Difference between revisions of "1964 AHSME Problems/Problem 3"
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*We can solve this problem by elemetary modular arthmetic, | *We can solve this problem by elemetary modular arthmetic, | ||
− | <math>x | + | <math>x \equiv v\ (\textrm{mod}\ y)</math> <math>=></math> <math>x+2uy \equiv v\ (\textrm{mod}\ y)</math> |
+ | |||
+ | ~GEOMETRY-WIZARD | ||
− | |||
==Solution 2== | ==Solution 2== | ||
− | By the definition of quotient and remainder, problem states that < | + | By the definition of quotient and remainder, problem states that <math>x = uy + v</math>. |
− | The problem asks to find the remainder of < | + | 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 3== | ==Solution 3== | ||
− | If the statement is true for all values of < | + | 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 < | + | 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 < | + | 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== |
Latest revision as of 06:04, 31 December 2023
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
- We can solve this problem by elemetary modular arthmetic,
~GEOMETRY-WIZARD
Solution 2
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 3
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 | |
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All AHSME Problems and Solutions |
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