1997 AMC 8 Problem 25

Attempt Problem 25 of the 1997 AMC 8 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 1997 AMC 8 solutions, or check the answer key.

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

25.

All of the even numbers from 22 to 9898 inclusive, excluding those ending in 0,0, are multiplied together. What is the rightmost digit (the units digit) of the product?
00
22
44
66
88
Answer: D
Concepts:units digitpairing and grouping
Small hint:
Group the factors by their units digit pattern
Big hint:
Only the units digits matter
Solution:
We only care about the units digit, which means that the tens digits don’t matter.
Then we have 1010 groups of 2⋅4⋅6⋅8=384. 2 \cdot 4 \cdot 6 \cdot 8 = 384.
The units digit of each group is 4.4. We now need to find the units digit of 410.4^{10}.
The units digit of 42=164^2 = 16 is 6.6. This means we only need to find the units digit of 65.6^5.
Note that every power of 66 always ends in a 66 (e.g. 6,6, 36,36, 216,216, …\ldots).
Thus, D is the correct answer.
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