Algebra Play | A

Question 1

Create your own calendar trick. For instance, choose a grid of a different size and shape.

Question diagram 1
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Solution
Understand the Question
  • In a standard monthly calendar, consecutive numbers in a row differ by 11 (day-to-day), and consecutive numbers in a column differ by 77 (week-to-week).
  • We can design calendar tricks by choosing a geometric shape or grid, expressing each date algebraically in terms of a single variable, and calculating their sum to find a simplified formula.

Step 1 · Cross-Shaped Grid Trick

Let cc be the number in the center of a cross-shaped grid of five dates:

  • Above: c−7c - 7
  • Below: c+7c + 7
  • Left: c−1c - 1
  • Right: c+1c + 1
  • Center: ccDiagram 1

Sum of the five numbers

Sum=(c−7)+(c−1)+c+(c+1)+(c+7)=c−7+c−1+c+c+1+c+7=5c+(−7−1+1+7)=5c+0=5c\begin{aligned} \text{Sum} &= (c - 7) + (c - 1) + c + (c + 1) + (c + 7) \\ &= c - 7 + c - 1 + c + c + 1 + c + 7 \\ &= 5c + (-7 - 1 + 1 + 7) \\ &= 5c + 0 \\ &= 5c \end{aligned}

Thus, the sum is always 55 times the center number.

Example: For a cross centered at 88 with numbers 1,7,8,9,151, 7, 8, 9, 15 1+7+8+9+15=401 + 7 + 8 + 9 + 15 = 40 5×8=405 \times 8 = 40

Step 2 · 3×33 \times 3 Square Grid Trick

Let aa be the top-left number in a 3×33 \times 3 square grid of nine dates:

  • Row 1: a,a+1,a+2a, a + 1, a + 2
  • Row 2: a+7,a+8,a+9a + 7, a + 8, a + 9
  • Row 3: a+14,a+15,a+16a + 14, a + 15, a + 16

The center number is a+8a + 8.Diagram 2

Sum of all nine numbers

Sum=a+(a+1)+(a+2)+(a+7)+(a+8)+(a+9)+(a+14)+(a+15)+(a+16)=(a+a+a+a+a+a+a+a+a)+(0+1+2+7+8+9+14+15+16)=9a+72=9(a+8)\begin{aligned} \text{Sum} &= a + (a + 1) + (a + 2) + (a + 7) + (a + 8) + (a + 9) + (a + 14) + (a + 15) + (a + 16) \\ &= (a + a + a + a + a + a + a + a + a) + (0 + 1 + 2 + 7 + 8 + 9 + 14 + 15 + 16) \\ &= 9a + 72 \\ &= 9(a + 8) \end{aligned}

Thus, the sum is always 99 times the center number (a+8a + 8).

Example: For a 3×33 \times 3 grid with top-left number 1010 and center 1818 10+11+12+17+18+19+24+25+26=16210 + 11 + 12 + 17 + 18 + 19 + 24 + 25 + 26 = 162 9×18=1629 \times 18 = 162

Step 3 · 1×31 \times 3 Horizontal Grid Trick

Let aa be the leftmost number in a horizontal row of three consecutive dates: a,a+1,a+2a, a + 1, a + 2. The middle number is a+1a + 1.Diagram 3

Sum of the three numbers

Sum=a+(a+1)+(a+2)=a+a+1+a+2=3a+3=3(a+1)\begin{aligned} \text{Sum} &= a + (a + 1) + (a + 2) \\ &= a + a + 1 + a + 2 \\ &= 3a + 3 \\ &= 3(a + 1) \end{aligned}

Thus, the sum is always 33 times the middle number (a+1a + 1).

Example: For dates 28,29,3028, 29, 30 with middle number 2929 28+29+30=8728 + 29 + 30 = 87 3×29=873 \times 29 = 87

Answer
  • Cross-shaped grid: Sum=5c\text{Sum} = 5c (55 times the center number)
  • 3×33 \times 3 grid: Sum=9(a+8)\text{Sum} = 9(a + 8) (99 times the center number)
  • 1×31 \times 3 horizontal grid: Sum=3(a+1)\text{Sum} = 3(a + 1) (33 times the middle number)
Common Mistakes
  • Row Offset Error: Forgetting that moving down one row in a weekly calendar adds +7+7 rather than +10+10.
  • Center vs. Starting Value: Confusing the top-left value aa with the center number a+8a + 8 in a 3×33 \times 3 grid when computing the multiplier.
  • Month Boundaries: Selecting grids that cross over the end or start of a month where the regular +1+1 / +7+7 pattern breaks.
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