Difficulty: Easy
Correct Answer: 11
Explanation:
Introduction / Context:This counting problem combines divisibility with a digit-presence condition. Numbers divisible by 5 end in 0 or 5; among them, we must count those that also have the digit 5 somewhere in their decimal representation.
Given Data / Assumptions:
Concept / Approach:Partition divisible-by-5 numbers into those ending with 5 and those ending with 0. The first group automatically contains a 5. The second group contains a 5 only if the tens digit is 5 (i.e., the number 50 within 1–100).
Step-by-Step Solution:
List multiples of 5 ending with 5: 5, 15, 25, 35, 45, 55, 65, 75, 85, 95 → 10 numbers.Check multiples of 5 ending with 0 that include a digit 5: only 50 qualifies (tens digit = 5).Total = 10 + 1 = 11.Verification / Alternative check:Scan all multiples of 5 from 5 to 100; the only candidate ending with 0 that contains a 5 is 50. Count remains 11.
Why Other Options Are Wrong:10 ignores 50; 12 overcounts by adding a non-qualifying multiple; 20 counts all multiples of 5 regardless of the digit condition; 9 misses more than one qualifying number.
Common Pitfalls:Forgetting to include 50 or accidentally including 100 (which has no digit 5).
Final Answer:11
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