Sums of digits

I am sure there are several ways to solve.

One way:

0-9 sum will be 45

Each of 10 increments will be 10 (beacuse of the 1's 2's 3's .... 9's of the 2 digit numbers) more than the previous one

a1= 45 a10 = 135

sum = n/2(a1+an)
= 10/2 (45+135)

= 900

20 x 9 x 10 / 2 = 900


Here is the explanation:


Answer: C[/quote]

from 1-9 : there will 1 set of (1-9)

after that each ten digits will have 1 set of (1-9) and 10 of that digit

so, we will have 10 x ( 1+2+3+...+9)
and also
10x1 + 10x2 + .... + 10x9 = 10 ( 1+2+3+...+9)

So in total we have 20 x ( 1+2+...+9)

20 x 9 x 10 / 2 = 900

I am sure I did not do it the fastest way...

1-9 is 45. And you have 10 1-9's so to speak.

So thats 450. Of course you are still missing the 1's from 11, 12, 13....and the 2's from 21,22,23. So there are 9 1's, 9 2's, 9 3's, etc. Which is really:

9(1+2+3...) or 9*45 which is 405. Thus so far we have 405+450. But, we are still missing 10, 20, 30, 40, 50...90. That, of course, also equals 45. So we have 45+405+450=900

Again, I don't think this is the fastest way, but it works.

Like the way of Logitech. Faster and clear. Thanks!