Difference between revisions of "2015 AMC 10B Problems/Problem 10"

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m (Solution)
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Since <math>-5>-2015</math>, the product must end with a <math>5</math>.
 
Since <math>-5>-2015</math>, the product must end with a <math>5</math>.
  
The multiplicands are the odd negative integers from <math>-1</math> to <math>-2013</math>. There are <math>\frac{2013+1}2=1007</math> of these numbers. Since <math>(-1)^{1007}=-1</math>, the product is negative.
+
The multiplicands are the odd negative integers from <math>-1</math> to <math>-2013</math>. There are <math>\frac{-2013+1}2+1=1006+1</math> of these numbers. Since <math>(-1)^{1007}=-1</math>, the product is negative.
  
 
Therefore, the answer must be <math>\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}</math>
 
Therefore, the answer must be <math>\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}</math>

Revision as of 21:56, 26 March 2018

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

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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