2020 AMC 12A Problem 4

Attempt Problem 4 of the 2020 AMC 12A 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 2020 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

4.

How many 44-digit positive integers (that is, integers between 10001000 and 9999,9999, inclusive) having only even digits are divisible by 5?5?
8080
100100
125125
200200
500500
Answer: B
Concepts:multiplication principledivisibilitydigits
Small hint:
A number divisible by 55 ends in 00 or 5,5, but the last digit must also be even
Big hint:
Count the choices for each digit: the leading digit cannot be 00
Solution:
To be divisible by 55 the last digit is 00 or 5,5, and to be even it must be 0.0. So the units digit is fixed.
The leading digit is a nonzero even digit: 2,4,6,82, 4, 6, 8 give 44 choices. Each of the two middle digits is any even digit 0,2,4,6,8,0, 2, 4, 6, 8, giving 55 choices each.
The total is 4⋅5⋅5⋅1=100.4 \cdot 5 \cdot 5 \cdot 1 = 100.
Thus, B is the correct answer.
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