Difference between revisions of "2016 AMC 10A Problems/Problem 4"
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==Solution 2== | ==Solution 2== | ||
Do note that the denominator of the answer will be a multiple of 5 and 8 (40) and that the answer will also be negative. The only answer choice that satisfies this is <math>\boxed{B}</math> | Do note that the denominator of the answer will be a multiple of 5 and 8 (40) and that the answer will also be negative. The only answer choice that satisfies this is <math>\boxed{B}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/VIt6LnkV4_w?t=195 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Revision as of 01:33, 5 May 2020
Problem
The remainder can be defined for all real numbers and with by where denotes the greatest integer less than or equal to . What is the value of ?
Solution
The value, by definition, is
Solution 2
Do note that the denominator of the answer will be a multiple of 5 and 8 (40) and that the answer will also be negative. The only answer choice that satisfies this is
Video Solution
https://youtu.be/VIt6LnkV4_w?t=195
~IceMatrix
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.