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    I came across this question while preparing for the senior team maths challenge in a few days, can anyone give me a hand?

    The question is:

    "Given any three-digit number,x, define A(x) to be: x- the sum of the squares of the digits of x.
    What is the maximum possible value of A(x)?"

    I got as far as defining  A(x)=100a+10b+c-(a^2+b^2+c^2) where the number x is written abc. However I don't know where to go from there, does it involve differentiating of some sort? Anyway any help would be appreciated.
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    Hint: Maximize 100a - a^2 and others separately.

    Spoiler:
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    Complex the square


    Differentiating could be used, but it's less "nice"

    PS. Good luck, and congratulations for beating Jersey
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    For "less "nice"" read "cambridge standard"
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    Less elementary
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    Sorry if I'm missing something silly here, but wouldn't the solution just be 9^2 + 9^2 + 9^2 = 243?
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    (Original post by marcusmerehay)
    Sorry if I'm missing something silly here, but wouldn't the solution just be 9^2 + 9^2 + 9^2 = 243?
    you are missing something silly here :P
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    its x minus the sum of the squares
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    Wow, I need my eyes testing.

    EDIT: I think I know what the answer should be just from a small amount of logic rather than any extensive calculation.

    Spoiler:
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    951?
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    Does anyone agree with my answer?

    Spoiler:
    Show
    951


    EDIT: I agree marcus
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    (Original post by Dominorum)
    Does anyone agree with my answer?

    Spoiler:
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    951


    EDIT: I agree marcus
    Agreed, although x could also be 950. Hence, the answer is A(951) (which is equal to A(950)).
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    one way to do it is to consider each digit separately in your equation (as simonm suggested), and differentiate (as simonm advised against )
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    (Original post by Dominorum)
    Does anyone agree with my answer?

    Spoiler:
    Show
    951


    EDIT: I agree marcus
    yep :>

    i found it quicky by just quickly checking each digit, but a more systematic approach:

    maximise 100a - a^2, i.e. 2a = 100, a = 50

    so pick the closest digit you can to 50, i.e. 9

    maximise 10b - b^2, i.e. 2b = 10, b = 5

    so pick 5

    maximise c - c^2, i.e. 2c = 1, c = 1/2

    so pick 0 or 1
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    ok i think this is the right answer (where  1 \leq a \leq 9 , 0 \leq b \leq 9 , 0 \leq c \leq 9

    A(x)= 100a + 10b + c - (a^2 +b^2+c^2) 

         

            =100a - a^2 + 10b - b^2 + c - c^2



            = a(100 - a) + b(10 - b) + c(1 - c)

    Max value of  A(x) occurs at  a(100 - a), b(10 - b) , c(1 - c) are all at a maximum

    Max value of 100a - a^2 occurs when  a = 50 technically but this is not in the given range of values for  a- sketching will show 9 is the max value of  a

    Max value of  b occurs when b=5
    \frac{d}{db}(10b -b^2) = 10-2b ( At max when 10-2b=0,  b=5)

    sketching again will show max value of  c - c^2 is at max when c=0 or 1

    when  a=9 , a(100-a)= 9 \times 91 = 819
    when  b=5 , b(10-b)= 5 \times 5 = 25
    when c= 0 or 1 c(c-1)= 0

    So Max value of  A(x)= 819 + 25= 844
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    (Original post by marcusmerehay)
    Wow, I need my eyes testing.

    EDIT: I think I know what the answer should be just from a small amount of logic rather than any extensive calculation.

    Spoiler:
    Show
    951?
    Yes the number could either be 951 of 950 but your question asked for max value of A(x) which is 844
 
 
 
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