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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