Playing with Numbers — Questions and Answers
Welcome to HSLC Guru. This page gives the complete question answers of ASSEB (Assam State School Education Board) Class 7 New Mathematics Chapter 6, “Playing with Numbers” (সংখ্যাৰ খেল) — every Work it Out box, the full Exercise 6, and extra practice questions, each solved step by step.
Summary
This chapter plays with numbers through interesting activities. It begins with arranging numbers on the fingers, then studies the properties of even and odd numbers and their patterns of the form 2n and 2n − 1, showing how addition and subtraction decide whether a result is even or odd.
The heart of the chapter is the magic square, whose rows, columns and both main diagonals add up to the same magic constant. We build 3 × 3 and 4 × 4 magic squares, generalise the 3 × 3 square using a centre value n (magic constant 3n), and meet famous Indian magic squares — the Lo Shu square, the Chautisa Yantra of Khajuraho, the Navagraha Yantra and the Kubera Yantra.
Finally the chapter explores magic triangles, magic circles and magic stars, the Virahanka–Fibonacci pattern found in nature, alphametic (cryptarithm) puzzles where letters stand for digits, and Sudoku — encouraging students to become mathematical investigators.
Summary: Complete ASSEB Class 7 New Mathematics Chapter 6 “Playing with Numbers” (সংখ্যাৰ খেল) question answers — arrangement of numbers, even and odd properties and patterns, 3 × 3 and 4 × 4 magic squares with their generalisation and magic constant, Indian magic squares (Lo Shu, Chautisa Yantra, Navagraha, Kubera Yantra), magic triangles, the Virahanka–Fibonacci pattern, alphametic/cryptarithm puzzles and Sudoku, with every Work it Out box and Exercise 6 solved.
Textbook Questions and Answers
6.1 Arrangement of Numbers
Counting on the fingers: thumb = 1, index = 2, middle = 3, ring = 4, little = 5; then counting back ring = 6, middle = 7, index = 8, thumb = 9; again index = 10, middle = 11, and so on.
Question: On which finger will the number 100 fall? Is there an easy method without actual counting?
Answer: The numbers on the thumb are 1, 9, 17, 25, … — each is 8 more than the previous and 1 more than a multiple of 8. The multiple of 8 nearest to 100 is 96, so 97 = 8 × 12 + 1 falls on the thumb. Then 97 thumb, 98 index, 99 middle, 100 ring — so 100 falls on the ring finger.
Activity: Find the finger positions of 45, 72, 88, 105, 125.
Answer: Divide the number by 8 and read the remainder — remainder 1 → thumb, 2 or 0 → index, 3 or 7 → middle, 4 or 6 → ring, 5 → little.
- 45 = 8 × 5 + 5 → remainder 5 → little finger
- 72 = 8 × 9 + 0 → remainder 0 → index finger
- 88 = 8 × 11 + 0 → remainder 0 → index finger
- 105 = 8 × 13 + 1 → remainder 1 → thumb
- 125 = 8 × 15 + 5 → remainder 5 → little finger
6.2 Properties of Being Even and Odd
Students A, B, C, D, E, F, G, H, I stand in a row and hold pens equal to their positions — A has 1 pen, B has 2 pens, C has 3 pens, and so on.
Question: Which students together have a total of 20 pens? In how many ways can students be chosen to make 20 pens?
Answer: Taking the students with an even number of pens: B(2) + D(4) + F(6) + H(8) = 20 pens. Taking those with an odd number of pens: A(1) + C(3) + G(7) + I(9) = 20 pens. So at least two different choices — {B, D, F, H} and {A, C, G, I} — give a total of 20 pens.
Work it Out 6.1
1. Can the sum of any 5 odd numbers be 60? Give reasons.
Answer: No. The sum of an odd number (here 5) of odd numbers is always odd, but 60 is even. Hence the sum of 5 odd numbers can never be 60.
2. Based on the pen-stand pictures —
(i) Is the sum of any two odd numbers and the sum of any two even numbers even or odd?
