 1, 3 or 5: that is 3 × 3 = 9 ways.
  3. So there are 9 such numbers.
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