Difference between revisions of "1964 AHSME Problems/Problem 3"
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==Solution 1== | ==Solution 1== | ||
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+ | *We can solve this problem by elemetary modular arthmetic, | ||
+ | <math>x \equiv v\ (\textrm{mod}\ y)</math> <math>=></math> <math>x+2uy \equiv v\ (\textrm{mod}\ y)</math> | ||
+ | |||
+ | ~GEOMETRY-WIZARD | ||
+ | |||
+ | ==Solution 2== | ||
By the definition of quotient and remainder, problem states that <math>x = uy + v</math>. | By the definition of quotient and remainder, problem states that <math>x = uy + v</math>. | ||
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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>. | 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 | + | ==Solution 3== |
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 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>. | ||
Latest revision as of 05:04, 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 | |
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.