Difference between revisions of "1979 AHSME Problems/Problem 22"

(Created page with "== Problem 22 == Find the number of pairs <math>(m, n)</math> of integers which satisfy the equation <math>m^3 + 6m^2 + 5m = 27n^3 + 9n^2 + 9n + 1</math>. <math>\textbf{(A)...")
 
(Problem 22)
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== Problem 22 ==
 
== Problem 22 ==
 
   
 
   
Find the number of pairs <math>(m, n)</math> of integers which satisfy the equation <math>m^3 + 6m^2 + 5m = 27n^3 + 9n^2 + 9n + 1</math>.
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Find the number of pairs <math>(m, n)</math> of integers which satisfy the equation <math>m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1</math>.
  
 
<math>\textbf{(A) }0\qquad
 
<math>\textbf{(A) }0\qquad
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\textbf{(C) }3\qquad
 
\textbf{(C) }3\qquad
 
\textbf{(D) }9\qquad
 
\textbf{(D) }9\qquad
\textbf{(E) }\infty </math>  
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\textbf{(E) }\infty </math>
  
 
==Solution==
 
==Solution==

Revision as of 11:23, 26 February 2017

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

Solution by e_power_pi_times_i

Notice that $m^3 + 6m^2 + 5m + 18n^2 = 27n^3 + 27n^2 + 9n + 1 = \frac{(3n+1)^3}{27}$. Then $27(m^3 + 6m^2 + 5m + 18n^2) = (3n+1)^3$, and $(3n+1)^3 | 27$. However, $(3n+1)^3$ will never be divisible by $3$, nor $27$, so there are $\boxed{\textbf{(A) } 0}$ integer pairs of $(m, n)$.

See also

1979 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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