2015 AMC 10B Problems/Problem 10

Revision as of 14:51, 12 January 2020 by Williamgolly (talk | contribs) (Solution)

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 $\frac{|-2013+1|}2+1=1006+1$ 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png