2015 AMC 10B Problem 10

We need two things: the sign and the last digit. Handle them separately.

1. List the numbers. The odd negative integers bigger than $-2015$ are $-1, -3, -5, \ldots, -2013$. The number $-2015$ itself is not included, because the problem says "strictly greater".
2. Count them. Ignoring signs, these are the odd numbers $1, 3, 5, \ldots, 2013$. From $1$ to $2014$ there are $2014$ whole numbers and half of them are odd, so there are $1007$ numbers.
3. The sign. Each pair of negatives multiplies to a positive. With $1007$ negative numbers, an odd count, one is left over after pairing, so the product is negative.
4. The last digit. Every number in the list is odd, so the product is odd. The list includes $-5$, so the product is a multiple of $5$. A multiple of $5$ ends in $0$ or $5$, and an odd one ends in $5$.

So the product is negative and ends in $5$.

The answer is $\boxed{\textbf{(C)}\ \text{It is a negative number ending with a 5.}}$

You never have to multiply anything out. The sign of a product depends only on how many negative numbers are in it: an odd count gives a negative result. The last digit comes from two quick facts: a multiple of $5$ ends in $0$ or $5$, and an odd number cannot end in $0$.

The integers in question are $-1, -3, -5, \dots, -2013$ (not $-2015$, since "strictly greater"). Writing them as $-(2m-1)$ for $m = 1, \dots, 1007$, there are $1007$ of them.

Sign: each factor is negative and there are $1007$ factors, an odd number, so the product is negative.

Units digit: the absolute value is $1 \cdot 3 \cdot 5 \cdots 2013$. This product is odd (all factors odd) and divisible by $5$ (it contains the factor $5$). An odd multiple of $5$ ends in $5$.

So the product is negative and ends in $5$: $\boxed{\textbf{(C)}\ \text{It is a negative number ending with a 5.}}$