2015 AMC 10B Problem 10

Attempt Problem 10 of the 2015 AMC 10B 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 2015 AMC 10B solutions, or check the answer key.

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

10.

What are the sign and units digit of the product of all the odd negative integers strictly greater than −2015?-2015?
It is a negative number ending with a 1.1.
It is a positive number ending with a 1.1.
It is a negative number ending with a 5.5.
It is a positive number ending with a 5.5.
It is a negative number ending with a 0.0.
Answer: C
Concepts:units digitparity
Small hint:
Count how many odd negative factors there are
Big hint:
An odd product with a factor of 55 has units digit 55
Solution:
There are 10071007 odd numbers greater than −2015.-2015.
Our product is of an odd number of negative numbers, so the result is negative.
Also, we multiply by −5-5 in there, so the product is a multiple of 5,5, making it end in 55 or 0.0. None of our factors are even, so the product can’t be even.
Therefore, the product must end in 5.5.
Thus, the correct answer is C .
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