Georgina listed the integers from \(1\) to \(13\):
\(1\), \(2\), \(3\), \(4\), \(5\), \(6\), \(7\), \(8\), \(9\), \(10\), \(11\), \(12\), \(13\)
She determined that the sum of all of the digits of the integers in this list is
\(1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+(1+2) + (1+3)= 55\)
She then challenged you to determine the sum of all of the digits of the integers from \(1\) to \(100\). What sum would you get?
In the integers from \(1\) to \(100\), each digit from \(0\) to \(9\) appears as the units digit of a number exactly ten times. For example, the digit \(1\) appears as the units digit in the numbers \(1\), \(11\), \(21\), \(31\), \(41\), \(51\), \(61\), \(71\), \(81\), \(91\), a total of \(10\) numbers. Therefore, the sum of all of the units digits is \[\begin{aligned} 10(1) &+ 10(2) + 10(3) + 10(4) + 10(5)+ 10(6) + 10(7) +10(8) +10(9) +10(0)\\ & = 10(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0)\\ & = 10(45)\\ &=450 \end{aligned}\]
Similarly, each digit from \(1\) to \(9\) appears as the tens digit of a number exactly ten times. For example, the digit \(1\) appears as the tens digit in the numbers \(10\), \(11\), \(12\), \(13\), \(14\), \(15\), \(16\), \(17\), \(18\), \(19\), a total of \(10\) numbers. The digit \(0\) appears as the tens digit of a number once. Therefore, the sum of all of the tens digits is \[\begin{aligned} 10(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + 0&=10(45)\\ &=450 \end{aligned}\]
The number \(100\) is the only number with a hundreds digit. We need to add \(1\) to our final sum.
Therefore, the sum of all of the digits of the integers from \(1\) to \(100\) is \(450 + 450 + 1 = 901\).