Answer: Both are even: odd + odd = even and even + even = even (e.g. 3 + 5 = 8; 4 + 6 = 10).
(ii) Is the sum of any three odd numbers and the sum of four even numbers even or odd?
Answer: The sum of three odd numbers is odd, while the sum of four even numbers is even (e.g. 1 + 3 + 5 = 9 odd; 2 + 4 + 6 + 8 = 20 even).
3. On the 78th Independence Day, 132 and 179 students from two schools took part in various competitions. Without calculating, tell whether the total number of competitors is even or odd.
Answer: 132 is even and 179 is odd; even + odd = odd. So the total number of competitors is odd.
4. Is the sum of any 9 odd numbers even or odd?
Answer: 9 is odd, so the sum of an odd number of odd numbers is odd.
6.2.1 Patterns of Even and Odd Numbers
The table below shows the values of 2n − 1 and 2n for different n —
| n | Value of 2n − 1 | Value of 2n |
|---|---|---|
| 1 | 2 × 1 − 1 = 1 | 2 × 1 = 2 |
| 2 | 2 × 2 − 1 = 3 | 2 × 2 = 4 |
| 3 | 5 | 6 |
| 4 | 7 | 8 |
| 5 | 9 | 10 |
So 2n − 1 is always odd (1, 3, 5, 7, 9, …) and 2n is always even (2, 4, 6, 8, 10, …). Thus the pattern of even numbers is 2n and the pattern of odd numbers is 2n − 1.
Examine the subtraction statements — odd − odd = even (7 − 3 = 4); even − even = even (8 − 4 = 4); even − odd = odd (8 − 3 = 5); odd − even = odd (7 − 4 = 3). All four statements are true.
Note: The sum of two consecutive natural numbers is always odd (3 + 4 = 7). For three consecutive natural numbers, if the first is odd the sum is even (3 + 4 + 5 = 12), and if the first is even the sum is odd (4 + 5 + 6 = 15). The sum of four consecutive natural numbers is always even (1 + 2 + 3 + 4 = 10).
Analyse the patterns —
- 6k, 4k + 2 (k = 1, 2, 3, …): both are always even.
- 4n + 1, 2n + 3 (n = 1, 2, 3, …): both are always odd.
- 5n + 4, 3n − 1 (n = 1, 2, 3, …): sometimes even, sometimes odd. When n is even, 5n + 4 is even and 3n − 1 is odd; when n is odd, 5n + 4 is odd and 3n − 1 is even.
- 8n + 1 (n = 1, 2, 3, …): always odd (9, 17, 25, …) — these are exactly the thumb numbers from the opening game.
Question: Find the 100th odd and even number.
Answer: nth even number = 2n, so the 100th even number = 2 × 100 = 200. nth odd number = 2n − 1, so the 100th odd number = 2 × 100 − 1 = 199. (Likewise the 50th even = 100 and the 50th odd = 99.)
6.3 Let’s Make a Magic Square
A magic square is a square arrangement of numbers in which the sum of each row, each column and both main diagonals is always the same. This sum is called the magic constant.
Making a 3 × 3 magic square with 1 to 9: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45. With three rows, each row must total 45 ÷ 3 = 15 (the magic constant).
There are 8 ways to get 15 with three numbers — 1 + 5 + 9, 1 + 6 + 8, 2 + 4 + 9, 2 + 5 + 8, 2 + 6 + 7, 3 + 4 + 8, 3 + 5 + 7, 4 + 5 + 6. Since 5 appears in four of them, it goes in the centre; 2, 4, 6, 8 each appear three times, so they go in the corners (2 + 5 + 8 = 15 on one diagonal, 4 + 5 + 6 = 15 on the other); and 1, 3, 7, 9 fill the edges. The result is —
| 4 | 9 | 2 |
| 3 | 5 | 7 |
| 8 | 1 | 6 |
Every row, column and diagonal totals 15 — a magic square with magic constant 15.
