1, 2, 3, 4, 5, 6, 7, 8, 9

Probability Level pending

Victor's mathematics teacher gave him a 3 × 3 3 \times 3 grid filled with the numbers 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 1, 2, 3, 4, 5, 6, 7, 8 and 9 9 in an unspecified order, with each number appearing only once.

Given any row

a b c \begin{array}{ | l | c | r | } \hline a & b & c \\ \hline \end{array}

or column

d e f \begin{array}{ | l | c | r | } \hline d \\ \hline e \\ \hline f \\ \hline \end{array}

in the grid, Victor is allowed to perform the following operations:

Row:

  • a + k a+k , b k b-k , c k c-k

  • a k a-k , b k b-k , c + k c+k

Column:

  • d + k d+k , e k e-k , f k f-k

  • d k d-k , e k e-k , f + k f+k

with k k being a non-negative real number and that the numbers in the grid must always be greater than 0 0 .

Given that it is possible to perform row and/or column operations such that all the numbers in the grid are equal to a positive real number N N , find the maximum value that N N can take.


The answer is 4.

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1 solution

Joel Tan
Aug 29, 2014

Since k k is nonnegative, the sum of all the numbers would decrease affer every operation. (Consider a row or column with sum a + b + c a+b+c where a , b , c a, b, c are the entries in that row/column. The sum after the operation will be a + k + b k + c k = a + b + c k < a + b + c a+k+b-k+c-k=a+b+c-k<a+b+c )

Note that the original sum is 1 + 2 + 3 + . . . + 9 = 45 1+2+3+...+9=45 . Clearly the sum must decrease in the end when all numbers are equal from the above, as at leaat one operation is needed.It is obvious that N N is an integer, so the largest possible is 4. I will post a possible arrangement in the comments.

7 3 6

1 9 8

5 2 4

Operate on first and third rows with k=1

6 4 5

1 9 8

2 4 3

Operate on second row with k=2

6 4 5

3 7 6

2 4 3

Operate on first column with k=2

4 4 5

1 7 6

4 4 3

Operate on second row with k=3

4 4 5

4 4 3

4 4 3

Finally, operate on last column with k=1

4 4 4

4 4 4

4 4 4

Joel Tan - 6 years, 9 months ago

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Joel Tan - 6 years, 9 months ago

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k k need not be an integer.

Victor Loh - 6 years, 9 months ago

Consider

1 8 9

3 6 7

2 5 4

Operate on first, second, third rows with k=4, 2, 1 respectively:

5 4 5

5 4 5

3 4 3

Operate on first and third columns with k=1 for both:

4 4 4

4 4 4

4 4 4

Hence 4 is the maximum.

Joel Tan - 6 years, 9 months ago

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