What is the sum of all digits which appear in all the integers from 10 to 100?

EXPLANATION

Correct Option (B)

To determine the sum of all digits appearing in integers from 10 to 100, the calculation is performed by segmenting the range into decades and including the final number.

The sum of individual digits from 0 to 9 is $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$ .
For each decade (e.g., 10-19, 20-29), the tens digit appears 10 times, and the units digits cycle through 0 to 9, contributing a sum of 45.

The sum of digits for each decade is calculated as follows:

10-19: $(1 \times 10) + 45 = 55$
20-29: $(2 \times 10) + 45 = 65$
30-39: $(3 \times 10) + 45 = 75$
40-49: $(4 \times 10) + 45 = 85$
50-59: $(5 \times 10) + 45 = 95$
60-69: $(6 \times 10) + 45 = 105$
70-79: $(7 \times 10) + 45 = 115$
80-89: $(8 \times 10) + 45 = 125$
90-99: $(9 \times 10) + 45 = 135$

The sum of digits for the number 100 is $1 + 0 + 0 = 1$ .

The total sum of all digits is the aggregate of these individual sums:

$55 + 65 + 75 + 85 + 95 + 105 + 115 + 125 + 135 + 1 = 856$ .

Incorrect Options:

Options (A) 855, (C) 910, and (D) 911 are incorrect as they do not correspond to the precisely calculated sum of all digits appearing in the integers from 10 to 100, which is 856.