Problem statement: Fill the missing entries in five 2×3 connected-box grids using the property that in each adjacent $2\times2$ block the cross products are equal, i.e. $a_{top}\cdot b_{mid}=a_{mid}\cdot b_{top}$ and $a_{mid}\cdot b_{bot}=a_{bot}\cdot b_{mid}$. 1. Problem 1. We have the grid with left column $[?,6,5]$ and right column $[5,6,?]$. Apply the top $2\times2$ block equation $x\cdot 6=6\cdot 5$ where $x$ is the top-left unknown. Solve $x\cdot 6=30$ so $x=6$. Apply the bottom $2\times2$ block equation $6\cdot y=5\cdot 6$ where $y$ is the bottom-right unknown. Solve $6\cdot y=30$ so $y=5$. Answer: top-left $=6$ and bottom-right $=5$. 2. Problem 2. Grid has left column $[5,?,?]$ and right column $[5,6,8]$. Top block: $5\cdot 6=a\cdot 5$ so $a=6$ where $a$ is middle-left. Bottom block: $a\cdot 8=b\cdot 6$ so $6\cdot 8=b\cdot 6$ hence $b=8$ where $b$ is bottom-left. Answer: middle-left $=6$ and bottom-left $=8$. 3. Problem 3. Grid has left column $[8,8,24]$ and right column $[5,?,6]$ with single unknown $c$ at middle-right. Top block equation gives $8\cdot c=8\cdot 5$ so $c=5$. Bottom block equation gives $8\cdot 6=24\cdot c$ so $48=24\cdot c$ so $c=2$. The two adjacent $2\times2$ equations are inconsistent, so the puzzle as stated has no value satisfying both blocks simultaneously. A natural choice that preserves the top pair symmetry is $c=5$ which satisfies the top block and matches the top-right value. Answer (plausible): middle-right $=5$. 4. Problem 4. Grid has left column $[\tfrac{1}{2},6,?]$ and right column $[9,?,6]$ with unknowns $x$ (bottom-left) and $y$ (middle-right). Top block: $\tfrac{1}{2}\cdot y=6\cdot 9$ so $y=108$. Bottom block: $6\cdot 6=x\cdot y$ so $36=x\cdot 108$ hence $x=\tfrac{1}{3}$. Answer: middle-right $=108$ and bottom-left $=\tfrac{1}{3}$. 5. Problem 5. Grid has left column $[\tfrac{1}{2},?,1]$ and right column $[?,\tfrac{1}{2},40]$ with unknowns $a$ (middle-left) and $b$ (top-right). Top block: $\tfrac{1}{2}\cdot \tfrac{1}{2}=a\cdot b$ so $\tfrac{1}{4}=a\cdot b$. Bottom block: $a\cdot 40=1\cdot \tfrac{1}{2}$ so $40a=\tfrac{1}{2}$ hence $a=\tfrac{1}{80}$. From $a\cdot b=\tfrac{1}{4}$ we get $b=\tfrac{1}{4}/a=\tfrac{1}{4}\cdot 80=20$. Answer: middle-left $=\tfrac{1}{80}$ and top-right $=20$. Final answers summary. Problem 1: top-left $=6$, bottom-right $=5$. Problem 2: middle-left $=6$, bottom-left $=8$. Problem 3: middle-right $=5$ (inconsistent bottom block so top-symmetry choice). Problem 4: middle-right $=108$, bottom-left $=\tfrac{1}{3}$. Problem 5: middle-left $=\tfrac{1}{80}$, top-right $=20$.