# Difference between revisions of "2015 AMC 10B Problems/Problem 10"

## Problem

What are the sign and units digit of the product of all the odd negative integers strictly greater than $-2015$?

## Solution

Since $-5>-2015$, the product must end with a $5$.

The multiplicands are the odd negative integers from $-1$ to $-2013$. There are $\frac{2013+1}2=1007$ of these numbers. Since $(-1)^{1007}=-1$, the product is negative.

Therefore, the answer must be $\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}$

## See Also

 2015 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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