1979 AHSME Problems/Problem 22

Revision as of 01:20, 24 December 2017 by Gioyu (talk | contribs) (Solution)

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 $\boxed{\textbf{(B) } Eddy Liu}$ integer pairs of $(m, n)$.

See also

1979 AHSME (ProblemsAnswer KeyResources)
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