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

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Notice that <math>m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1 = \frac{(3n+1)^3}{27}</math>. Then <math>27(m^3 + 6m^2 + 5m) = (3n+1)^3</math>, and <math>(3n+1)^3 | 27</math>. However, <math>(3n+1)^3</math> will never be divisible by <math>3</math>, nor <math>27</math>, so there are <math>\boxed{\textbf{(A)} 0 }</math> integer pairs of <math>(m, n)</math>.
 
Notice that <math>m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1 = \frac{(3n+1)^3}{27}</math>. Then <math>27(m^3 + 6m^2 + 5m) = (3n+1)^3</math>, and <math>(3n+1)^3 | 27</math>. However, <math>(3n+1)^3</math> will never be divisible by <math>3</math>, nor <math>27</math>, so there are <math>\boxed{\textbf{(A)} 0 }</math> integer pairs of <math>(m, n)</math>.
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Solution 2:
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Take equation is equivalent to <math>m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1</math>. Taking mod 3, we get
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<math>m(m+1)(m+2)=1 /mod 3</math>
  
 
== See also ==
 
== See also ==

Revision as of 01:25, 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: Take equation is equivalent to $m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1$. Taking mod 3, we get $m(m+1)(m+2)=1 /mod 3$

See also

1979 AHSME (Problems • Answer Key • Resources)
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

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