Difference between revisions of "1967 AHSME Problems/Problem 24"
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== Solution == | == Solution == | ||
− | <math>\ | + | We have <math>y = \frac{501 - 3x}{5}</math>. Thus, <math>501 - 3x</math> must be a positive multiple of <math>5</math>. If <math>x = 2</math>, we find our first positive multiple of <math>5</math>. From there, we note that <math>x = 2 + 5k</math> will always return a multiple of <math>5</math> for <math>501 - 3x</math>. Our first solution happens at <math>k=0</math>. |
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+ | We now want to find the smallest multiple of <math>5</math> that will work. If <math>x = 2 + 5k</math>, then we have <math>501 - 3x = 501 - 3(2 + 5k)</math>, or <math>495 - 15k</math>. When <math>k = 32</math>, the expression is equal to 15<math>, and when </math>k = 33<math>, the expression is equal to </math>0<math>, which will no longer work. | ||
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+ | Thus, all integers from </math>k = 0<math> to </math>k = 32<math> will generate an </math>x = 2 + 5k<math> that will be a positive integer, and which will in turn generate a </math>y<math> that is also a positive integer. So, the answer is </math>\fbox{A}$. | ||
== See also == | == See also == |
Revision as of 23:57, 12 July 2019
Problem
The number of solution-pairs in the positive integers of the equation is:
Solution
We have . Thus, must be a positive multiple of . If , we find our first positive multiple of . From there, we note that will always return a multiple of for . Our first solution happens at .
We now want to find the smallest multiple of that will work. If , then we have , or . When , the expression is equal to 15k = 330$, which will no longer work.
Thus, all integers from$ (Error making remote request. No response to HTTP request)k = 0k = 32x = 2 + 5ky\fbox{A}$.
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.