Why $6$? Is this meant to be intuitive?

Question

What is this question asking?

Do you just simply count to number of 1's and 0's and then choose that integer, 6?

$$y=10^k+10^{k+1}+10^{k+2}$$

In the equation above, $k$ is an integer. For what value of $k$ can $y$ be expressed as an integer whose digits consist of twice as many $0$'s as $1$'s?

(original photo of question)

Answer 1

Each of the powers of $10$ will contribute a single $1$ digit, so there will be $3\ 1$'s in total. You therefore need $6$ zeros, which you get if $k=6$, because $10^6$ ends in $6\ 0$'s

Answer 2

Hint:

In decimal form $y$ starts with $111\dots$ and the other digits are zeros.

Answer 3

Hint $\ $ Factor $ $ it: $\,\ y \,=\, (10^2\!+10+1)10^k =\, 111\cdot 10^k =\, 111\overbrace{00\cdots00}^{\large k}$

Remark $\ $ Factorization often simplifies matters, esp. when the common factor is complicated (here an exponential). So, as a first start, one should always pull out obvious (complicated) factors to see if that lends any insight to the problem at hand. Let's consider another example

$$\begin{align} 0\, &=\, e^{(n+2)x}\!-3e^{(n+1)x}+2e^{nx}\\ &=\, e^{nx}(e^{2x}-3 e^x + 2)\\ &=\, e^{nx}(e^{x}\!-2)(e^x\! - 1) \end{align}\qquad$$

and now the equation is very easy to solve. Notice how factoring out the $\,e^{nx}\,$ allowed us to more easily recognize the easily factorable quadratic cofactor