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Senior maths challenge question

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,xx, define A(x)A(x) to be: x- the sum of the squares of the digits of x.
What is the maximum possible value of A(x)A(x)?"

I got as far as defining A(x)=100a+10b+c−(a2+b2+c2) 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.
Reply 1
Avatar for SimonM
SimonM
18
Hint: Maximize 100a−a2100a - a^2 and others separately.

Spoiler



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

PS. Good luck, and congratulations for beating Jersey :smile:
Reply 2
Avatar for Dominorum
Dominorum
OP
For "less "nice"" read "cambridge standard"
Reply 3
Avatar for SimonM
SimonM
18
Less elementary
Sorry if I'm missing something silly here, but wouldn't the solution just be 9^2 + 9^2 + 9^2 = 243?
Reply 5
Avatar for Dominorum
Dominorum
OP
its x minus the sum of the squares
Wow, I need my eyes testing. :colondollar:

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

Spoiler

Reply 7
Avatar for Dominorum
Dominorum
OP
Does anyone agree with my answer?

Spoiler



EDIT: I agree marcus
ok i think this is the right answer (where 1≤a≤9,0≤b≤9,0≤c≤9 1 \leq a \leq 9 , 0 \leq b \leq 9 , 0 \leq c \leq 9

A(x)=100a+10b+c−(a2+b2+c2)[br][br]=100a−a2+10b−b2+c−c2[br][br]=a(100−a)+b(10−b)+c(1−c)A(x)= 100a + 10b + c - (a^2 +b^2+c^2) [br] [br] =100a - a^2 + 10b - b^2 + c - c^2[br][br] = a(100 - a) + b(10 - b) + c(1 - c)

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

Max value of 100a−a2100a - a^2 occurs when a=50 a = 50 technically but this is not in the given range of values for a a- sketching will show 99 is the max value of a a

Max value of b b occurs when b=5b=5
ddb(10b−b2)=10−2b\frac{d}{db}(10b -b^2) = 10-2b ( At max when 10−2b=0,b=5)10-2b=0, b=5)

sketching again will show max value of c−c2 c - c^2 is at max when c=0c=0 or 11

when a=9,a(100−a)=9×91=819 a=9 , a(100-a)= 9 \times 91 = 819
when b=5,b(10−b)=5×5=25 b=5 , b(10-b)= 5 \times 5 = 25
when c=0c= 0 or 1c(c−1)=01 c(c-1)= 0

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

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

Spoiler



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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