How do you find the sum of digits appearing in 1, 2, 3, 4, ..., 999?

Question

Do not find the actual sum like 1+2+3+4...+997+998+999
Rather find the sum of the digits appearing in the number.
For example the contribution of 86 is 8+6

Answer 1

Everyone is missing that you want the sum of the sum of the digits, not the numbers!

You can think of it this way:
000
001
002
003
004
005
...
997
998
999

All you want is the sum of all the *digits* shown, not their actual value.

You have 1000 numbers (I'm including 000 and all leading zeroes, just to make it complete, but it doesn't affect the sum).

In the units column, you have 0,1,2,3,4,...,8,9 repeating in that pattern 100 times. So each digit will appear exactly 100 times.
100(0+1+2+...+8+9) = 100 * 45 = 4500
Your total is 4500 from the units column

In the tens column you have ten 0s, ten 1s, ten 2s ... ten 9s. This pattern repeats 10 times... So again each digit appears exactly 100 times:
100(0+1+2+...+8+9) = 4500

And finally in the hundreds column you one hundred 0s, one hundred 1s, ..., one hundred 9s. Once again each digit appears 100 times = 4500

So the sum of the digits appearing will be:
3 x 4500 = 13,500

Edit: Indeed there are other ways (as Geezah has shown) to get to this same number.

Answer 2

I see a number of people failed to read the first line of the question detail! Anyway, here's one way to do it:

You have single-digit numbers, doube-digit numbers, and triple-digit numbers. Consider how many times each of the nine non-zero digits show up in each of these groups.

The sum of 1 through 9 is just 9(10)/2 = 45.

For the double digits, notice that the sum of all the digits from 10-19 is (1+0) + (1+1) + (1+2) + ... + (1+9). In other words, it's 10*1 (for the ten leading digits) plus 45 (for the sum of 0 through 9. Likewise, the digits in 20-29 add up to 10*2 + 45, and the digits of 30-39 add up to 10*3 + 45, etc. So the sum from all the double-digit numbers is:
(10+45) + (20+45) + (30+45) + ... + (90+45). =
10(1+2+3+...+9) + 9*45 =
10*45 + 9*45 = 19*45 = 855

Combining this with the single-digit numbers gives 900.

Now for the three-digit numbers (100 through 999), this is the same as sticking a single digit in front of a two-digit number. We know that the sum of the individual digits in "00" through "99" is 900. So if we were to write the numbers 100 through 199 in a column, and split the column into two smaller columns with "1"s all down the left and double-digit on the right, we know that the right adds up to 900 and the left adds up to 100*1 = 100. So similarly, the 3-digit numbers add up to this:
(100*1 + 900) + (100*2 + 900) + (100*3 + 900) + ... + (100*9 + 900) =
100(1+2+...+9) + 9*900 =
100(45) + 9*900 = 12600

Adding this to the 900 from before brings the total to 13,500.

Answer 3

The above answers all give the way to add the numbers. But that's not what you want. Your question clearly states that you want to add the digits themselves. This is an interesting twist on the question, for which you deserve a star.

But how to do it? Hmmm... let's look for patterns.
The sum of the digits in 1..9 is easy (45, same as sum of 1..9).
The digits in the next 10 numbers, 10..19, will sum to 45 + 10*1.
Similarly the digits in 20..29 will sum to 45 + 10*2.
This should cover you up to 99.

Then you need to apply the same pattern-searching again for the groups 100..199, 200..299 etc.

I hope that helps for now. If I get any better/further ideas, I'll edit my answer.

Edit: well done Geezah, you got there before me.

Answer 4

For the sum of the digits from 1 to 9 we have 1 + 2 + ... + 9 =45. For 10 to 19, we have 10(1) + 45. For 20 to 29 we have 10(2) + 45. And so on, until for 990 to 999, we have 10(99) + 45.

Our complete sum is 45(1000) + 10(1 + 2 + ... + 99) = 45000 + 10*(99*100)/2 = 45000 + 49500 = 54000.

Answer 5

take the middle number and multiply it by the number of terms? just a guess but it works for
1+2+3+4+5+6+7 = 28
4 x 7 = 28

and 1+2+3+4+5+6+7+8+9 = 45
5 x9

and 1+2+3+4+5+6+7+8+9+10+11 = 66
6 x 11 = 66

damn to slow, hes got a fancy way of sayin it:P.. oh well

Answer 6

sum = 1/2 (1 + 999)) 999)
= 499500