1994 AMC 12 Problem 15

Attempt Problem 15 of the 1994 AMC 12 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 1994 AMC 12 solutions, or check the answer key.

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

15.

For how many nn in {1,2,3,…,100}\{1,2,3,\ldots,100\} is the tens digit of n2n^2 odd?
1010
2020
3030
4040
5050
Answer: B
Concepts:squares modulo 100decimal digits
Small hint:
The parity of the tens digit of n2n^2 depends only on the units digit of nn
Big hint:
Check the squares of the ten possible units digits
Solution:
Write n=10a+b.n=10a+b. Modulo 100,100, n2≡20ab+b2. n^2\equiv20ab+b^2. The term 20ab20ab changes the tens digit by an even amount, so only the tens digit of b2b^2 matters. Among b=0,1,…,9,b=0,1,\ldots,9, it is odd only for b=4b=4 and b=6.b=6. Each units digit occurs 1010 times from 11 through 100,100, giving 2⋅10=20.2\cdot10=20. Thus the correct answer is B.
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