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Difference between revisions of "1977 Canadian MO Problems/Problem 3"
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== Problem ==
 
== Problem ==
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<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.
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<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 ==
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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.
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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^4 \ge 7^4.</math> Since we want to minimize <math>b,</math> we check to see if <math>a=7</math> works.
  
  
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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>.
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When <math>a=7,</math> <math>b^2+b+1=7^3.</math> Solving this quadratic, <math>b = 18 </math>.
  
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{{alternate solutions}}
 
  
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== See Also ==
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{{Old CanadaMO box|num-b=2|num-a=4|year=1977}}
  
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* [[1977 Canadian MO Problems]]
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[[Category:Intermediate Number Theory Problems]]
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* [[1977 Canadian MO]]
 
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[[Category:Olympiad Number Theory Problems]]
 

Latest revision as of 00:45, 19 August 2012

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, $a^4 \ge 7^4.$ Since we want to minimize $b,$ we check to see if $a=7$ works.


When $a=7,$ $b^2+b+1=7^3.$ 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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