Magic square - 2.0

Probability Level pending

The figure below shows a 3 × 3 3\times3 magic square.

If e = 2017 e=2017 , then find the maximum value of the sum of the nine numbers ( a , b , c , d , e , f , g , h , i a,b,c,d,e,f,g,h,i ).

Note: a + b + c = d + e + f = g + h + i = a + d + g = b + e + h = c + f + i = a + e + i = c + e + g a+b+c=d+e+f=g+h+i=a+d+g=b+e+h=c+f+i=a+e+i=c+e+g


The answer is 18153.

Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

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1 solution

Áron Bán-Szabó
Jul 29, 2017

If the sum of three numbers in the same row/column/diagonal is S S , then

( a + b + c ) + ( c + f + i ) + ( i + g + h ) + ( a + d + g ) = 4 S = 2 a + 2 c + 2 g + 2 i + b + d + f + h = 2 ( a + i ) + 2 ( c + g ) + ( b + h ) + ( d + f ) = 2 ( S − e ) + 2 ( S − e ) + ( S − e ) + ( S − e ) = 6 ( S − e ) = 6 S − 6 e (a+b+c)+(c+f+i)+(i+g+h)+(a+d+g)=4S=2a+2c+2g+2i+b+d+f+h=2(a+i)+2(c+g)+(b+h)+(d+f)=2(S-e)+2(S-e)+(S-e)+(S-e)=6(S-e)=6S-6e

4 S = 6 S − 6 e 4 S + 6 e = 6 S 6 e = 2 S e = 2 S 6 = S 3 \begin{aligned} 4S & = 6S-6e \\ 4S+6e & = 6S \\ 6e & = 2S \\ e & = \dfrac{2S}{6} \\ & =\dfrac{S}{3} \end{aligned}

So S = 3 ∗ S 3 = 3 ∗ e = 3 ∗ 2017 = 6051 S=3*\dfrac{S}{3}=3*e=3*2017=6051 and the sum of the nine numbers is 3 S = 18153 3S=\boxed{18153} .

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