Difference between revisions of "2016 AMC 12A Problems/Problem 3"

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(Solution)
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<cmath>=\frac{3}{8}-\frac{2}{5}</cmath>
 
<cmath>=\frac{3}{8}-\frac{2}{5}</cmath>
 
<cmath>=\boxed{\textbf{(B)}-\frac{1}{40}}</cmath>
 
<cmath>=\boxed{\textbf{(B)}-\frac{1}{40}}</cmath>
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==See Also==
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{{AMC12 box|year=2016|ab=A|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 22:46, 3 February 2016


Problem

The remainder function can be defined for all real numbers $x$ and $y$ with $y\ne 0$ by $\text{rem}(x,y)=x-y\Big\lfloor\frac{x}{y}\Big\rfloor$, where $\Big\lfloor\frac{x}{y}\Big\rfloor$ denotes the greatest integer less than or equal to $\frac{x}{y}$. What is the value of $\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)$?

$\textbf{(A)}\ -\frac{3}{8}\qquad\textbf{(B)}\ -\frac{1}{40}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{3}{8}\qquad\textbf{(E)}\ \frac{31}{40}$

Solution

\[\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)\] \[=\frac{3}{8}-\left(-\frac{2}{5}\right)\Big\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\Big\rfloor\] \[=\frac{3}{8}+\left(\frac{2}{5}\right)\Big\lfloor -\frac{15}{16}\Big\rfloor\] \[=\frac{3}{8}+\left(\frac{2}{5}\right)\left(-1\right)\] \[=\frac{3}{8}-\frac{2}{5}\] \[=\boxed{\textbf{(B)}-\frac{1}{40}}\]

See Also

2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AMC 12 Problems and Solutions

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