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

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 10 Problems and Solutions

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2015 AMC 10B Problems/Problem 10

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