1964 AHSME Problems/Problem 3

Revision as of 04:39, 31 December 2023 by Geometry-wizard (talk | contribs) (Solution $1$=)

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. $v$ ($mod y$) $=>$ $x$ + $2uy$ \equiv. $v$ ($mod y$).

% Solution by GEOMETRY-WIZARD$==Solution 2==

By the definition of quotient and remainder, problem states that$ (Error compiling LaTeX. Unknown error_msg)x = uy + v$.

The problem asks to find the remainder of$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)(x, y, u, v)$, then it must be true for a specific set of$(x, y, u, v)$.

If you let$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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 (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 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