This MCQ module is based on: 6.3 Some Explorations with Numbers — Magic Squares
TOPIC 21 OF 31
6.3 Some Explorations with Numbers — Magic Squares
🎓 Class 7
Mathematics
CBSE
Theory
Ch 6 — Number Play
⏱ ~18 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 6.3 Some Explorations with Numbers — Magic Squares
Targeting Class 7 level in Number Theory, with Basic difficulty.
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6.3 Some Explorations with Numbers — Magic Squares
Let's now explore a famous arrangement of numbers where rows, columns and diagonals all sum to the same total. Such grids, called magic squares?, have fascinated mathematicians across the world for over two thousand years.
A 3 × 3 Magic Square
Look at the square below. Each small cell holds one of the numbers 1 to 9, with no repeats. Add any row, any column, or any of the two diagonals — the total is always 15.
| 2 | 7 | 6 |
| 9 | 5 | 1 |
| 4 | 3 | 8 |
Fig. 6.3 — A 3 × 3 magic square using 1 – 9. Magic sum = 15.
Try to make your own magic square by randomly filling the grid with numbers 1 – 9. You will quickly see it is hard! There are a huge number of arrangements (9! = 362{,}880) but only a few form a magic square. Instead we will reason systematically.
What must the magic sum be?
Consider any 3 × 3 magic square using 1 – 9 once each. The three rows together contain all 9 numbers, so their total is
\[1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.\]Each row has the same sum \(S\); there are 3 rows, so \(3S = 45\) and therefore \(S = 15\). The same reasoning works for columns.
Observation 1
In a magic square using 1 – 9, the magic sum must be 15. Each row, column and diagonal will always add to 15.
Where must 5 sit?
Which number goes in the centre? Let us classify cells into three types — corner (4 cells), edge-middle (4 cells), and centre (1 cell).
Fig. 6.4 — Three types of positions in a 3 × 3 grid.
Can 1 go in the centre? If it does, the other two numbers in the same row (or column or diagonal) must sum to 14. Pairs from 2 – 9 with sum 14 are only: (5, 9) and (6, 8) — just 2 pairs. But the centre sits in four lines (row, column, two diagonals), so we need 4 distinct pairs. Impossible. So 1 cannot be the centre.
Can 9 go in the centre? Same reasoning — pairs summing to 6 from the remaining numbers are (1, 5) and (2, 4). Only 2 pairs — impossible.
Observation 2
The centre must be 5. Only 5 has 4 suitable pairs summing to 10: (1, 9), (2, 8), (3, 7), (4, 6).
Where do 1 and 9 go?
Can 1 sit in a corner? A corner lies on 3 lines (one row, one column, one diagonal). So 1 needs 3 pairs of other numbers (from 2 – 9 excluding 9 in the paired diagonal cell) summing to 14. We can check: 1 + 5 + 9, 1 + 6 + 8, and 1 + 8 + 6 are too few distinct pairs needed for the 3 lines through a corner. In fact 1 has only two pairs summing to 14 using available cells — so 1 cannot be at a corner. By symmetry, 9 cannot be at a corner either.
Observation 3
The numbers 1 and 9 cannot occur in any corner and cannot sit in the middle of the grid either. Therefore each of them must be in one of the four edge-middle cells.
Figure it Out
1. How many different magic squares can be made using the numbers 1 – 9?
2. Create a magic square using the numbers 2 – 10. What strategy would you use for this? Compare it with the magic squares made using 1 – 9.
3. Take a magic square, and
(a) increase each number by 1;
(b) double each number.
In each case, is the resulting grid also a magic square? How do the magic sums change?
4. What other operations can be performed on a magic square to yield another magic square?
5. Discuss ways of creating a magic square using any set of 9 consecutive numbers (like 2 – 10, 3 – 11, 9 – 17, etc.).
2. Create a magic square using the numbers 2 – 10. What strategy would you use for this? Compare it with the magic squares made using 1 – 9.
3. Take a magic square, and
(a) increase each number by 1;
(b) double each number.
In each case, is the resulting grid also a magic square? How do the magic sums change?
