i need the answer!! PLEASE help!!?

Question

if the sum of 2000 consecutive integers is 1000 then the sum of the digits of these 2000 integers is__________"
any ideas people... I dont even understand this thing!! but if you know how to do it and can eplain the process and the answer it would be greatly appreciated

TO NONEDESCRIPT:
would x be 1? do u no what the answer is or are u just pulling my leg??
thanx to everyone for trying... please continue!!

YES YES YEs that last answer helped soooo much!!! gosh ur like a genious!!! lol

Answer 1

First figure out x + (x+1) + (x+2) + ... + (x+1999) = 1000
2000x + (1+2+3+..+1999) = 1000

[If you add 1+2+3+..+1999 with 1999+1998+1997+..+1 you get
2000+2000+2000+...+2000 then you have to divide by two
So, 1+2+3+..+1999 = 1999*2000/2]

2000x+(1999*2000/2) = 1000
2x + 1999 = 1
x = -1998/2
x = -999

So, the consecutive integers are:
-999, -998, -997, ..., 999, 1000

Now you have to figure out the digits.

Let's first figure out the sum of the digits from 1 to 999

If you have one 1 in a number from 1 to 999, it will be of the form xx1, x1x, or 1xx. The other digits would have to be 0 and 2-9 which are 9 numbers for one x and 9 numbers for the other x. This give 3*9*9 different ways to have just one 1 in a number from 1 to 999 = 243 instances of a number from 1 to 999 with only one 1.

If you have two 1's in a number from 1 to 999, it will be of the form x11, 1x1, or 11x. The x can be 0 or 2-9 which are 9 digits, so the total numbers with two 1's is 3*9=27

If you have three 1's in a number from 1 to 999, it has to be 111.

So... If you add up all those 1's, it is 1*(243+2*27+3*1) = 1 * 300
But the same logic holds for 2 through 9, so the sum of the digits from 1 to 999 are:
1*300+2*300+3*300+...+9*300 =
(1+2+3+...+9)*300=
45*300 = 13500

Now, that was for 1 through 999. The same holds true for -999 to -1. And the sum of the digits of 1000 = 1.

13500 + 13500 + 1 = 27001

So, the answer is 27001

Answer 2

you do solve for x but an easy way to do it is start with x-1000 then x-999 .... all the way to x+999. That gives you the 2000 consecutive integers and when you add them up you see that adding (x-999)+(x+999) the 999s cancel out and you are left with 2x. you do that for all the numbers and you end up with (2*999)x because each pair gives you 2x and there are 999 pairs. and you also have x and x-1000 left over. In the end you will get the equation
(2*999)x + x + x-1000 = 1000 and you want to solve for x
2000x = 2000
x = 1
so you know the 2000 consecutive integers start with -999, -998, ... all the way to 1000.
Then you just add up the digits of those 2000 numbers, that is the only part i'm not sure about because you have negative numbers.
I hope that somewhat helps.

Answer 3

It is say that 2000 integers (meaning numbers) added up = 1000.

so for example we can say this "the sum of 5 intergers is 10"
2+2+2+2+2 =10

As far as the sum of the digits of these 2000 intergers is....i have no clue what they are asking

Answer 4

isit .5? ifnot i dont getit