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1977 Canadian MO Problems/Problem 3: Difference between revisions

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== Problem ==
== Problem ==
<math>\displaystyle N</math> is an integer whose representation in base <math>\displaystyle b</math> is <math>\displaystyle 777.</math> Find the smallest positive integer <math>\displaystyle b</math> for which <math>\displaystyle N</math> is the fourth power of an integer.
<math>N</math> is an integer whose representation in base <math>b</math> is <math>777.</math> Find the smallest positive integer <math>b</math> for which <math>N</math> is the fourth power of an integer.


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


{{alternate solutions}}


== See Also ==
{{Old CanadaMO box|num-b=2|num-a=4|year=1977}}
 
* [[1977 Canadian MO Problems]]
* [[1977 Canadian MO]]


[[Category:Olympiad Number Theory Problems]]
[[Category:Olympiad Number Theory Problems]]

Revision as of 21:49, 17 November 2007

Problem

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

Solution

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


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


1977 Canadian MO (Problems)
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 4
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