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

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

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2015 AMC 10B Problems/Problem 10

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