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Difference between revisions of "1977 Canadian MO Problems/Problem 3"

 
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== Solution ==
 
== Solution ==
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Rewriting <math>\displaystyle N</math> in base <math>\displaystyle 10,</math> <math>\displaystyle N=7(b^2+b+1)=a^4</math> for some integer <math>\displaystyle a.</math> Because <math>\displaystyle a^4|7</math> and <math>\displaystyle 7</math> is prime, <math>\displaystyle a \ge 7^4.</math> Since we want to minimize <math>\displaystyle b,</math> we check to see if <math>\displaystyle a=7^4</math> works.
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Rewriting <math>\displaystyle N</math> in base <math>\displaystyle 10,</math> <math>\displaystyle N=7(b^2+b+1)=a^4</math> for some integer <math>\displaystyle a.</math> Because <math>\displaystyle 7\mid a^4</math> and <math>\displaystyle 7</math> is prime, <math>\displaystyle a \ge 7^4.</math> Since we want to minimize <math>\displaystyle b,</math> we check to see if <math>\displaystyle a=7^4</math> works.
  
  
 
When <math>\displaystyle a=7^4,</math> <math>\displaystyle b^2+b+1=7^3.</math> Solving this quadratic, <math>\displaystyle b = 18 </math>.
 
When <math>\displaystyle a=7^4,</math> <math>\displaystyle b^2+b+1=7^3.</math> Solving this quadratic, <math>\displaystyle b = 18 </math>.
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== See Also ==
 
== See Also ==

Revision as of 23:32, 24 July 2006

Problem

$\displaystyle N$ is an integer whose representation in base $\displaystyle b$ is $\displaystyle 777.$ Find the smallest positive integer $\displaystyle b$ for which $\displaystyle N$ is the fourth power of an integer.

Solution

Rewriting $\displaystyle N$ in base $\displaystyle 10,$ $\displaystyle N=7(b^2+b+1)=a^4$ for some integer $\displaystyle a.$ Because $\displaystyle 7\mid a^4$ and $\displaystyle 7$ is prime, $\displaystyle a \ge 7^4.$ Since we want to minimize $\displaystyle b,$ we check to see if $\displaystyle a=7^4$ works.


When $\displaystyle a=7^4,$ $\displaystyle b^2+b+1=7^3.$ Solving this quadratic, $\displaystyle b = 18$.

See Also

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