4. What other operations can be performed on a magic square to yield another magic square?
5. Discuss ways of creating a magic square using any set of 9 consecutive numbers (like 2 – 10, 3 – 11, 9 – 17, etc.).
1. There are essentially one magic square (up to rotation/reflection), giving 8 different-looking versions.
2. Add 1 to each entry of the 1 – 9 square. Magic sum goes from 15 to \(15 + 3 = 18\).
3(a) Adding 1 to each cell: still magic; new sum = \(15 + 3 = 18\). (b) Doubling each: still magic; new sum = \(2 \times 15 = 30\).
4. Any operation applied uniformly — add any constant, multiply by any constant, swap rows/columns symmetrically, reflect/rotate — preserves the magic property.
5. Build the base square for 1 – 9, then add (smallest number − 1) to every cell. For 9 consecutive numbers \(k, k+1, \ldots, k+8\) the magic sum is \(3(k+4)\).
2. Add 1 to each entry of the 1 – 9 square. Magic sum goes from 15 to \(15 + 3 = 18\).
3(a) Adding 1 to each cell: still magic; new sum = \(15 + 3 = 18\). (b) Doubling each: still magic; new sum = \(2 \times 15 = 30\).
4. Any operation applied uniformly — add any constant, multiply by any constant, swap rows/columns symmetrically, reflect/rotate — preserves the magic property.
5. Build the base square for 1 – 9, then add (smallest number − 1) to every cell. For 9 consecutive numbers \(k, k+1, \ldots, k+8\) the magic sum is \(3(k+4)\).
Generalising to any 3 × 3 Magic Square
We can describe how the numbers of a magic square relate to each other using algebra. Let the centre be \(m\). Any pair of opposite cells (corner-to-corner across the centre) must sum to \(2m\) because the diagonal total is \(3m\). If we label the cells like this:
| m + a | m − a − b | m + b |
| m − a + b | m | m + a − b |
| m − b | m + a + b | m − a |
Fig. 6.5 — Generalised 3 × 3 magic square with centre \(m\) and parameters \(a, b\).
Every row, column and diagonal sums to \(3m\). You can verify this by adding the expressions along any line. For the 1 – 9 square, \(m = 5\), and suitable \(a, b\) give the entries 1, 2, ..., 9.
Try it out:
1. Using the generalised form, find a magic square if the centre is 23.
2. What is the expression for the sum of the 3 terms of any row, column or diagonal?
3. Write the result stated by:
(a) adding 1 to every term in the generalised form.
(b) doubling every term in the generalised form.
4. Create a magic square whose magic sum is 60.
5. Is it possible to get a magic square by filling nine non-consecutive numbers?
1. Using the generalised form, find a magic square if the centre is 23.
2. What is the expression for the sum of the 3 terms of any row, column or diagonal?
3. Write the result stated by:
(a) adding 1 to every term in the generalised form.
(b) doubling every term in the generalised form.
4. Create a magic square whose magic sum is 60.
5. Is it possible to get a magic square by filling nine non-consecutive numbers?
Work through the hints
1. Centre 23: magic sum \(= 3 \times 23 = 69\). Pick \(a = 1, b = 3\) to get: 24, 19, 26, 25, 23, 21, 20, 27, 22.
2. Sum of any row/column/diagonal \(= 3m\).
3(a) Adds 3 to the magic sum (becomes \(3m + 3\)). (b) Doubles the magic sum.
4. Magic sum 60 ⇒ \(m = 20\). Use centre 20 with \(a = 1, b = 3\): entries 21, 16, 23, 22, 20, 18, 17, 24, 19.
5. Yes. Any 9 numbers of the form \(m, m \pm a, m \pm b, m \pm (a+b), m \pm (a-b)\) will work, so they need not be consecutive — e.g., 4, 9, 2, 3, 5, 7, 8, 1, 6 scaled or shifted.
2. Sum of any row/column/diagonal \(= 3m\).
3(a) Adds 3 to the magic sum (becomes \(3m + 3\)). (b) Doubles the magic sum.
