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 "1979 AHSME Problems/Problem 22"

(→‎Problem 22)
(The first solution is total nonsense!)
 
(13 intermediate revisions by 3 users not shown)
Line 10: Line 10:
  
 
==Solution==
 
==Solution==
−
Solution by e_power_pi_times_i
+
The equation is equivalent to <math>m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1</math>. Taking mod 3, we get
−
 
+
<math>m(m+1)(m+2)=1 (\bmod 3)</math>. However, <math>m(m+1)(m+2)</math> is always divisible by <math>3</math> for any integer <math>m</math>. Thus, the answer is <math>\boxed{\textbf{(A)} 0}</math>
−
Notice that <math>m^3 + 6m^2 + 5m + 18n^2 = 27n^3 + 27n^2 + 9n + 1 = \frac{(3n+1)^3}{27}</math>. Then <math>27(m^3 + 6m^2 + 5m + 18n^2) = (3n+1)^3</math>, and <math>(3n+1)^3 | 27</math>. However, <math>(3n+1)^3</math> will never be divisible by <math>3</math>, nor <math>27</math>, so there are <math>\boxed{\textbf{(A) } 0}</math> integer pairs of <math>(m, n)</math>.
+
Solution by mickyboy789
  
 
== See also ==
 
== See also ==

Latest revision as of 09:26, 19 August 2021

Problem 22

Find the number of pairs $(m, n)$ of integers which satisfy the equation $m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1$.

$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }3\qquad \textbf{(D) }9\qquad \textbf{(E) }\infty$

Solution

The equation is equivalent to $m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1$. Taking mod 3, we get $m(m+1)(m+2)=1 (\bmod 3)$. However, $m(m+1)(m+2)$ is always divisible by $3$ for any integer $m$. Thus, the answer is $\boxed{\textbf{(A)} 0}$ Solution by mickyboy789

See also

1979 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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

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=1979_AHSME_Problems/Problem_22&oldid=160570"