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Difference between revisions of "1966 AHSME Problems/Problem 26"

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== Solution ==
 
== Solution ==
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<math>\fbox{C}</math>
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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>.
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 +
~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 $m$ be a positive integer and let the lines $13x+11y=700$ and $y=mx-1$ intersect in a point whose coordinates are integers. Then m can be:

$\text{(A) 4 only} \quad \text{(B) 5 only} \quad \text{(C) 6 only} \quad \text{(D) 7 only} \\ \text{(E) one of the integers 4,5,6,7 and one other positive integer}$

Solution

Substitute the second equation into the first one, we have $13x+11mx-11=700$.

So $(13+11m)x=711$. So $13+11m$ is a factor of $711$. $711=3*3*79$, so the factors of $711$ are: $1, 3, 9, 79, 237, 711$.

Clearly, because $m\ge1$, so $13+11m\ge24$. So we only need to check whether $m$ is an integer when $11m+13=79, 237, 711$.

When $11m+13=79$, $m=6$.

Checking the other two choices, $m$ dosen't yield to be an integer. So $m=6$ is the only option. Select $\boxed{C}$.

~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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