# Math

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

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1. so you have,

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

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2. 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

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3. I misread it, thought it said ODD numbers

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

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