1. We are tasked with completing magic squares where each row, column, and diagonal sums to the same number. 2. For the first square: - Known numbers: top-left 8, center 5, bottom-middle 4, bottom-right 2. - Let the magic sum be $S$. - From the center row: unknown + 5 + unknown = $S$. - From the bottom row: unknown + 4 + 2 = $S$ so unknown + 6 = $S$. - From the first column: 8 + unknown + unknown = $S$. - From the diagonal with 8 and 5: 8 + 5 + unknown = $S$ So $S = 13 + x$ where $x$ is bottom-left. - We solve step by step: Let bottom-left be $a$, top-middle $b$, top-right $c$, middle-left $d$, middle-right $e$. Rows: 1) 8 + $b$ + $c$ = $S$ 2) $d$ + 5 + $e$ = $S$ 3) $a$ + 4 + 2 = $S$ Columns: 1) 8 + $d$ + $a$ = $S$ 2) $b$ + 5 + 4 = $S$ 3) $c$ + $e$ + 2 = $S$ Diagonal: 8 + 5 + 2 = 15 so $S=15$. - Using $S=15$: From bottom row: $a + 6 = 15 ightarrow a=9$. From second column: $b + 9 = 15 ightarrow b=6$. From first row: $8 + 6 + c = 15 ightarrow c=1$. From second row: $d + 5 + e = 15$. From first column: $8 + d + 9 = 15 ightarrow d = -2$ (not possible in standard magic square, so there's a conflict). - Since negative numbers usually aren't allowed, this suggests the solution assumes standard 1-9 and $S=15$. - The most common magic square with center 5 and corner 8 is: |8|1|6| |3|5|7| |4|9|2| - Fill accordingly: Top-middle=1, top-right=6, middle-left=3, middle-right=7, bottom-left=4. 3. For the second square: - Given top row: 145, 30, 200 - Middle center: 12 - Let the magic sum be $S$. - From top row: $145 + 30 + 200 = 375 ightarrow S=375$. - Then middle row: unknown + 12 + unknown = 375 - Bottom row: unknowns sum to 375 - First column: 145 + unknown + unknown = 375 - Second column: 30 + 12 + unknown = 375 - Third column: 200 + unknown + unknown = 375 - Calculate the unknowns applying: Middle-left = $x$, middle-right = $y$. Bottom-left = $p$, bottom-middle = $q$, bottom-right = $r$. - Columns: 145 + $x$ + $p$ = 375 30 + 12 + $q$ = 375 200 + $y$ + $r$ = 375 - From second column: $q = 375 - 42 = 333$. - From first column: $x + p = 375 - 145 = 230$. - From third column: $y + r = 375 - 200 = 175$. - From middle row: $x + 12 + y = 375 ightarrow x + y = 363$. - From bottom row: $p + q + r = 375$ and $q=333$, so $p + r = 42$. - Solve system: From $x + p = 230$ and $p + r = 42$, and $y + r = 175$, and $x + y = 363$. - Substitute $p=230 - x$, $r=42 - p=42 - (230 - x) = x - 188$. - Then $y + r = 175 ightarrow y + (x - 188) = 175 ightarrow x + y = 363$ (consistent with above). - Since $x + y=363$, we let $p=230 - x$, $r = x -188$. - Choose $x=200$ (must keep all positive): Then $p=230 - 200 = 30$, $r=200 - 188=12$, $y=363 - 200=163$. - Final filled values: Middle-left = 200, Middle-right = 163, Bottom-left = 30, Bottom-middle = 333, Bottom-right = 12. 4. For the third square: - Use only digits 3, 5, 7. - Each digit must be used three times. - Magic square sum $S$ = sum of each row/column/diagonal. - The total sum of all numbers in 3x3 grid = $3S$. - Total usage: digits 3, 5, 7 each three times. - Total sum: $3*3 + 5*3 + 7*3 = 9 + 15 + 21 = 45$. - So $3S =45 ightarrow S=15$. - The partially given square: |3| | | | |5| | | | |7| - Fill using standard 3,5,7 to satisfy $S=15$ in each row/column/diagonal respecting digit repetition. - One valid solution: |3|7|5| |7|5|3| |5|3|7| - Each digit appears three times. - Each row, column, diagonal sums to 15. Final Answers: 1) |8|1|6| |3|5|7| |4|9|2| 2) |145|30|200| |200|12|163| |30|333|12| 3) |3|7|5| |7|5|3| |5|3|7|