777 base

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


The answer is 18.

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2 solutions

Let N = k 4 N=k^{4} for some positive integer k k . Then, we get 7 b 2 + 7 b + 7 = k 4 7b^{2}+7b+7=k^{4} 7 k 4 7 k 7\mid k^{4} \Rightarrow 7\mid k Substituting in k = 7 k=7 , we get the answer b = 18 b=18 . (You can check the discriminant as well, but I left it out.)

How did you conclude its 7?, i didn't get it.

Shivam Singh - 4 years, 10 months ago

Good solution (I did it by brute force, translating 777 in various bases into base 10 until I had a number that I knew was a fourth power - I am good at rapid fire calculations, so it did not take l.ong)

Thomas Sutcliffe - 3 years, 4 months ago

We have 7 b 2 + 7 b + 7 = k 4 or 7 ( b 2 + b + 1 ) = k 4 then 7 k 4 7 k 7b^2+7b+7=k^4\text{ or }7(b^2+b+1)=k^4\text{ then } 7|k^4 \rightarrow 7|k

Tran Hieu - 5 years, 4 months ago

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Sorry for the long and convoluted solution...

A Former Brilliant Member - 5 years, 4 months ago
Les Schumer
Jul 27, 2020

Let N = k 4 N = k^4 for some positive integer k k . Then, we get 7 b 2 + 7 b + 7 = k 4 7 ( b 2 + b + 1 ) = k 4 \\ 7b^2 + 7b + 7 = k^4\\ 7(b^2 + b + 1) = k^4\\ k 4 k^4 has 7 7 as a factor, so is of a form 7 4 x 4 = > 7 4 x 4 = 7 ( b 2 + b + 1 ) = > 7 3 ( b 2 + b + 1 ) = > ( 7 3 1 ) b ( b + 1 ) = > 342 b ( b + 1 ) 7^4*x^4 \\=> 7^4*x^4 = 7(b^2 + b + 1) \\=> 7^3\; |\; (b^2 + b + 1) \\ => (7^3-1)\;|\;b(b+1)\\ => 342\;|\;b(b+1)\\ By observation, 342 = 18 19 342=18*19 so we have b = 18 b=\boxed{18}

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