Work it Out 6.2
1. Construct magic squares —
(i) Using consecutive numbers from 2 to 10.
Answer: Add 1 to every number of the square above. Centre = 6, magic constant = 3 × 6 = 18.
| 5 | 10 | 3 |
| 4 | 6 | 8 |
| 9 | 2 | 7 |
(ii) Using consecutive numbers from 37 to 45.
Answer: Centre = 41, magic constant = 3 × 41 = 123. Add 36 to every number of the original square.
| 40 | 45 | 38 |
| 39 | 41 | 43 |
| 44 | 37 | 42 |
2. Take any 3 × 3 magic square (magic constant 15) and — (a) increase each number by 10, (b) decrease each number by 3, (c) multiply each number by 3.
(i) In each case, is the new grid a magic square?
Answer: Yes, in all three cases the new grid is still a magic square, because every row, column and diagonal changes by the same amount.
(ii) Does the new magic constant relate to the original one?
Answer: Yes. (a) increase by 10 → new constant = 15 + 3 × 10 = 45; (b) decrease by 3 → 15 − 3 × 3 = 6; (c) multiply by 3 → 15 × 3 = 45.
(iii) Is 3 times the middle number the magic constant? If the magic constant is 60, what is the middle number?
Answer: Yes, the magic constant is 3 times the middle number (5 × 3 = 15). So if the magic constant is 60, the middle number = 60 ÷ 3 = 20.
6.3.1 Generalisation of the 3 × 3 Magic Square
Denoting the centre number of a magic square made from 9 consecutive numbers by ‘n’, the general form (magic constant 3n) is —
| n − 3 | n + 4 | n − 1 |
| n + 2 | n | n − 2 |
| n + 1 | n − 4 | n + 3 |
1. Using the general form, construct a magic square with centre number 27. What is the magic constant?
Answer: With n = 27, the magic constant = 3 × 27 = 81.
| 24 | 31 | 26 |
| 29 | 27 | 25 |
| 28 | 23 | 30 |
2. Is a new magic square formed if each term of the general form is (A) increased by 2, (B) multiplied by 3?
Answer: Yes in both cases. (A) Adding 2 gives centre n + 2 and magic constant 3(n + 2) = 3n + 6; (B) multiplying by 3 gives centre 3n and magic constant 9n. In each case the rows, columns and diagonals still add up equally.
3. Use the general form to make a magic square with magic constant 75.
Answer: Centre number n = 75 ÷ 3 = 25.
| 22 | 29 | 24 |
| 27 | 25 | 23 |
| 26 | 21 | 28 |
6.3.2 The World’s First 4 × 4 Magic Square
The world’s first recorded magic square was found in a 10th-century inscription at the Parshvanath Jain temple in Khajuraho, India, and is known as the ‘Chautisa Yantra’. It contains the numbers 1 to 16.
| 7 | 12 | 1 | 14 |
| 2 | 13 | 8 | 11 |
| 16 | 3 | 10 | 5 |
| 9 | 6 | 15 | 4 |
Question: What do you get on adding the four numbers of each row, column or diagonal? Why is it called Chautisa Yantra?
Answer: The sum of the four numbers in every row, column and diagonal is 34 (e.g. 7 + 12 + 1 + 14 = 34; 7 + 2 + 16 + 9 = 34; 7 + 13 + 10 + 4 = 34). The word ‘chautisa’ means 34, hence the name Chautisa Yantra. Moreover, four numbers taken at the corners of various geometric figures — square, rectangle, parallelogram, trapezium — also add up to 34 (e.g. 2 + 13 + 8 + 11 = 34).
