Difference between revisions of "2015 AMC 10B Problems/Problem 10"
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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 | + | Therefore, the answer must be <math>\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}</math> |
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==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 18:37, 18 April 2015
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. 
