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

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The equation is equivalent to <math>m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1</math>. Taking mod 3, we get
 
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 (mod 3)</math> or <math>0=1 (mod 3)</math> which is a contradiction. Thus, the answer is <math>(A) 0</math>
 
<math>m(m+1)(m+2)=1 (mod 3)</math> or <math>0=1 (mod 3)</math> which is a contradiction. Thus, the answer is <math>(A) 0</math>
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Solution by mickyboy789
  
 
== See also ==
 
== See also ==

Revision as of 01:26, 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 $m(m+1)(m+2)=1 (mod 3)$ or $0=1 (mod 3)$ which is a contradiction. Thus, the answer is $(A) 0$ Solution by mickyboy789

See also

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