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Difference between revisions of "1967 AHSME Problems/Problem 24"

(Created page with "== Problem == The number of solution-pairs in the positive integers of the equation <math>3x+5y=501</math> is: <math>\textbf{(A)}\ 33\qquad \textbf{(B)}\ 34\qquad \textbf{(C)}\ ...")
 
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== Solution ==
 
== Solution ==
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<math>\fbox{A}</math>
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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 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 $3x+5y=501$ is:

$\textbf{(A)}\ 33\qquad \textbf{(B)}\ 34\qquad \textbf{(C)}\ 35\qquad \textbf{(D)}\ 100\qquad \textbf{(E)}\ \text{none of these}$

Solution

We have $y = \frac{501 - 3x}{5}$. Thus, $501 - 3x$ must be a positive multiple of $5$. If $x = 2$, we find our first positive multiple of $5$. From there, we note that $x = 2 + 5k$ will always return a multiple of $5$ for $501 - 3x$. Our first solution happens at $k=0$.

We now want to find the smallest multiple of $5$ that will work. If $x = 2 + 5k$, then we have $501 - 3x = 501 - 3(2 + 5k)$, or $495 - 15k$. When $k = 32$, the expression is equal to 15$, and when$k = 33$, the expression is equal to$0$, which will no longer work.

Thus, all integers from$ (Error compiling LaTeX. Unknown error_msg)k = 0$to$k = 32$will generate an$x = 2 + 5k$that will be a positive integer, and which will in turn generate a$y$that is also a positive integer. So, the answer is$\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

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