Difference between revisions of "1964 AHSME Problems/Problem 3"
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==Solution 1== | ==Solution 1== | ||
− | + | By the definition of quotient and remainder, problem states that <math>x = uy + v</math>. | |
− | <math>x</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>. | ||
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==Solution 2== | ==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>. | |
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− | If you let < | ||
− | When you plug in < | + | 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== | ==See Also== |
Revision as of 05:39, 31 December 2023
Contents
Problem
When a positive integer is divided by a positive integer
, the quotient is
and the remainder is
, where
and
are integers.
What is the remainder when
is divided by
?
Solution 1
By the definition of quotient and remainder, problem states that .
The problem asks to find the remainder of when divided by
. Since
is divisible by
, adding it to
will not change the remainder. Therefore, the answer is
.
Solution 2
If the statement is true for all values of , then it must be true for a specific set of
.
If you let and
, then the quotient is
and the remainder is
. The problem asks what the remainder is when you divide
by
. In this case, the remainder is
.
When you plug in and
into the answer choices, they become
, respectively. Therefore, the answer is
.
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.