i need to know fast (by 5 oclock tommorow eastern time) a faster way to do this problem. We need an alternate way other than adding 200 times. Help would be appreciated, thank you!!!!!! :Find the sum of the digits of the first 200 even natural numbers.

2. Originally Posted by jarny
i need to know fast (by 5 oclock tommorow eastern time) a faster way to do this problem. We need an alternate way other than adding 200 times. Help would be appreciated, thank you!!!!!! :Find the sum of the digits of the first 200 even natural numbers.
(n^2 + n)/2 will come in handy; there are other ways to solve this too.

S_n = n(2a_1+(n-1)d)/2

a_1 is the first term; d is the difference between the two terms

Sn=200(2*2+(199)2)/2

= 40,200

OR:

(n^2 + 200)/2 = 20,100, but you want the even numbers;

Thus, multiply by 2 = 40,200

3. Hello, jarny!

Here's a rather "primitive" approach . . .

Find the sum of the digits of the first 200 even natural numbers.

In a set of any ten consecutive natural numbers,
. . the even ones end in: $\displaystyle 2,4,6,8,0$
Their sum is: $\displaystyle 20$

Among the first 200 natural numbers, there are 20 such sets-of-ten.
. . Hence, the unit-digits add up to: $\displaystyle 20\times20 = \boxed{400}$

Now consider the ten-digits.

In the first hundred numbers, they are:
. . $\displaystyle 0,0,0,0,0,\;1,1,1,1,1,\;2,2,2,2,\;\hdots\;9,9,9,9, 9$
Their sum is: $\displaystyle 5 \times (1 + 2 + 3 + \hdots + 9) \:=\:5(45) \:=\:225$

Since this happens again in the second hundred numbers,
. . the ten-digits total: $\displaystyle 2 \times 225 \:=\:\boxed{450}$

Now, the hundred-digits.
. . They are all $\displaystyle 0$ in the first 99 numbers.
. . They are all $\displaystyle 1's$ in the next 100 numbers.
. . Finally, the "200" has a $\displaystyle 2$.
Hence, the hundred-digits add up to: $\displaystyle 100 + 2 \:=\:\boxed{102}$

Therefore, the total of all the digits is: $\displaystyle 400 + 450 + 102 \:=\:\boxed{952}$

4. ## whos right

can somebody help? i dont know whos right, the first one i dont know but soroban, did you do the first 200 or 200 even ones and would it go up to 400 then?

5. ## trying to find solution but can't

Soroban, you went up to 200 but it would be up to four hundred right? so it would be 0 as the third digit in the first 49 a 0, next 50 a 1, third 50 a 2, fourth 50 a 3, then last one 4 right? could you reexplain the work then?

6. When you said you wanted the sum of the digits, do you mean the actual digits, such as 2, 4, 6, 8 ,0, 2, 4, 6, 8, ... and so forth, or the actual numbers? To me it seems more realistic you'd want to know the numbers, in which case 40,200 is correct.

That is, 2 + 4 + 6 + ... + 400 = 40,200

7. Hello, jarny!

You're absolutely right . . .

The first 200 even numbers: $\displaystyle 2,\,4,\,6,\,8,\,\hdots\,400$

I'll try again . . . *blush*

In a set of any ten consecutive natural numbers,
. . the even ones end in: $\displaystyle 2,4,6,8,0$
Their sum is: $\displaystyle 20$

Among the first 400 natural numbers, there are 40 such sets-of-ten.
. . Hence, the unit-digits add up to: $\displaystyle 40\times20 = \boxed{800}$

Now consider the ten-digits.

In the first hundred numbers, they are:
. . $\displaystyle 0,0,0,0,0,\;1,1,1,1,1,\;2,2,2,2,\;\hdots\;9,9,9,9, 9$
Their sum is: $\displaystyle 5 \times (1 + 2 + 3 + \hdots + 9) \:=\:5(45) \:=\:225$

Since this happens again in the 100's, 200's and 300's,
. . the ten-digits total: $\displaystyle 4 \times 225 \:=\:\boxed{900}$

Now, the hundred-digits.
. . They are all $\displaystyle 0$ in the first 99 numbers.
. . They are all $\displaystyle 1$ in the next hundred numbers.
. . They are all $\displaystyle 2$ in the next hundred numbers.
. . They are all $\displaystyle 3$ in the next hundred numbers.
. . Finally, the "400" has a $\displaystyle 4$.
Hence, the hundred-digits add up to: $\displaystyle 100 + 200 + 300 + 4 \:=\:\boxed{604}$

Therefore, the total of all the digits is: $\displaystyle 800 + 900 + 604 \:=\:\boxed{2304}$

Hope I got it right this time . . .

8. Originally Posted by jarny
Soroban, you went up to 200 but it would be up to four hundred right? so it would be 0 as the third digit in the first 49 a 0, next 50 a 1, third 50 a 2, fourth 50 a 3, then last one 4 right? could you reexplain the work then?
Hello,

you are right but I presume it's too late now, but nevertheless...:

Do the same method as Soroban has explained.
You'll get:
1. Sum of all one-digits: 40 * 20 = 800
2. Sum of all ten-digits: 4 * 225 = 900
3. Sum of all hundred-digits: 100 * (1+2+3) +4

Total: 2104

EB