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1977 Canadian MO Problems/Problem 3: Difference between revisions

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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>.
{{alternate solutions}}


== See Also ==
== See Also ==
* [[1977 Canadian MO Problems]]
* [[1977 Canadian MO]]
[[Category:Olympiad Number Theory Problems]]

Revision as of 21:20, 25 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, $\displaystyle b = 18$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

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