The numbers $1, 2,..., 9$ are arranged in a $3 \times 3$ square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.

If the top left and the top right entries of the grid are $6$ and $2$, respectively, then the bottom middle entry is

We have a 3×3 magic square where numbers 1 through 9 are arranged so that each number appears exactly once and every row, column, and diagonal adds up to the same value.

Given information:

Top left = 6

Top right = 2

Find: Bottom middle entry

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In any magic square, all rows, columns, and diagonals have the same sum.

Since we're using numbers 1, 2, 3, 4, 5, 6, 7, 8, 9:

Total sum = $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$

Since we have 3 rows, and each row has the same sum:

Magic sum = $45 ÷ 3 = 15$

This means every row, column, and diagonal adds up to 15.

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In a 3×3 magic square using consecutive numbers, the center element is always the middle number.

Since we're using 1, 2, 3, 4, 5, 6, 7, 8, 9:

Center element = 5

The center appears in 4 sums (1 row + 1 column + 2 diagonals), and mathematical properties of magic squares require the center to be the median value.

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Let's set up our grid:

6	?	2
?	5	?
?	?	?

Since the top row must sum to 15:

$6 \ +$ Top middle $+\ 2 = 15$

$\Rightarrow$ Top middle $= 7$

6	7	2
?	5	?
?	?	?

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Now we look at the middle column:

$7 + 5 \ +$ Bottom middle $= 15$

$\Rightarrow$ Bottom middle $= 3$

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Let's check our answer:

6	7	2
?	5	?
?	3	?

Top row: $6 + 7 + 2 = 15$

Middle column: $7 + 5 + 3 = 15$

Therefore, the bottom middle entry is 3.

When we solve magic square problems we calculate the magic sum (total ÷ number of rows), use the center element property, work systematically with known values, and always verify our intermediate steps.