6.3.3 History of the Magic Square and Indian Contributions
The third book of the Chinese mythological ‘Five Scriptures’ is called ‘I-Qiu’. According to legend, during the reign of Emperor Yu a large turtle came out of the Lo River carrying a grid on its shell — this grid is the ‘Lo Shu Magic Square’ —
| 4 | 9 | 2 |
| 3 | 5 | 7 |
| 8 | 1 | 6 |
Indian mathematicians built magic squares with 3 × 3, 4 × 4 and 5 × 5 grids. A collection of nine such magic squares is the Navagraha Yantra, in which each planet represents a day of the week: Sun (from 1) — Sunday, Moon (from 2) — Monday, Mars (from 3) — Tuesday, Mercury (from 4) — Wednesday, Jupiter (from 5) — Thursday, Venus (from 6) — Friday, Saturn (from 7) — Saturday.
Question: What is the magic constant of the Kubera Yantra below?
| 27 | 20 | 25 |
| 22 | 24 | 26 |
| 23 | 28 | 21 |
Answer: Each row totals 27 + 20 + 25 = 72; each column and diagonal also totals 72. So the magic constant of the Kubera Yantra is 72.
Magic triangle: On a triangle made from the numbers 1 to 9, the four numbers on each side add up to the same total (20). Make magic triangles using 11 to 19 and 20 to 28 as well.
Answer: Adding 10 to every number (11 to 19) makes each side total 20 + 4 × 10 = 60 — e.g. the sides (15 + 17 + 16 + 12), (12 + 19 + 11 + 18) and (18 + 13 + 14 + 15) each equal 60. Similarly, adding 19 to every number (20 to 28) makes each side total 20 + 4 × 19 = 96.
Magic circle and magic star: In a magic circle the numbers 1 to 9 are arranged so that the three numbers on each line add up to 15. A magic star is made of two triangles; the numbers 1 to 12 are arranged so that the four numbers on each side add up to the same total (26) — e.g. 10 + 7 + 8 + 1 = 26, 1 + 11 + 12 + 2 = 26, 2 + 5 + 9 + 10 = 26, and so on.
6.4 The Virahanka–Fibonacci Pattern
The petals of many flowers follow the pattern 1, 2, 3, 5, 8, 13, 21, 34, … — the Virahanka pattern. The Fibonacci pattern is 1, 1, 2, 3, 5, 8, 13, 21, 34, …
What is the special feature of this pattern?
Answer: After the first two numbers, each number is the sum of the previous two — 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, and so on. This pattern was discovered by Virahanka in the study of Sanskrit and Prakrit poetry, centuries before Fibonacci popularised it in Europe.
6.5 Numbers in Disguise (Alphametic Problems)
When numbers are represented by letters and then added, the puzzle is called an alphametic (or cryptarithm) problem. Each letter can take a value from 0 to 9; a two-digit number is $\overline{AB} = 10A + B$ and $\overline{AA} = 11A$.
Example 1: A + A = B0. Find A and B.
Answer: Adding a digit to itself must give 0 in the units place; this happens only for A = 5 (5 + 5 = 10). So A = 5, B = 1.
Example 2: A + A + A = BA. Find A and B.
Answer: Adding a digit three times must give the same digit in the units place; this happens for A = 5 (5 + 5 + 5 = 15). So A = 5, B = 1.
Example 3: AA + BB = CBC. Find A, B and C.
Answer: The greatest sum of two two-digit numbers is 198, so C = 1 and AA + BB = 1B1. In the units place A + B must end in 1 (2 + 9, 3 + 8, 4 + 7, 5 + 6). To make the tens place 9, taking A = 9, B = 2 gives 99 + 22 = 121. So A = 9, B = 2, C = 1.
Work it Out 6.3
Solve the following alphametic problems —
(i) UT + UT = VTT. Find U, T, V.
Answer: In the units place, 2T must end in T, so T = 0. In the tens place, 2U must end in 0, so U = 5 (carry 1), giving V = 1. That is 50 + 50 = 100. U = 5, T = 0, V = 1.
(ii) UT + AT = AVV. Find A, U, T, V.
Answer: The sum is a three-digit number with hundreds digit A; two two-digit numbers give at most a carry of 1, so A = 1. Testing gives 85 + 15 = 100 = AVV with A = 1, V = 0. U = 8, T = 5, A = 1, V = 0.