4. Magic sum 60 ⇒ \(m = 20\). Use centre 20 with \(a = 1, b = 3\): entries 21, 16, 23, 22, 20, 18, 17, 24, 19.
5. Yes. Any 9 numbers of the form \(m, m \pm a, m \pm b, m \pm (a+b), m \pm (a-b)\) will work, so they need not be consecutive — e.g., 4, 9, 2, 3, 5, 7, 8, 1, 6 scaled or shifted.
The First-ever 4 × 4 Magic Square — Chautisa Yantra
The first known 4 × 4 magic square was inscribed in the Parshvanatha Jain temple at Khajuraho, India, and is called the Chautisa Yantra. Its magic sum is 34 — hence the name chautisa (Hindi for "thirty-four").
| 7 | 12 | 1 | 14 |
| 2 | 13 | 8 | 11 |
| 16 | 3 | 10 | 5 |
| 9 | 6 | 15 | 4 |
Fig. 6.6 — The Chautisa Yantra (Khajuraho, 10th-century).
Check: every row adds to 34, every column adds to 34, and so do both diagonals! Beyond this, the four 2 × 2 corner blocks, the central 2 × 2 block, and many "broken diagonals" also sum to 34 — a deeply structured pattern.
History — Magic Squares
The earliest magic square, the Chinese Lo Shu Square, dates to over 2000 years ago. Its legend tells of a turtle emerging from a flood carrying a 3 × 3 grid on its shell. Indian mathematicians wrote extensively about magic squares — the 3 × 3, 4 × 4, and larger squares — describing their construction methods. Magic squares appear in temples in Khajuraho, Gwalior, Parali (Tamil Nadu), in Central Asia, Japan, Europe and Africa. The "Nāsagara Yantra" in India uses 3 × 3 magic squares linked to planets (Mercury, Venus, Moon, Jupiter, Sun, Mars, Ketu, Saturn, Rahu).
Activity: Build Your Own Magic Square
Activity: Calendar Magic
L4 AnalyseMaterials: A wall calendar, pencil, ruler.
Predict: Take any 3 × 3 block of dates from a calendar (e.g., 7, 8, 9 | 14, 15, 16 | 21, 22, 23). Will the centre-row/column/diagonal sums be equal? Is this a magic square?
- Pick a 3 × 3 grid of consecutive dates.
- Compute sums of each row, each column, and the two diagonals.
- Note which sums are equal and which are not.
- Explain why the centre cell is the average of the block.
All three rows sum to the same value! Reason: rows are consecutive weeks, so each row sum differs from the previous by \(3 \times 7 = 21\), but actually the middle row is \(3 \times\) centre. The four corners + centre show a pattern: for any such block, the magic sum is \(3 \times\) centre. Similarly columns sum uniformly to \(3 \times\) centre. Diagonals too! So any 3 × 3 calendar block is a magic square with magic sum \(3m\) where \(m\) = middle date.
Competency-Based Questions
Scenario: A local museum displays the 10th-century Chautisa Yantra tile from Khajuraho. A student visiting the museum wonders why exactly the number 34 is central to the puzzle, and whether she could build her own 3 × 3 magic square whose magic sum equals her age (12).
Q1. Show that a 3 × 3 magic square filled with 9 distinct positive whole numbers must have its centre equal to one-third of its magic sum.
L3 ApplyThe two diagonals and the middle row and middle column all pass through the centre \(m\). Summing these four lines: each of the 8 non-centre cells appears exactly once and the centre \(m\) appears 4 times, giving total \(4S = (\text{sum of all 9 cells}) + 3m = 3S + 3m\), so \(S = 3m\), i.e., \(m = S/3\).
Q2. Can the student build a 3 × 3 magic square with magic sum 12 using 9 distinct positive integers? If yes, give an example. If no, explain.
L4 AnalyseMagic sum 12 ⇒ centre = 4. Using the generalised form with \(m = 4\) and small \(a, b\): try \(a = 1, b = 2\) → entries 5, 1, 6, 3, 4, 5, 2, 7, 3 — duplicates! Try \(a = 1, b = 3\) gives 5, 0, 7, 2, 4, 6, 1, 8, 3. That has 0 — not positive. So with 9 distinct positive integers and magic sum 12, it's impossible (we need a spread of 8 around centre 4, but that forces 0 or negatives). Minimum magic sum for 1–9 is 15.
