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>\frac{3}{8}-\left(-\frac{2}{5}\right)\lfloor{\frac{3}{8} | + | The value, by definition, is <cmath>\begin{align*} |
| + | \text{rem}\left(\frac{3}{8},-\frac{2}{5}\right) | ||
| + | &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\right\rfloor \\ | ||
| + | &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{3}{8}\times\frac{-5}{2}\right\rfloor \\ | ||
| + | &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\\ | ||
| + | &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\ | ||
| + | &= \frac{3}{8}-\frac{2}{5}\\ | ||
| + | &= \boxed{\textbf{(B) } -\frac{1}{40}}. | ||
| + | \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== | ||
| − | {{AMC10 box|year=2016|ab=A|num-b=2|num-a=4}} | + | {{AMC10 box|year=2016|ab=A|num-b=3|num-a=5}} |
| + | {{AMC12 box|year=2016|ab=A|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 09:57, 16 June 2023
Contents
Problem
The remainder can be defined for all real numbers 
and 
with 
by ![\[\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor\]](http://latex.artofproblemsolving.com/9/5/c/95c21b1ce04c5f3c86f558991d39aa35a13bca21.png)
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. 
