Nines in Binary

This question is easy to understand once you add 1 to everything! Here is how you solve:

Add 1 to everything you see above (Adding in binary is not that hard, the only thing different is that 1 + 1 becomes 0 in binary. Don't forget the carry overs!):

1 0 in base 10 is 101 0 in binary.

1 00 in base 10 is 110001 00 in binary.

1 000 in base 10 is 1111101 000 in binary.

1 0,000 in base 10 is 1001110001 0000 in binary.

Do you see what I see? The number of trailing zeroes in both base 10 and 2 are the same! All we need now is a little bit of logic:

• 10 is 2(mod 4), and when you times 10, you only get an extra multiple of 2, therefore:

• 100 is 4(mod 8), and it is not 0(mod 8) because it does not contain 3 multiples of 2!

• 1000 is 8(mod 16), as it only has 3 multiples of 2, not 4.

And so on......

See that this pattern will keep going on forever? I hope you understand it better than the other solutions!

Hence, by induction, this pattern will keep going on forever, so the answer is Yes.

The answer is "Yes". We'll show this, and a bit more...

Note that $\overbrace{99\ldots 9}^{n \,\, 9\text{s}} = 10^n - 1$ and $N \text{ in binary ends with } 00\overbrace{11\ldots 1}^{n \,\, 1\text{s}} \iff N \equiv 2^n-1 \pmod{2^{n+2}}$ Therefore, we need to check whether $10^n - 1 \equiv 2^n - 1 \pmod{2^{n+2}}$ for every natural number $n.$

To see this, note that $\begin{aligned} \frac{\left(10^n - 1\right) - \left(2^n - 1\right)}{2^{n+2}} &= \frac{10^n - 2^n}{2^{n+2}} \\ &= \frac{5^n - 1}{2^2} \\ &= \frac{(5-1)\sum_{k=0}^{n-1} 5^k}{2^2} \\ &= \sum_{k=0}^{n-1} 5^k \end{aligned}$ which is an integer. Therefore, we have shown that $\overbrace{99\ldots 9}^{n \,\, 9\text{s}} \text{ in binary ends with } 00\overbrace{11\ldots 1}^{n \,\, 1\text{s}}$