Calendar Magic - STD 8 - Chap: 3
📅 Calendar Magic: 3×3 Square | Prove Difference of Products = 48
Click any date as center of a 3×3 block | Left×Right − Top×Bottom = 48 (always!)
📆 Select Month & Year
🔢 Selected 3×3 Square
Left Number: -
Right Number: -
Top Number: -
Bottom Number: -
Left × Right = -
Top × Bottom = -
Difference (L×R − T×B) = -
Click "Next Step" to begin the proof
💡 Algebraic Proof:
\[ \text{Let the center number be } n \] \[ \text{In a calendar: right } +1,\ \text{down } +7 \] \[ \text{Left} = n-1,\quad \text{Right} = n+1,\quad \text{Top} = n-7,\quad \text{Bottom} = n+7 \] \[ \text{Left} \times \text{Right} = (n-1)(n+1) = n^2 - 1 \] \[ \text{Top} \times \text{Bottom} = (n-7)(n+7) = n^2 - 49 \] \[ \therefore \text{Difference} = (n^2 - 1) - (n^2 - 49) = 48 \]
✅ Therefore, for any 3×3 block, the difference is ALWAYS 48!
\[ \text{Let the center number be } n \] \[ \text{In a calendar: right } +1,\ \text{down } +7 \] \[ \text{Left} = n-1,\quad \text{Right} = n+1,\quad \text{Top} = n-7,\quad \text{Bottom} = n+7 \] \[ \text{Left} \times \text{Right} = (n-1)(n+1) = n^2 - 1 \] \[ \text{Top} \times \text{Bottom} = (n-7)(n+7) = n^2 - 49 \] \[ \therefore \text{Difference} = (n^2 - 1) - (n^2 - 49) = 48 \]
✅ Therefore, for any 3×3 block, the difference is ALWAYS 48!