Magic square - 1.0

Probability Level pending

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

Which of the following statements is/are true?

\quad A : It is possible that ( a , b , c , d , e , f , g , h , i a,b,c,d,e,f,g,h,i ) are 2 , 2 , 3 , 3 , 3 , 4 , 4 , 4 , 4 2,2,3,3,3,4,4,4,4 in some order.

\quad B : It is possible that ( a , b , c , d , e , f , g , h , i a,b,c,d,e,f,g,h,i ) are 0 , 1 , 2 , 4 , 5 , 7 , 8 , 10 , 11 0,1,2,4,5,7,8,10,11 in some order.

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

Only A Both of A and B Neither A nor B Only B

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1 solution

Áron Bán-Szabó
Jul 29, 2017

\quad A : The sum of the numbers is 29 29 , which is not divisible by 3 3 . Since the 2 , 2 , 3 , 3 , 3 , 4 , 4 , 4 , 4 2,2,3,3,3,4,4,4,4 numbers are intgeres, it is not possible.

\quad B : Suppose yes. Then a + b + c = d + e + f = g + h + i = a + d + g = b + e + h = c + f + i = 0 + 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 3 = 16 a+b+c=d+e+f=g+h+i=a+d+g=b+e+h=c+f+i=\dfrac{0+1+2+4+5+7+8+10+11}{3}=16 . It is celar that since 10 + 11 > 16 10+11>16 , 10 10 and 11 11 can't be in the same row or in the same column. There are two ways to add two numbers to 11 11 to get 16 16 : 16 = 11 + 0 + 5 16=11+0+5 and 16 = 11 + 1 + 4 16=11+1+4 . There are two ways to add two numbers to 10 10 to get 16 16 : 16 = 10 + 1 + 5 16=10+1+5 and 16 = 10 + 2 + 4 16=10+2+4 . But 2 2 and 4 4 can't be in the same row/column, which is contradiction. So it is impossible.

Therefore neither of them is possible.

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