the sum of 5 consecutive whole numbers is 2005. what is the sum of all the digits of these 5 numbers

I’m trying to study and I have to 4 choices none of these are the answer

I misread it, thought it said ODD numbers

Well, let's get the clown car rolling and solve this math puzzle in a hilarious way! To sum up the digits of those five consecutive whole numbers that add up to 2005, we'll need a bit of mathematical trickery.

Let's assume the middle number is "x." Then, the five consecutive whole numbers would be x - 2, x - 1, x, x + 1, and x + 2.

Now, if we add up all these numbers, we get:

(x - 2) + (x - 1) + x + (x + 1) + (x + 2) = 5x

Since the sum of these five numbers equals 2005, we can set up the equation:

5x = 2005

After some mathematical juggling, we find that x = 401. Ta-da!

Now, let's sum up the digits of these five numbers. The numbers are 399, 400, 401, 402, and 403. The sum of all the digits is:

3 + 9 + 9 + 4 + 0 + 0 + 4 + 0 + 1 + 4 + 0 + 2 + 4 + 0 + 3 = 43!

So, the sum of all the digits of these five numbers is a joyful 43. Keep smiling and clowning around with your mathematical adventures!

To find the sum of the digits of the five consecutive whole numbers whose sum is 2005, we first need to find the five numbers.

Let's label the unknown number as "x". According to the given information, the five consecutive whole numbers can be represented as: x, x+1, x+2, x+3, and x+4.

The sum of these five numbers is given as 2005:
x + (x+1) + (x+2) + (x+3) + (x+4) = 2005

We can simplify the equation:
5x + 10 = 2005
5x = 1995
x = 399

Now that we have found the value of "x" (which is 399), we can find the five consecutive numbers:
x = 399
x+1 = 400
x+2 = 401
x+3 = 402
x+4 = 403

To find the sum of the digits of these five numbers, we can add the digits of each number individually and then sum them up:

399: 3 + 9 + 9 = 21
400: 4 + 0 + 0 = 4
401: 4 + 0 + 1 = 5
402: 4 + 0 + 2 = 6
403: 4 + 0 + 3 = 7

Now sum up the individual sums:
21 + 4 + 5 + 6 + 7 = 43

Therefore, the sum of the digits of the five consecutive numbers is 43.

so you have,

(n-2)+(n-1)+n+(n+1)+(n+2)=2005
5n=2005
n=401
nos are,
399,400,401,402,403

n-4 + n-2 + n + n+2 + n+4 = 2005

5 n = 2005

n = 401
so the list

397
399
401
403
405
so
2*3 + 3*4 + 2*9 + 25
= 61