Sum of sum

We note that the sum of digits of 4-digit numbers is the sum of sum of individual unit, ten, hundred, and thousand digits. Let $n=\overline{d_3d_2d_1d_0}$ , then we have:

$\sum_{n=1}^{1000} f(n) = \sum_{n=1}^{1000} d_{0n} + \sum_{n=1}^{1000} d_{1n} + \sum_{n=1}^{1000} d_{2n} + \sum_{n=1}^{1000} d_{3n}$

• For the unit digit $d_{0n}$ , we have $1+2+3+4+5+6+7+8+9 = 45$ for every 10, since we have 100 $\times$ 10 in 1000, $\displaystyle \implies \sum_{n=1}^{1000} d_{0n} = 100 \times 45 = 4500$ .
• Similarly, for the ten digit $d_{1n}$ , we have 450 for every 100, since we have 10 $\times$ 100 in 1000, $\displaystyle \implies \sum_{n=1}^{1000} d_{1n} = 10 \times 450 = 4500$ .
• For the hundred digit $d_{2n}$ , we have 4500 for every 1000, since we have 1 $\times$ 1000 in 1000, $\displaystyle \implies \sum_{n=1}^{1000} d_{2n} = 1 \times 4500 = 4500$ .
• For the thousand digit $d_{3n}$ , we have only $f(1000) = 1$ , $\displaystyle \implies \sum_{n=1}^{1000} d_{3n} = 1$ .

Therefore, $\displaystyle \sum_{n=1}^{1000} f(n) = 4500+4500+4500+1 = 13501$ $\implies \boxed{\text{None of the other choices.}}$

The question asks for the sum of the sum of the digits of integers from 1 to 1000.

This sum won't change, if we add the 0. So let's write all numbers from 0 to 999 (1000 numbers) with three digits (e.g. 0 = 000, 7 = 007, 12 = 012, 789 = 999) as a first step.

Since all the 10 digits from 0 to 9 occur equally likely (each of them 1000 / 10 = 100 times at each place (units, tens, hundreds)), therefore the sum of the digits for one place is:

$100 × (0 + 1 +2 + ... + 8 + 9) = 100 × \frac {10 × (0 + 9)}{2} = 4500$

The sum of sum of digits for integers from 0 to 999 (or from 1 to 999) is three times as much as this (3 places):

3 × 4500 = 13500

And the sum from 1 to 1000 is just our sum from 1 to 999 and the sum of the digits of 1000 (1 + 0 + 0 + 0 = 1):

$13500 + 1 = \boxed {13501}$

Hence, our answer is "None of the other choices".

class MyClass {

public static void main(String[ ] args) {


     int sum=0;


      for(int n=1;n<=1000;n++){


      int b=Integer.toString(n).length();


                   for(int a=0;a<=b-1;a++){


                       sum+=(int)(n/(Math.pow(10,a)))%10;


                   } 


      }


      System.out.println(""+sum);


}

}

OUTPUT 13501