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

(Created page with "==Problem== What are the sign and units digit of the product of all the odd negative integers strictly greater than <math>-2015</math>? ==Solution== Since <math>-5>-2015</math...")
 
(Solution)
Line 6: Line 6:
 
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=1007</math> of these numbers. Since <math>(-1)^{1007}=-1</math>, the product is negative.
  
Therefore, the answer must be <math>\boxed{\text{(\textbf B) 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>
 +
 
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2015|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:37, 18 April 2015

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

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

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

Invalid username
Login to AoPS