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The value of a three digit number increases by 108 when the first digit is placed at
the last (the 2nd value becomes the 1st and the 3rd value becomes the 2nd).

The value of the number increases by 189 from the main number when the last number is
placed at the first (the 1st value becomes the 2nd and the 2nd value becomes the 3rd).. Determine the summation of all the possible numbers.

e.g. 123
After replacing the 1st digit, the number is 231
(231 - 123 = 108)
After replacing the last digit, the number is 312.
(312 - 123 = 189)

First of all suppose that the three digits of the number are $x$, $y$ and $z$ ($1le xle , 0le y, zle9$). Then the number is $100x+10y+z$.
The number formed by putting first digit ($x$) at the last is $100y+10z+x$. So by given condition:

$$100y+10z+x=100x+10y+z+108$$
$$Rightarrow -99x+90y+9z=108$$
$$Rightarrow -11x+10y+z=12 ldots(i)$$

Also the number formed when last digit ($z$) is put at the first, is $100z+10x+y$, hence by given condition:

$$100z+10x+y=100x+10y+z+189$$
$$Rightarrow -90x-9y+99z=189$$
$$Rightarrow -10x-y+11z=21 ldots(ii)$$

Multiplying $(i)$ by $11$,
$$-121x+110y+11z=132 ldots(iii)$$

Subtracting $(iii)$ from $(ii)$,
$$111x-111y=-111$$
$$Rightarrow x-y=-1$$
$$Rightarrow y = x+1 ldots(iv)$$
Now putting in $(i)$,
$$-x+z=2$$
$$Rightarrow z = x+2 ldots(v)$$

So the sum of digits is $x+y+z=3x+3$ where $1le xle7$ (Why??!!). Find possible values by placing the values of $x$ in the sum.

First note that any number $ge 100$ with a $0$ in the middle won't work, since moving the first digit to last will decrease the number. Having the $0$ at the end similarly fails when we try to move the last digit to first.

This leaves numbers with all digits nonzero or with an initial zero and two nonzero digits following. If $n$ is any number of the former type and it satisfies the conditions given I, the problem, so does $n-111$ (why?), and we can step down to a number in ${000,001,...,110}$. The numbers from $100$ to $110$ have $0$ digits in the wrong place, and numbers from $000$ to $099$ are always multiplied by $10$ when the first ($0$) digit is moved to last. This means the only possible solution in ${000,001,...,110}$ is $012$ and thus the only numbers to enter the requested sum itself are $equiv 12bmod 111$. Dropping $012$ from the list as it is conventionally considered the two-digit number $12$, we are left with $123,234,...,789$ (why can't we go higher than $789$?). The sum of these numbers can than be worked out accordingly.