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. $$\newline$$ Given information: $\newline$ Top left = 6 $\newline$ Top right = 2 $\newline$ Find: Bottom middle entry $$\newline$$

$$\newline$$ In any magic square, all rows, columns, and diagonals have the same sum. $$\newline$$ Since we're using numbers 1, 2, 3, 4, 5, 6, 7, 8, 9: $$\newline$$ Total sum = $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$ $$\newline$$ Since we have 3 rows, and each row has the same sum: $$\newline$$ Magic sum = $45 ÷ 3 = 15$ $$\newline$$ This means every row, column, and diagonal adds up to 15. $$\newline$$

$$\newline$$ In a 3×3 magic square using consecutive numbers, the center element is always the middle number. $$\newline$$ Since we're using 1, 2, 3, 4, 5, 6, 7, 8, 9: $$\newline$$ Center element = 5 $$\newline$$ 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. $$\newline$$

$$\newline$$ Let's set up our grid: $$\newline$$

6	?	2
?	5	?
?	?	?

$$\newline$$ Since the top row must sum to 15: $$\newline$$ $6 \ +$ Top middle $ +\ 2 = 15$ $$\newline$$ $\Rightarrow$ Top middle $= 7$ $$\newline$$

6	7	2
?	5	?
?	?	?

$$\newline$$

$$\newline$$ Now we look at the middle column: $$\newline$$ $7 + 5 \ +$ Bottom middle $= 15$ $$\newline$$ $\Rightarrow$ Bottom middle $= 3$ $$\newline$$

$$\newline$$ Let's check our answer: $$\newline$$

6	7	2
?	5	?
?	3	?

$$\newline$$ Top row: $6 + 7 + 2 = 15$ $$\newline$$ Middle column: $7 + 5 + 3 = 15$ $$\newline$$ Therefore, the bottom middle entry is 3. $$\newline$$ 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.

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