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 "1960 AHSME Problems/Problem 25"

m (→‎See Also)
m (→‎Solution)
 
Line 19: Line 19:
 
If <math>a</math> and <math>b</math> are both even, then <math>a-b</math> is even.  If <math>a</math> and <math>b</math> are both odd, then <math>a-b</math> is even as well.  If <math>a</math> is odd and <math>b</math> is even (or vise versa), then <math>a+b+1</math> is even.  Therefore, in all cases, <math>8</math> can be divided into all numbers with the form <math>m^2-n^2</math>.
 
If <math>a</math> and <math>b</math> are both even, then <math>a-b</math> is even.  If <math>a</math> and <math>b</math> are both odd, then <math>a-b</math> is even as well.  If <math>a</math> is odd and <math>b</math> is even (or vise versa), then <math>a+b+1</math> is even.  Therefore, in all cases, <math>8</math> can be divided into all numbers with the form <math>m^2-n^2</math>.
  
−
This can be confirmed by setting <math>m=3</math> and <math>n=1</math>, making <math>m^2-n^2=9-1=8</math>.  Since <math>8</math> is not a multiple of <math>3</math> and is less than <math>16</math>, we can confirm that the answer is <math>\boxed{\textbf{(D)}}</math>
+
This can be confirmed by setting <math>m=3</math> and <math>n=1</math>, making <math>m^2-n^2=9-1=8</math>.  Since <math>8</math> is not a multiple of <math>3</math> and is less than <math>16</math>, we can confirm that the answer is <math>\boxed{\textbf{(D)}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 01:21, 18 June 2018

Problem

Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$. The largest integer which divides all possible numbers of the form $m^2-n^2$ is:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$

Solution

First, factor the difference of squares. \[(m+n)(m-n)\] Since $m$ and $n$ are odd numbers, let $m=2a+1$ and $n=2b+1$, where $a$ and $b$ can be any integer. \[(2a+2b+2)(2a-2b)\] Factor the resulting expression. \[4(a+b+1)(a-b)\] If $a$ and $b$ are both even, then $a-b$ is even. If $a$ and $b$ are both odd, then $a-b$ is even as well. If $a$ is odd and $b$ is even (or vise versa), then $a+b+1$ is even. Therefore, in all cases, $8$ can be divided into all numbers with the form $m^2-n^2$.

This can be confirmed by setting $m=3$ and $n=1$, making $m^2-n^2=9-1=8$. Since $8$ is not a multiple of $3$ and is less than $16$, we can confirm that the answer is $\boxed{\textbf{(D)}}$.

See Also

1960 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1960_AHSME_Problems/Problem_25&oldid=95268"