Fourier base - not as you know it!

Number Theory Level pending

A natural number $n$ is called foury if there exsits a natural number $b$ such that in base $b$ all the digits of $n$ are fours, e.g. $4$ is foury and so is $624$ as $624_{10}=4444_{5}$ . Clearly, if $n$ is foury then $n\equiv0\pmod 4$ . There exists a smallest foury number $N$ such that for all $n\geq N$ if $n\equiv0\pmod 4$ then $n$ is foury.

A natural number $n$ is called unfoury if in any natural base $b$ , none of the digits of $n$ is four. E.g. $1$ is unfoury. There exists a largest unfoury number $M$ .

Find $N$ and $M$ . Your answer will be of the form $N.M$ (with a decimal point between them). E.g. if you think that $N=4$ and $M=1$ then you should enter $4.1$ .

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The answer is 24.8.

1 solution

Dennis Gulko
Apr 2, 2014

Suppose that $N$ is foury. Then there exists a natural number $b\geq 5$ (otherwise $4$ is not a digit in base $b$ ) such that $N=44...4_b$ ( $k$ digits, all equal to $4$ ), i.e.: $N=\sum_{i=0}^{k-1} 4b^i=4\sum_{i=0}^{k-1} b^i=4\frac{b^k-1}{b-1}$ So, for $k=1$ we have $N=4$ . The next foury number will be for $b=5$ and $k=2$ , i.e. $N=4\cdot 6=24$ . So there are no foury numbers in between.

Now let $n\geq 24$ such that $n\equiv 0\pmod 4$ . Then $n=4m$ for $m\geq 6$ . Let $b=m-1$ and $k=2$ . Then $4\frac{b^2-1}{b-1}=4(b+1)=4m=n$ , i.e. $n=44_b$ is foury.

So $N=24$ .

As to the second part, for any number $k\geq 9$ we have $k=14_{k-4}$ , so there are no unfoury numbers $>8$ . On the other hand, $8=10_8=11_7=12_6=13_5$ is unfoury, as for $b<5$ , $4$ is not a digit. Hence $M=8$ .

So the answer is $N.M=\boxed{24.8}$

Fourier base - not as you know it!

The answer is 24.8.

1 solution

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