What do we get if we add the digits of the integers from 1 to 99?

Question

The function $S(n)$ is defined for positive integers $n$ by $S(n) = \text{sum of the digits of }n.$ For example, $S(723) = 7+2+3=12$ . The sum $S(1) + S(2) + S(3) + \dots + S(99)$ equals

$746$ ,

$862$ ,

$900$ ,

$924$ .

Suggestion

Could we lay out the $S$ values in a table?

How often does the digit $1$ appear in the numbers from $1$ to $99$ ?

What about the other digits?

Solution

Approach 1

We can lay out the digits we have to add in a table and then add up each column.

$0+0$	$0+1$	$0+2$	$0+3$	$\dots$	$0+9$
$1+0$	$1+1$	$1+2$	$1+3$	$\dots$	$1+9$
$2+0$	$2+1$	$2+2$	$2+3$	$\dots$	$2+9$
$\dots$	$\dots$	$\dots$	$\dots$	$\dots$	$\dots$
$9+0$	$9+1$	$9+2$	$9+3$	$\dots$	$9+9$
$1+2+\dots+9=45$	$45+10\times1$	$45+10\times2$	$45+10\times3$	$\dots$	$45+10\times9$

Adding up the column totals we have $45\times10 + 10\times1 + 10\times2 +\dots+10\times9 = 45\times10+10\times45 = 900$ so the answer is (c).

Approach 2

Since zeroes do not contribute to the sum, we can rewrite this as $S(1) + S(2) + S(3) + \dots + S(99) = S(00) + S(01) + \dots + S(99).$

Then in the $100$ two-digit numbers $00, \dots, 99$ , each digit $0,\dots ,9$ appears $20$ times ( $10$ times as the first digit, $10$ times as the second).

So $S(1) + S(2) + S(3) + \dots + S(99) = 20 \times (0 + 1 + \dots + 9) = 20 \times 10 \times \frac{(0 + 9)}{2} = 900,$ and the answer is (c).