# 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 $-2015$? $\textbf{(A) }$ It is a negative number ending with a 1. $\textbf{(B) }$ It is a positive number ending with a 1. $\textbf{(C) }$ It is a negative number ending with a 5. $\textbf{(D) }$ It is a positive number ending with a 5. $\textbf{(E) }$ It is a negative number ending with a 0.

## Solution

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

The multiplicands are the odd negative integers from $-1$ to $-2013$. There are $\frac{|-2013+1|}2+1=1006+1$ of these numbers. Since $(-1)^{1007}=-1$, the product is negative.

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

Alternatively, we can calculate the units digit of each group of 5 odd numbers using modular arithmetic

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