Find the least whole number only consisting of the digit 1 such that it is divisible by 3333...3.(100 3's) Find the least whole number only consisting of the digit 1 such that it is divisible by 3333...3.(100 3's).
My approach: we see that 111 is divisible by 3. Hence 100 3's would divide 300 1's. Is my analogy correct?
 A: Here is a proper proof for three hundred $1$s.
First use the digit sum test for multiples of $3$ to verify that the number of $1$s is a multiple of $3$, else the number is not divisible by $3$ let alone $333...3$.
Next consider that the number must also be a multiple of $111...1$ with one hundred $1$s.  If you try, let us say, $120$ $1$s for your dividend, the last $20$ of those $1$s will be a remainder, no good.  To avoid that kind of outcome you need a dividend having a number of $1$s that's a multiple of $100$ to go with being a multiple of $3$ proven above.
Ergo $300$ $1$s is minimal.
A: $3n$ ones is $\dfrac{10^{3n}-1}9$, while $n$ threes is $\dfrac{10^{n}-1}3$.
We have
$$\frac39\frac{10^{3n}-1}{10^n-1}=\frac{10^{2n}+10^n+1}3,$$ which is indeed an integer number (because $10\bmod3=1$). This is somewhat "by chance", it wouldn't work with other digits.
And there is no guarantee that it is the smallest solution.
A: Analogy is not proof. You are basing your claim on the fact that


*

*$n$ threes will always divide $3n$ ones

*There is no smaller number of threes that will divide $3n$ ones.


Neither of the two statements is proven.
