Difference between revisions of "2016 AMC 10A Problems/Problem 4"

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==Problem==
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The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?
 
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?
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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>
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==Solution 1==
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The value, by definition, is <cmath>\begin{align*}
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\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)
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&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\right\rfloor \\
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&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{3}{8}\times\frac{-5}{2}\right\rfloor \\
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&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\\
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&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\
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&= \frac{3}{8}-\frac{2}{5}\\
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&= \boxed{\textbf{(B) } -\frac{1}{40}}.
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\end{align*}</cmath>
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==Solution 2==
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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>
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==Video Solution (HOW TO THINK CREATIVELY!!!)==
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https://youtu.be/pu_Hqo5sfWs
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/VIt6LnkV4_w?t=195
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https://youtu.be/CrW0Yx5tqV0
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~savannahsolver
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==See Also==
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{{AMC10 box|year=2016|ab=A|num-b=3|num-a=5}}
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{{AMC12 box|year=2016|ab=A|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 10:57, 16 June 2023

Problem

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

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

Solution 1

The value, by definition, is \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*}

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 $\boxed{B}$

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

2016 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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