Prime Time | A

Question 6

Rules

Fill the grid with prime numbers only so that the product of each row is the number to the right of the row and the product of each column is the number below the column.

Question diagram 1
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Solution
Understand the Question
  • Each cell of the 3×33 \times 3 grid must be filled with a prime number (2,3,5,7,…2, 3, 5, 7, \dots).
  • The product of the three numbers in each row must equal the target number at the right of that row.
  • The product of the three numbers in each column must equal the target number below that column.
  • We solve each grid by finding the prime factorisation of each target number and identifying which factor belongs to each cell using the intersections of rows and columns.

(i) First grid

Step 1 · Fill the First Grid

Diagram 1

Let the grid cells be:

abcdefghi\begin{array}{|c|c|c|} \hline a & b & c \\ \hline d & e & f \\ \hline g & h & i \\ \hline \end{array}

1. Column 2: Product is 125=5×5×5125 = 5 \times 5 \times 5.

Since all factors must be prime:

b=5,e=5,h=5b = 5, \quad e = 5, \quad h = 5

2. Row 1: Product is 105105.

a×5×c=105a×c=105÷5=21=3×7\begin{aligned} a \times 5 \times c &= 105 \\ a \times c &= 105 \div 5 \\ &= 21 = 3 \times 7 \end{aligned}

So, aa and cc are 33 and 77 in some order.

3. Row 2: Product is 2020.

d×5×f=20d×f=20÷5=4=2×2\begin{aligned} d \times 5 \times f &= 20 \\ d \times f &= 20 \div 5 \\ &= 4 = 2 \times 2 \end{aligned}

Therefore:

d=2,f=2d = 2, \quad f = 2

4. Row 3: Product is 3030.

g×5×i=30g×i=30÷5=6=2×3\begin{aligned} g \times 5 \times i &= 30 \\ g \times i &= 30 \div 5 \\ &= 6 = 2 \times 3 \end{aligned}

So, gg and ii are 22 and 33 in some order.

5. Column 1: Product is 2828.

a×d×g=28a×2×g=28a×g=28÷2=14\begin{aligned} a \times d \times g &= 28 \\ a \times 2 \times g &= 28 \\ a \times g &= 28 \div 2 \\ &= 14 \end{aligned}

Since a∈{3,7}a \in \{3, 7\} and g∈{2,3}g \in \{2, 3\}, the only combination giving a product of 1414 is:

a=7,g=2a = 7, \quad g = 2

This gives:

c=3,i=3c = 3, \quad i = 3

6. Check Column 3: Product is 1818.

3×2×3=183 \times 2 \times 3 = 18

This confirms all values in the first grid are correct.

Answer

(i) 753252253\begin{array}{|c|c|c|} \hline 7 & 5 & 3 \\ \hline 2 & 5 & 2 \\ \hline 2 & 5 & 3 \\ \hline \end{array}

(ii) Second grid

Step 1 · Fill the Second Grid

Diagram 2

Let the grid cells be:

ABCDEFGHI\begin{array}{|c|c|c|} \hline A & B & C \\ \hline D & E & F \\ \hline G & H & I \\ \hline \end{array}

1. Row 1: Product is 8=2×2×28 = 2 \times 2 \times 2.

Since all factors must be prime:

A=2,B=2,C=2A = 2, \quad B = 2, \quad C = 2

2. Column 1: Product is 3030.

2×D×G=30D×G=30÷2=15=3×5\begin{aligned} 2 \times D \times G &= 30 \\ D \times G &= 30 \div 2 \\ &= 15 = 3 \times 5 \end{aligned}

So, DD and GG are 33 and 55 in some order.

3. Column 2: Product is 7070.

2×E×H=70E×H=70÷2=35=5×7\begin{aligned} 2 \times E \times H &= 70 \\ E \times H &= 70 \div 2 \\ &= 35 = 5 \times 7 \end{aligned}

So, EE and HH are 55 and 77 in some order.

4. Column 3: Product is 2828.

2×F×I=28F×I=28÷2=14=2×7\begin{aligned} 2 \times F \times I &= 28 \\ F \times I &= 28 \div 2 \\ &= 14 = 2 \times 7 \end{aligned}

So, FF and II are 22 and 77 in some order.

5. Row 2: Product is 105105.

D×E×F=105D \times E \times F = 105

Testing possible values from the factor pairs:

  • If D=3D = 3, then E×F=105÷3=35E \times F = 105 \div 3 = 35.
  • Since E∈{5,7}E \in \{5, 7\} and F∈{2,7}F \in \{2, 7\}, choosing E=5E = 5 and F=7F = 7 gives 5×7=355 \times 7 = 35.

Therefore:

D=3,E=5,F=7D = 3, \quad E = 5, \quad F = 7

This gives the remaining cells:

G=5,H=7,I=2G = 5, \quad H = 7, \quad I = 2

6. Check Row 3: Product is 7070.

5×7×2=705 \times 7 \times 2 = 70

This confirms all values in the second grid are correct.

Answer

(ii) 222357572\begin{array}{|c|c|c|} \hline 2 & 2 & 2 \\ \hline 3 & 5 & 7 \\ \hline 5 & 7 & 2 \\ \hline \end{array}

Common Mistakes
  • Using Composite Numbers: Entering numbers like 4,6,8,94, 6, 8, 9 instead of prime numbers. Always ensure every factor placed in a cell is prime.
  • Not Cross-Checking Intersections: Placing factors arbitrarily in a row without verifying if they satisfy the corresponding column product.

More questions in A

Q1

Let us now play the 'idli-vada' game with different pairs of numbers:

(a) 2 and 5, (b) 3 and 7, (c) 4 and 6.

We will say 'idli' for multiples of the smaller number, 'vada' for multiples of the larger number and 'idli-vada' for common multiples. Draw a figure similar to Fig. 5.1 if the game is played up to 60.

Q2

Co-prime art

Observe the following thread art. The first diagram has 12 pegs and the thread is tied to every fourth peg (we say that the thread-gap is 4). The second diagram has 13 pegs and the thread-gap is 3. What about the other diagrams? Observe these pictures, share and discuss your findings in class.

In some diagrams, the thread is tied to every peg. In some, it is not. Is it related to the two numbers (the number of pegs and the thread-gap) being co-prime?

Q3

Make such pictures for the following:

a. 15 pegs, thread-gap of 10

b. 10 pegs, thread-gap of 7

c. 14 pegs, thread-gap of 6

d. 8 pegs, thread-gap of 3

Q4

Below are some boxes with four numbers in each box. Within each box try to say how each number is special compared to the rest. Share with your classmates and find out who else gave the same reasons as you did. Did anyone give different reasons that may not have occurred to you?

Q5

A prime puzzle

The figure on the left shows the puzzle. The figure on the right shows the solution of the puzzle. Think what the rules can be to solve the puzzle.

Q6

Rules

Fill the grid with prime numbers only so that the product of each row is the number to the right of the row and the product of each column is the number below the column.

Q7

Q. Solve the prime puzzles by filling the grids with prime numbers only so that the product of each row is the number to the right of the row and the product of each column is the number below the column.

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