Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1967 AHSME Problems/Problem 24"
m (→‎See also)
 
(One intermediate revision by one other user not shown)
Line 11: Line 11:
 
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>.   
  
−
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.
 
  
−
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}$.
+
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.
 +
 
 +
 
 +
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 ==
−
{{AHSME box|year=1967|num-b=23|num-a=25}}   
+
{{AHSME 40p box|year=1967|num-b=23|num-a=25}}   
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:40, 16 August 2023

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 $k = 0$ 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 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1967_AHSME_Problems/Problem_24&oldid=197010"