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 $-2015$ ?
$\textbf{(A) }$ It is a negative number ending with a 1.
$\textbf{(B) }$ It is a positive number ending with a 1.
$\textbf{(C) }$ It is a negative number ending with a 5.
$\textbf{(D) }$ It is a positive number ending with a 5.
$\textbf{(E) }$ It is a negative number ending with a 0.

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.

Video Solution 1

https://youtu.be/r5QASm5hnII

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Video Solution

https://youtu.be/UoU-cvet1pg

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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

Contents

Problem

Solution

Video Solution 1

Video Solution

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