Difference between revisions of "2015 AMC 10B Problems/Problem 10"
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==Problem== | ==Problem== | ||
What are the sign and units digit of the product of all the odd negative integers strictly greater than <math>-2015</math>? | What are the sign and units digit of the product of all the odd negative integers strictly greater than <math>-2015</math>? | ||
| + | <br> | ||
| + | <math>\textbf{(A) }</math> It is a negative number ending with a 1. <br> | ||
| + | <math>\textbf{(B) }</math> It is a positive number ending with a 1. <br> | ||
| + | <math>\textbf{(C) }</math> It is a negative number ending with a 5. <br> | ||
| + | <math>\textbf{(D) }</math> It is a positive number ending with a 5. <br> | ||
| + | <math>\textbf{(E) }</math> It is a negative number ending with a 0. <br> | ||
| + | |||
==Solution== | ==Solution== | ||
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> | ||
| + | |||
| + | Alternatively, we can calculate the units digit of each group of 5 odd numbers using modular arithmetic. | ||
| + | |||
| + | ==Video Solution 1== | ||
| + | https://youtu.be/r5QASm5hnII | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/UoU-cvet1pg | ||
| + | |||
| + | ~savannahsolver | ||
==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}} | ||
Latest revision as of 16:13, 2 August 2022
Problem
What are the sign and units digit of the product of all the odd negative integers strictly greater than 
?
It is a negative number ending with a 1.
It is a positive number ending with a 1.
It is a negative number ending with a 5.
It is a positive number ending with a 5.
It is a negative number ending with a 0.
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 
Alternatively, we can calculate the units digit of each group of 5 odd numbers using modular arithmetic.
Video Solution 1
~Education, the Study of Everything
Video Solution
~savannahsolver
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. 
