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(C) To find the sum of all digits of numbers from $1$ to $100$,we consider numbers from $00$ to $99$ (since $00$ to $09$ are the same as $0$ to $9$ an

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$901$

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(C) To find the sum of all digits of numbers from $1$ to $100$,we consider numbers from $00$ to $99$ (since $00$ to $09$ are the same as $0$ to $9$ and $100$ contributes $1+0+0=1$).There are $100$ numbers,each having $2$ digits,making a total of $200$ digit positions.Each digit ($0$ through $9$) appears an equal number of times in the units and tens places.Number of times each digit appears $= \frac{200}{10} = 20$ times.Sum of digits from $0$ to $9 = 0+1+2+3+4+5+6+7+8+9 = 45$.Total sum of digits for numbers $00$ to $99 = 20 	imes 45 = 900$.Finally,adding the sum of digits of $100$ $(1+0+0 = 1)$:Total sum $= 900 + 1 = 901$.

The sum of all the digits of the numbers from $1$ to $100$ is

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