Difference between revisions of "2015 AMC 10B Problems/Problem 10"
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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+1=1006+1</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 20:59, 26 March 2018
Problem
What are the sign and units digit of the product of all the odd negative integers strictly greater than 
?
Solution
Since 
, the product must end with a 
.
The multiplicands are the odd negative integers from 
to 
. There are 
of these numbers. Since 
, the product is negative.
Therefore, the answer must be 
See Also
| 2015 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
