$7^3\equiv 1\pmod 9\Longrightarrow 7^{1996}\equiv 7\pmod 9$. Note that the number modulo $9$ doesn't change on doing the process described. So, if the left number must have all different digits, then it cannot be $7\pmod 9$ and hence, it has two equal digits.

Goutham wrote: Note that the number modulo $9$ doesn't change on doing the process described. Can you explain further this part, please.

Obel1x wrote: Can you explain further this part, please. Sum of the digits of a number $\equiv$ the remainder of that number $\pmod 9$.

Obel1x wrote: Goutham wrote: Note that the number modulo $9$ doesn't change on doing the process described. Can you explain further this part, please. Suppose we have $2354$, we can see that $2+354$ is the same modulo $9$. The proof is easy. Suppose a number is $\overline{ab}$ where $a$ is the leading digit and $b$ is the number formed by the remaining digits. Then we can note that $\overline{ab}=10^k\times a+b\equiv a+b\pmod 9$ where $k$ is the one less than the number of digits in $\overline{ab}$. So, we have the result. Thanks to AHP for explaining as well.

Quote: Starting with the number $7^{1996}$, remove its first digit and then add that digit to the rest of the number. This process continues until the remaining number has $10$ digits. Show that $2$ of the digits of the remaining number are equal. First, note that $7^3 \equiv 1 \bmod9,$ so $7^{1996} \equiv (7^3)^{665} \cdot 7^1 \equiv 1^{665} \cdot 7^1 = 1 \cdot 7 \equiv 7 \bmod9.$ Claim A number $n$ is congruent to the sum of its digits mod $9$ since a number $\overline{a_k a_{k-1} a_{k-2} \cdots a_2 a_1}$ is equal to $a_k \cdot 10^{k-1} + a_{k-1} \cdot 10^{k-2} + a_{k-2} \cdot 10^{k-3} \cdots a_2 \cdot 10^1 + a_1 \cdot 10^0,$ which is congruent to $a_k \cdot 1^{k-1} + a_{k-1} \cdot 1^{k-2} + a_{k-2} \cdot 1^{k-3} \cdots a_2 \cdot 1^1 + a_1 \cdot 1^0 \equiv a_k + a_{k-1} + a_{k-2} + \cdots + a_2 + a_1 \bmod9$ since $10 \equiv 1 \bmod9.$ Suppose we have a number $a+b$ where $a$ is the leading digit multiplied by $10^{k-1}$ where $k$ is the number of digits that the number has and $b$ is the number formed by the remaining digits. By the claim, we know that $a \cdot 10^{k-1} + b \equiv a \cdot 1^{k-1} + b \equiv a + b \bmod 9,$ so a step of the process will not affect the number mod $9.$ Thus the remaining $10$ digit number is also congruent to $7^{1996} \equiv 7 \bmod9.$ Assume, for the sake of contradiction, that the ten-digit number has no $2$ equal digits. Since there are only $10$ unique digits in base $10,$ the ten-digit number has $1$ of each digit. Then its remainder when divided by $9$ will be $1+2+3+4+5+6+7+8+9+0 \equiv \dfrac{9(10)}{2} \equiv 45 \equiv 0 \bmod9.$ However, if the ten-digit number is congruent to $0 \bmod 9,$ then it cannot be a result of the process, since we already know that the ten-digit number must be congruent to $7 \bmod 9.$ Thus, by contradiction, the ten-digit number must have at least two digits equal to each other, as desired. $\blacksquare$

The important point to note here was that the number obtained after each process was invariant modulo $9$.