All divisible?

Relevant wiki: Proof of Mathematical Induction

The crucial observation is that any two successive terms have a difference of the form $1083\underbrace{0000\ldots0}_{n \text{ number of 0's}}$ , which is divisible by $19$ .

Therefore, let the terms in the series be expressed as $a_{n}=a_{n-1}+1083\times 10^{n},\: \: n\geqslant 2,$ with $a_{1}=12008.$ We need to prove that $a_{n}$ is divisible by $19$ for all $n\in \mathbb{N}$ .

Proof by induction is used:

Initial Step:

For $n=1$ :

$a_{1}=12008=19\times 632$ .

$\therefore$ The statement is true for $n=1$ .

Inductive Step:

Assume the statement is true for some $n=k$ , $n\in \mathbb{N}$ , i.e. $a_{k}$ is divisible by $19$ .

Then, for $n=k+1$ :

$a_{k+1}=a_{k}+1083\times 10^{k+1}.$

Since $1083=57\times 19$ , $1083$ is divisible by $19$ , therefore $1083\times 10^{k+1}$ is divisible by $19$ ; Also since $a_{k}$ is divisible by $19$ , we can conclude that $a_{k+1}$ is divisible by $19$ .

$\therefore$ We have proven that if the statement is true for some $n=k$ , then it is also true for $n=k+1$ .

Thus, by the Principle of Mathematical Induction, the statement is true for all $n\in \mathbb{N}$ . QED

We have:

$\quad120\underbrace{333333\ldots3}_{n \text{ number of 3's}}08 \\ =12\cdot 10^{n+3}+\dfrac{10^{n+2}-1}{3}-25\\ = \dfrac{ (360+ 1 ) \cdot 10^{n+2} - (1+75 ) } { 3 } \\ =\dfrac{19\cdot\left(19\cdot10^{n+2}-4\right)}{3}$
is divisible by $19$ .

After dividing 120 by 19 the remainder is 6.

63 when divided by 19 again the remainder is 6.

the remainder will continue to be 6 as long as 3 is the next digit in the number.

when we are done with 3s, we are left with 08 and 608 is divisible by 19.

So irrespective of number of threes the number is always divisible by 19.

We note that the only difference between successive numbers is the number, n, of 3's between 120 and 08. To shift the first 3 digits to the left, we merely subtract 8 and multiply the number by 10. To construct the next term, we add 308. For a number with n 3's, this procedure generates the next number in the sequence, with n+1 3's. It is trivial to demonstrate that 12008 is divisible by 19, so multiplication by 10 yields a number that is also divisible by 19. Subtracting 8, multiplying by 10 and adding 308 is equal to adding 228 to 10 times the original number, 12008. It is also clear that 228 is divisible by 19. So the next number is also divisible by 19. In general, 10 times a number divisible by 19 plus another number divisible by 19 yields a number also divisible by 19. Any number constructed this way is divisible by 19, and starting with the base number 12008 also satisfies the description of the form specified in the problem. Therefore, the answer is TRUE.