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 (→‎Solution)
(→‎Solution)
 
Line 20: Line 20:
  
 
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>.
 +
 +
==solution 2(logic)==
 +
 +
we can factorize it into <cmath>(m+n)(m-n)</cmath> the we knows both of them are even and if we have 2 consecutive even number atlesast one of the me is divisible by 4 so our answer is 4*2 =8
  
 
==See Also==
 
==See Also==

Latest revision as of 13:05, 26 August 2025

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 $\boxed{\textbf{(D)}}$.

solution 2(logic)

we can factorize it into \[(m+n)(m-n)\] the we knows both of them are even and if we have 2 consecutive even number atlesast one of the me is divisible by 4 so our answer is 4*2 =8

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=258952"