I'm taking an online course in Discrete Maths. The course starts by explaining the need for proofs, by presenting a proof to an example problem. The way the tutor presents the proof does not make sense to me. Can anyone explain, and if not, can you provide a better proof?
The problem:
Find a number such that, after removing the first digit, it becomes the original number divided by 57
The tutor gives one solution as 7125. And then starts the proof like this:
If we think of $b...z$ to be the remaining sequence of $k$ digits (after removing the first one) we can use $x$ to represent it, like this:
$ x = b...z$
And lets call the first digit which we removed, $a$. The digit $a$ is a value $0 \lt a \le 9$, and in the original number, it has been "shifted left" by $k$ places, i.e. it is that single digit multiplied by $10 ^ k$.
And the tutor goes through the following 3 steps:
$\,\,\,\,\,\,\,\,a \times 10^k + x = 57 \times x $
$\,\,\,\,\,\,\,\,a \times 10^k =\,\,\,\, 56 \times x \,\,\,\,= \,\,\,\,\, 7 \times 8 \times x $
$\,\,\,\,\,\,\,\,a \times (2^k \times 5^k) = 7 \times 8 \times x$
I understand it so far, but then the tutor states "here 7 cannot hide", while circling the expression $(2^k \times 5^k)$. I've taken "cannot hide" to mean 7 is obviously never a factor of. Is that correct? And if so, how can you tell this simply by glancing at the expression?
Anyway, the tutor then concludes therefore that $a$ "must be divisible by 7", and because $a$ is a single digit, makes the following step:
$\,\,\,\,\,\,\,\,7 \times (2^k \times 5^k) = 7 \times 8 \times x$
$\,\,\,\,\,\,\,\,10^k = 8 \times x$
So, going back to his solution 7125: $x$ is 125, and therefore $k$ is 3 (3 digits long), and it is true that 125 = 1000 $\div$ 8. Thus proving that $a$ is indeed 7.
But how did he make that leap about 7 not being a factor of $(2^k \times 5^k)$?
As said in the other comments and answers the reason why $2^k \cdot 5^k$ is not divisible by $7$ if the fundamental theorem of arithmetic. (If a number is divisible by $7$ it has $7$ as one of its prime factors, but the prime factors of $2^k \cdot 5^k$ are only $2$ and $5$)
The proof shows that the first digit of the number has to be $7$, so you are left with $$ 2^k \cdot 5^k=8 x \rightarrow \frac{2^k \cdot 5^k}{8}=x $$ you know that the LHS has to be divisible by $8=2^3$ (because $x$ is an integer), so you have $k \ge 3$. This means that the smallest numbers that satisfies your theorem is $7125$, but you can choose $k$ bigger than $3$ and still obtain a valid result. In particular, because your number can be written as $$ 7 \cdot 10^k +x=\left(7+\frac{1}{8} \right) \cdot 10^k $$ and $\left(7+\frac{1}{8} \right) \cdot 10^3=7125$ if you choose $k>3$ your number becames $7125 \cdot 10^{k-3}$