Difference between revisions of "2015 AMC 10B Problems/Problem 10"
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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 | + | Alternatively, we can calculate the units digit of each group of 5 odd numbers using modular arithmetic. |
+ | |||
+ | ==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 11:54, 18 July 2021
Contents
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
~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.