1966 AHSME Problems/Problem 26: Difference between revisions
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== Solution == | == Solution == | ||
<math> | Substitute the second equation into the first one, we have <math>13x+11mx-11=700</math>. | ||
So <math>(13+11m)x=711</math>. So <math>13+11m</math> is a factor of <math>711</math>. <math>711=3*3*79</math>, so the factors of <math>711</math> are: <math>1, 3, 9, 79, 237, 711</math>. | |||
Clearly, because <math>m\ge1</math>, so <math>13+11m\ge24</math>. So we only need to check whether <math>m</math> is an integer when <math>11m+13=79, 237, 711</math>. | |||
When <math>11m+13=79</math>, <math>m=6</math>. | |||
Checking the other two choices, <math>m</math> dosen't yield to be an integer. So <math>m=6</math> is the only option. Select <math>\boxed{C}</math>. | |||
~hastapasta | |||
== See also == | == See also == | ||
{{AHSME box|year=1966|num-b=25|num-a=27}} | {{AHSME box|year=1966|num-b=25|num-a=27}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
[[Category:Introductory Number Theory Problems]] | |||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:26, 2 May 2022
Problem
Let 
be a positive integer and let the lines 
and 
intersect in a point whose coordinates are integers. Then m can be:
Solution
Substitute the second equation into the first one, we have 
.
So 
. So 
is a factor of 
. 
, so the factors of are: 
.
Clearly, because 
, so 
. So we only need to check whether is an integer when 
.
When 
, 
.
Checking the other two choices, dosen't yield to be an integer. So is the only option. Select 
.
~hastapasta
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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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