Each digit of a 9-digit number is 1. It is multiplied by itself. What is the sum of the digits of the resulting number ?

To solve this elegantly, we should not try to multiply the 9-digit number. Instead, we use 'Inductive Reasoning' [12] to find the hidden pattern using smaller examples.\n\nLet's test numbers with fewer digits (Repunits [34]):\n\n1. **1-digit number (1):** \n $1 \\times 1 = 1$. \n Sum of digits = **1** (which is $1^2$).\n\n2. **2-digit number (11):** \n $11 \\times 11 = 121$. \n Sum of digits = $1 + 2 + 1 = 4$ (which is $2^2$).\n\n3. **3-digit number (111):** \n $111 \\times 111 = 12321$. \n Sum of digits = $1 + 2 + 3 + 2 + 1 = 9$ (which is $3^2$).\n\n**The Pattern:**\nFor a number made of $n$ ones, the sum of the digits of its square is always **$n^2$**.\n\n**Applying to the Question:**\nWe have a **9-digit number** ($n=9$).\nUsing our pattern, the sum of digits will be $9^2$.\n$9 \\times 9 = 81$.\n\n(This works perfectly because for $n=9$, the square is $12345678987654321$, where no carry-over disturbs the simple palindromic pattern.)