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

Revision as of 10:23, 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 $(m, n)$ .

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

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

@@ Line 1: / Line 1: @@
 == 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>.
+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
@@ Line 7: / Line 7: @@
 \textbf{(C) }3\qquad
 \textbf{(D) }9\qquad
 \textbf{(E) }\infty </math>
 ==Solution==

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

Revision as of 10:23, 26 February 2017

Problem 22

Solution

See also