Difference between revisions of "2016 AMC 10A Problems/Problem 4"
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<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math> | <math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math> | ||
− | ==Solution== | + | ==Solution 1== |
The value, by definition, is <cmath>\begin{align*} | The value, by definition, is <cmath>\begin{align*} | ||
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&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\ | &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\ | ||
&= \frac{3}{8}-\frac{2}{5}\\ | &= \frac{3}{8}-\frac{2}{5}\\ | ||
− | &= \boxed{\textbf{(B) } -\frac{1}{40}} | + | &= \boxed{\textbf{(B) } -\frac{1}{40}}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/pu_Hqo5sfWs | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/VIt6LnkV4_w?t=195 | ||
+ | |||
+ | https://youtu.be/CrW0Yx5tqV0 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 10:57, 16 June 2023
Contents
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 1
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 (HOW TO THINK CREATIVELY!!!)
https://youtu.be/pu_Hqo5sfWs
~Education, the Study of Everything
Video Solution
https://youtu.be/VIt6LnkV4_w?t=195
~savannahsolver
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.