2015 AMC 10B Problems/Problem 10

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

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
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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