Prove that there is not any integer $b$ such that when its initial digit is deleted , the integer becomes $ b/35 $

Question

Prove that there is no integer b such that when its initial digit is deleted, the integer becomes $ b/35 $?

Answer 1

Let $b$ be $10^nx+y$, where $x \in \{1,2,\ldots,9\}$. Deleting the first digit, we obtain $y = b/35$. This means we have $$b = 10^n x+ b/35 \implies 34b = 35 \times 10^n x \implies 17 \vert x$$ which is not possible.

Answer 2

If you mean that when you delete the first digit it becomes $b/35$ then: Take $b=a_1 a_2 \dots a_n $ where the $a_i $ are digits If $a_2 a_3 \dots a_n = b/35$ this is equivalent to: $35*(a_2 a_3 \dots a_n)=b=a_1 a_2 \dots a_n $ Name $c=(a_2 a_3 \dots a_n) $ and it means that $b=(a_1 *10^n) +c=(35*c)$ and then $(a_1*10^n)=34c$ then $17|a_1 *5^n$ and it is absurd because $a_i \in \{0,1, \dots ,9 \}$ and $5^n$ is only divided by powers of $ 5$ and $1$, then $b $ can't exist.