If N = 1 + 11 + 111 + 1111 + …… + 111111111, then what is the sum of the digits of N?
45
Stacking 1, 11, 111, …, 111111111 right-aligned, the columns from units leftward contain 9, 8, 7, 6, 5, 4, 3, 2, 1 ones respectively. So:
\[N=123{,}456{,}789\]
Sum of digits \(=1+2+3+\cdots+9=\dfrac{9\cdot 10}{2}=\mathbf{45}\).
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