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>. |
Revision as of 16:36, 2 May 2020
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
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.