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

Revision as of 00:30, 24 December 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 $m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1 = \frac{(3n+1)^3}{27}$ . Then $27(m^3 + 6m^2 + 5m) = (3n+1)^3$ , 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 $(m, n)$ .

Solution 2: 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

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 17: / Line 17: @@
 Solution 2:
 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> or <math>0=1 (\bmod 3)</math> which is a contradiction. Thus, the answer is <math>(A) 0</math>
+<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>
 Solution by mickyboy789

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

Revision as of 00:30, 24 December 2017

Problem 22

Solution

See also