Sum of f ( x ) f(x)

If f ( x ) f(x) = sum of all the digits of x x , where x x is a natural number, then what is the value of:

f ( 101 ) + f ( 102 ) + f ( 103 ) + . . . + f ( 200 ) f(101) + f(102) + f(103) + ... + f(200)

1000 1111 999 1001 901

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1 solution

This question is primarily about recognizing the pattern.

In unit's place, we'll have all the numbers from 0 to 9 ten times each. In ten's place again, we'll have all the numerals from 0 to 9 ten times each. In hundred's place, we'll have 99 one's and 1 two's.

Hence, the sum will be = 10 ( 0 + 1 + . . . + 9 ) 10(0 + 1 + ... + 9) + 10 ( 0 + 1 + . . . + 9 ) 10(0 + 1 + ... + 9) + 99 99 x 1 1 + 1 1 x 2 2 = 1001 1001

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