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The main question is:

$N$ is a natural number. If the leftmost digit is deleted, the number gets reduced to $\frac N{57}$. Find the sum of its digits.

My approach:

Let there be $n$ digits in $N$. Thus, $N$ can be written as

$$x_1+10x_2+10^2x_3+\dots+10^{n-1}x_n$$

Deleting $10^{n-1}x_n$, we get,

$$\frac{N}{57}=x_1+10x_2+\dots+10^{n-2}x_{n-1}$$

I know these steps are child's play, but I can't get any further than this. Please, any hint or solution will be appreciated.

P.S. I'm not sure if the tags are correct, please verify.

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$$N=\frac N{57} +k\cdot 10^m$$ Where $k\in \{1,2,...,9\}$, $m\in \Bbb Z^+$

$$56N=57k \cdot 10^m$$

$56=7 \cdot 8$

Since neither $57$ nor $10^m$ contain the factor $7$, and $k$ has only one digit, so the only possible value is: $k=7$

$$8N=57 \cdot 10^m$$

Thus $m\geq 3$

Since $1000/8=125$

$$10^m/8 = 125 *10^{m-3}$$

So we have:

$$N=57 \cdot 125 \cdot 10^{m-3}$$

$$N=7125 *10^{m-3}$$

$$7+1+2+5=15$$

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  • $\begingroup$ I did not understand why $m$ has to be greater than or equal to 3? $\endgroup$ – Akshar Gandhi Aug 10 '16 at 16:15
  • $\begingroup$ @AksharGandhi Because $10^m$ must be a multiple of $8$ for the equation to make sense. And $10^m$ has at most has the factor $2^m$, so $m$ must be at least $3$ $\endgroup$ – lEm Aug 10 '16 at 16:16
  • $\begingroup$ @AksharGandhi As $8$ and $57$ are coprime, we must have $8\mid 10^m$. $\endgroup$ – Hagen von Eitzen Aug 10 '16 at 16:16
  • $\begingroup$ $m$ needs to be at least $3$ because when you remove a digit, you need to have at least $2$ digits in the number in order for it to be divisible by $57$ $\endgroup$ – Aruka J Aug 10 '16 at 17:01
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Let first digit be $x_k$ in position $k+1$ from the right, let the whole number be $57M$. Then $57M = M+x_k10^k$, $M < 10^k$.

$x_k10^k = 56M$. Obviously, $x_k = 7$ and $8M=10^k$ hence $M = 125*10^{k-3}$. Finally, $57M = 7125*10^{k-3}$ and its sum of digits is $15$.

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  • $\begingroup$ Will you please elaborate the answer? Everything just went above my head. $\endgroup$ – Akshar Gandhi Aug 10 '16 at 16:06
  • $\begingroup$ We can write $N = x_0 + x_110^1 + ... + x_{k-1}10^{k-1} + x_k10^k$. Then, we know that $x_0 + x_110^1+ ... + x_{k-1}10^{k-1} = {N \over 57}$, which is again a natural number, let's call it $M = {N \over 57}$. We have $57M = M + x_k10^k$ or $56M = x_k10^k$. In the last equality, left part is divisible by $7$, $10^k$ isn't divisible by $7$ thus $x_k$ is divisible by $7$ thus $x_k = 7$. Finally, $8M = 10^k$ so $k \ge 3$ (for right part to be divisible by $8$) and $8M = 1000* 10^{k-3}$, thus $M = 125*10^{k-3}$ and $N = 57M = 7125*10^{k-3}$. $\endgroup$ – Abstraction Aug 10 '16 at 16:16
  • $\begingroup$ Thanks a lot, I got it now. $\endgroup$ – Akshar Gandhi Aug 10 '16 at 16:19

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