Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1964 AHSME Problems/Problem 3"
Line 1: Line 1:
−
We can replace the letters with any numbers that satisfy this condition. It must be true for any numbers we use. Let <math>x</math> = <math>5</math> <math>y</math> = <math>2</math> <math>u</math> = <math>2</math> and <math>v</math> = <math>1</math>. Plus in you numbers and get <math>13</math> ÷ <math>2</math> <math>=</math> <math>6</math> remainder <math>1</math>. Since <math>v</math> = <math>1</math>, our answer is <math>E</math>.
+
==Problem==
  
−
Solution by superagh
+
When a positive integer <math>x</math> is divided by a positive integer <math>y</math>, the quotient is <math>u</math> and the remainder is <math>v</math>, where <math>u</math> and <math>v</math> are integers.
 +
What is the remainder when <math>x+2uy</math> is divided by <math>y</math>?
 +
 
 +
<math>\textbf{(A)}\ 0 \qquad
 +
\textbf{(B)}\ 2u \qquad
 +
\textbf{(C)}\ 3u \qquad
 +
\textbf{(D)}\ v \qquad
 +
\textbf{(E)}\ 2v </math>   
 +
 
 +
 
 +
==Solution 1==
 +
 
 +
By the definition of quotient and remainder, problem states that <math>x = uy + v</math>.
 +
 
 +
The problem asks to find the remainder of <math>x + 2uy</math> when divided by <math>y</math>.  Since <math>2uy</math> is divisible by <math>y</math>, adding it to <math>x</math> will not change the remainder.  Therefore, the answer is <math>\boxed{\textbf{(D)}}</math>.
 +
 
 +
==Solution 2==
 +
If the statement is true for all values of <math>(x, y, u, v)</math>, then it must be true for a specific set of <math>(x, y, u, v)</math>.
 +
 
 +
If you let <math>x=43</math> and <math>y = 8</math>, then the quotient is <math>u = 5</math> and the remainder is <math>v = 3</math>.  The problem asks what the remainder is when you divide <math>x + 2uy = 43 + 2 \cdot 5 \cdot 8 = 123</math> by <math>8</math>.  In this case, the remainder is <math>3</math>.
 +
 
 +
When you plug in <math>u=5</math> and <math>v = 3</math> into the answer choices, they become <math>0, 5, 10, 3, 6</math>, respectively.  Therefore, the answer is <math>\boxed{\textbf{(D)}}</math>.
 +
 
 +
==See Also==
 +
{{AHSME 40p box|year=1964|num-b=2|num-a=4}}
 +
 
 +
[[Category:Introductory Number Theory Problems]]
 +
 
 +
{{MAA Notice}}

Revision as of 03:30, 23 July 2019

Problem

When a positive integer $x$ is divided by a positive integer $y$, the quotient is $u$ and the remainder is $v$, where $u$ and $v$ are integers. What is the remainder when $x+2uy$ is divided by $y$?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 2u \qquad \textbf{(C)}\ 3u \qquad \textbf{(D)}\ v \qquad \textbf{(E)}\ 2v$


Solution 1

By the definition of quotient and remainder, problem states that $x = uy + v$.

The problem asks to find the remainder of $x + 2uy$ when divided by $y$. Since $2uy$ is divisible by $y$, adding it to $x$ will not change the remainder. Therefore, the answer is $\boxed{\textbf{(D)}}$.

Solution 2

If the statement is true for all values of $(x, y, u, v)$, then it must be true for a specific set of $(x, y, u, v)$.

If you let $x=43$ and $y = 8$, then the quotient is $u = 5$ and the remainder is $v = 3$. The problem asks what the remainder is when you divide $x + 2uy = 43 + 2 \cdot 5 \cdot 8 = 123$ by $8$. In this case, the remainder is $3$.

When you plug in $u=5$ and $v = 3$ into the answer choices, they become $0, 5, 10, 3, 6$, respectively. Therefore, the answer is $\boxed{\textbf{(D)}}$.

See Also

1964 AHSC (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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_3&oldid=107811"