1964 AHSME Problems/Problem 3

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

We can solve this problem by elemetary modular arthmetic,

$x \equiv v\ (\textrm{mod}\ y)$ \Rightarrow $x+2uy \equiv v\ (\textrm{mod}\ y)$

$Solution$ $by$ $GEOMETRY-WIZARD$

Solution 2

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 3

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 $\boxed{\textbf{(D)}}$ .

1964 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1964 AHSME Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also