The sum of the digits used to write the sum 10 + 11 + 12 + 13 is 10.
What is the sum of the digits used to write the sum of the integers from 1 to 110, inclusive?
1+0+1+1+1+2+1+3 = 10
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 9*10/2 = 45
1+0 + 1+1 + 1+2 + 1+3 + 1+4 + 1+5 + 1+6 + 1+7 + 1+8 + 1+9 = 10 + 9*10/2 = 55
2+0 + 2+1 + 2+2 + 2+3 + 2+4 + 2+5 + 2+6 + 2+7 + 2+8 + 2+9 = 2*10 + 9*10/2 = 65
Sum of digit of next 10 numbers (30-39) = 3*10 + 9*10/2 = 75
Sum of digit of next 10 numbers (40-49) = 4*10 + 9*10/2 = 85
Sum of digit of next 10 numbers (50-59) = 5*10 + 9*10/2 = 95
Sum of digit of next 10 numbers (60-69) = 6*10 + 9*10/2 = 105
Sum of digit of next 10 numbers (70-79) = 7*10 + 9*10/2 = 115
Sum of digit of next 10 numbers (80-89) = 8*10 + 9*10/2 = 125
Sum of digit of next 10 numbers (90-99) = 9*10 + 9*10/2 = 135
Sum of digit of next 10 numbers (100-109) = 1+0+0 + 1+0+1 + .... + 1+0+9 = 10 + 9*10/2 = 55
Sum of digits of 110 = 2
Sum of digits used to write the sum of the integers from 1 to 110, inclusive = 45 + 55 + 65 + 75 + 85 + 95 + 105 + 115 + 125 + 135 + 55 + 2 = 10/2 (45+135) = 900 + 55 + 2 = 957
IMO D