$2^n$ base $5$ contains more than ones and zeros?

Solution 1:

We assume it can be written as $N=5^{a_i} + 5^{a_{i-1}} +・・・+ 5^{a_1} + 1$

Since this equal to $2^n$. Let $n=4m$,

$2^{n}-1$ $=16^{m}-1$ $=(3*5+1)^m-1 $ $=(3*5)^m+・・・+m(m-1)/2*(3*5)^2+5*3m$

It must be $3m=5^{b_1}=1$, or $5^{b_1}+1$. $9m(m-1)/2$ must be $5^{b_2},5^{b_2}+1$. So possible case are two.

$(i)$ $5^{b_1}*3(m-1)/2=5^{b_2}+1$

this leads $-5^{b_2-b_1}=2,b_1=0$. This is false.

$(ii)$ $9m(m-1)/2=5^{b_2}+1$

$⇔(5^{b_1}+1)*(5^{b_1}-2)=2(5^{b_2}+1)$

$-2 \not \equiv2 \mod 5 $ This is a contradiction.

Therefore it was proved there are no $2^n$ of base 5 which is made of only $0$ and $1$.