Difference between revisions of "1967 AHSME Problems/Problem 24"
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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>. | 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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− | Thus, all integers from < | + | 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 <math>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}</math>. | ||
== See also == | == See also == |
Revision as of 23:58, 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 , and when , the expression is equal to , which will no longer work.
Thus, all integers from to will generate an that will be a positive integer, and which will in turn generate a that is also a positive integer. So, the answer is .
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 |
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