Q3. Evaluate the claim: "If I multiply every entry of the Chautisa Yantra by 5, the resulting grid is no longer a magic square."
L5 EvaluateIncorrect. Multiplying every entry by the same constant \(k\) multiplies every row/column/diagonal sum by \(k\). So each sum becomes \(34k\), still identical across all lines — the grid is still magic. The new magic sum becomes \(34 \times 5 = 170\).
Q4. Design a 3 × 3 magic square using only odd numbers, with a magic sum of 33. List all 9 entries and verify one row, one column, and one diagonal.
L6 CreateMagic sum 33 ⇒ centre \(m = 11\). Pick \(a = 2, b = 6\) (both even keep all entries odd). Grid: \(\begin{pmatrix} 13 & 3 & 17 \\ 15 & 11 & 7 \\ 5 & 19 & 9 \end{pmatrix}\). Row 1: \(13 + 3 + 17 = 33\). Col 2: \(3 + 11 + 19 = 33\). Diagonal: \(13 + 11 + 9 = 33\). All odd, all distinct.
Assertion–Reason Questions
Assertion (A): In every 3 × 3 magic square using 1 – 9, the number 5 must sit at the centre.
Reason (R): The centre number must have four distinct pairs summing to (magic sum − centre).
Reason (R): The centre number must have four distinct pairs summing to (magic sum − centre).
Answer: (a). Only 5 has four pairs from 1–9 summing to 10 [(1,9),(2,8),(3,7),(4,6)]. R directly forces A.
Assertion (A): The magic sum of a 3 × 3 magic square made from numbers 11, 12, ..., 19 is 45.
Reason (R): Shifting all entries by a constant shifts the magic sum by 3 times that constant.
Reason (R): Shifting all entries by a constant shifts the magic sum by 3 times that constant.
Answer: (a). 11–19 is obtained by adding 10 to each entry of 1–9. New magic sum = \(15 + 3 \times 10 = 45\). R explains A exactly.
Assertion (A): The Chautisa Yantra earned its name because all its rows and columns add to 40.
Reason (R): The Chautisa Yantra is a 4 × 4 magic square from Khajuraho.
Reason (R): The Chautisa Yantra is a 4 × 4 magic square from Khajuraho.
Answer: (d). A is false — the magic sum is 34 (hence "chautisa" = 34), not 40. R is true.
Frequently Asked Questions — Number Play
What is Part 2 — Some Explorations with Numbers & Magic Squares | Class 7 Maths Ch 6 | MyAiSchool in NCERT Class 7 Mathematics?
Part 2 — Some Explorations with Numbers & Magic Squares | Class 7 Maths Ch 6 | MyAiSchool is a key concept covered in NCERT Class 7 Mathematics, Chapter 6: Number Play. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 2 — Some Explorations with Numbers & Magic Squares | Class 7 Maths Ch 6 | MyAiSchool step by step?
To solve problems on Part 2 — Some Explorations with Numbers & Magic Squares | Class 7 Maths Ch 6 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 6: Number Play?
The essential formulas of Chapter 6 (Number Play) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 2 — Some Explorations with Numbers & Magic Squares | Class 7 Maths Ch 6 | MyAiSchool important for the Class 7 board exam?
Part 2 — Some Explorations with Numbers & Magic Squares | Class 7 Maths Ch 6 | MyAiSchool is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 2 — Some Explorations with Numbers & Magic Squares | Class 7 Maths Ch 6 | MyAiSchool?
Common mistakes in Part 2 — Some Explorations with Numbers & Magic Squares | Class 7 Maths Ch 6 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 2 — Some Explorations with Numbers & Magic Squares | Class 7 Maths Ch 6 | MyAiSchool?
End-of-chapter NCERT exercises for Part 2 — Some Explorations with Numbers & Magic Squares | Class 7 Maths Ch 6 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.
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