I'm having a hard time with this proof. Perhaps I'm not understanding the question correctly. Is what they're saying to detatch the first digit then add it to the total, e.g $567 \rightarrow 5+67 = 72$ until we get a sum that is $10$ digits long. But this is different from the digital sum so I don't see how the ideas from the digital sum apply to this operation. And beyond that, in the proof, they say $7^3\equiv1$ mod $9$ implies $7^{1996}\equiv 7^1$ mod $9$. How does this follow. Multiplying in modular arithmetic allows us to get to $7^4\equiv 7$ mod $9$ from $7^3\equiv1$ mod $9$, and I see that $4|1996$ but I'm still unsure of how this implication is made.
Thanks
- Solution
Claim:$\;$After each transition, the new number is congruent to the old one, mod $9$.
Let say at a given stage, we have the number $x$, with leading digit $a$.
Then $x$ has the form $x=a10^k+b$, for some positive integers $k,b$.
The next transition yields the number $y=b+a$, hence \begin{align*} x&=a10^k+b\\[4pt] &\equiv a10^k+b\;(\text{mod}\;9)\\[4pt] &\equiv a(1)^k+b\;(\text{mod}\;9)\\[4pt] &\equiv a+b\;(\text{mod}\;9)\\[4pt] &\equiv y\;(\text{mod}\;9)\\[4pt] \end{align*} as claimed.
Thus, at the end, the final $10$-digit number is congruent to the starting number, mod $9$.
But any nonnegative integer is congruent, mod $9$ to the sum of its digits.
If the final $10$-digit number had $10$ distinct digits, it would be congruent, mod $9$, to $$0 + 1 + 2 +\cdots + 9=45$$ which is congruent to $0$, mod $9$, contradiction, since $7^{1996}$ is not a multiple of $9$.
Note:
There is a subtle point here which the author may have missed . . .
How do we know that the process ends with a $10$-digit number (with nonzero leading digit)?
One way to show this is to compute $7^{1996}$ mod $10^{\,10}$, and see that the result is an actual $10$-digit number. Then argue that the number of steps is insufficient to change, at any stage, the digit which is tenth from the right. But computing $7^{1996}$ mod $10^{\,10}$ is a lot of work.
Without that verification, the first transition which yields a final number less than $10^{10}$ might actually yield a number with less than $10$ digits, in which case, it's conceivable that the final number has all distinct digits.