(iii) ABC + ABC = CDDB. Find A, B, C, D.
Answer: The sum is four-digit, so C = 1. Units: 2 × C = 2 = B, so B = 2; tens: 2 × B = 4 = D, so D = 4; hundreds: 2 × A must end in 4 with carry 1, so A = 7. That is 721 + 721 = 1442. A = 7, B = 2, C = 1, D = 4.
Sudoku
In Sudoku a 9 × 9 square is split into nine 3 × 3 squares. Each row, each column and each 3 × 3 square must contain the numbers 1 to 9 exactly once (no number repeats). The teacher’s completed Sudoku is shown below —
| 1 | 6 | 2 | 7 | 3 | 8 | 4 | 9 | 5 |
| 7 | 3 | 8 | 4 | 9 | 5 | 1 | 6 | 2 |
| 4 | 9 | 5 | 1 | 6 | 2 | 7 | 3 | 8 |
| 6 | 2 | 7 | 3 | 8 | 4 | 9 | 5 | 1 |
| 3 | 8 | 4 | 9 | 5 | 1 | 6 | 2 | 7 |
| 9 | 5 | 1 | 6 | 2 | 7 | 3 | 8 | 4 |
| 2 | 7 | 3 | 8 | 4 | 9 | 5 | 1 | 6 |
| 8 | 4 | 9 | 5 | 1 | 6 | 2 | 7 | 3 |
| 5 | 1 | 6 | 2 | 7 | 3 | 8 | 4 | 9 |
Following the same rule, complete the given partly-filled Sudoku (starting 5, 6, …) by writing 1 to 9 exactly once in every row, column and 3 × 3 square.
Exercise 6
1. In a Ludo game there are 4 players A, B, C and D. A throws the dice first, then B, then C, then D, then A again, and so on. Who throws the dice on the 103rd time? Give the reason.
Answer: Every 4 throws complete one cycle (A, B, C, D). 103 ÷ 4 gives remainder 3 (103 = 4 × 25 + 3). Remainder 1 → A, 2 → B, 3 → C, 0 → D. So C throws on the 103rd time.
2. A school has 50 classrooms, each with two fans, numbered 1 to 100 (room one: 1 and 2, room two: 3 and 4, …). One day the fans in 25 rooms failed. Can the sum of the numbers on the fans of those 25 rooms be 1000?
Answer: The two fans of any room carry (2k − 1) and 2k, whose sum 4k − 1 is always odd. So the total of 25 rooms is the sum of 25 odd numbers — an odd number of odd numbers, hence odd. But 1000 is even, so the sum cannot be 1000.
3. A 3 × 2 grid is given. The sum of each row and column is marked even (e) or odd (o). Fill the six boxes with even/odd so that the row and column sums match the given conditions.
Answer: Taking the row sums as odd, even, even and the column sums as even, odd, one matching filling is — Row 1: even, odd; Row 2: even, even; Row 3: even, even. Then the row sums are odd, even, even and the column sums are even, odd, exactly as required. (Several correct fillings are possible.)
4. Construct a 3 × 3 magic square with magic constant 0 (zero) without repeating numbers. Use negative numbers if necessary.
Answer: Taking the centre n = 0 in the general form, the numbers −4 to 4 give a magic square with magic constant 3 × 0 = 0 —
| −3 | 4 | −1 |
| 2 | 0 | −2 |
| 1 | −4 | 3 |
5. Write true/false for the following —
- (a) The sum of an even number of odd numbers is even. — True
- (b) The sum of an odd number of even numbers is even. — True (any count of even numbers sums to even)
- (c) The sum of an even number of even numbers is odd. — False (it is even)
- (d) The sum of an odd number of odd numbers is even. — False (it is odd)
6. What will be the sum of the numbers from 1 to 75 — even or odd?
Answer: Sum = 75 × 76 ÷ 2 = 75 × 38 = 2850, which is even. (From 1 to 75 there are 38 odd numbers; an even count of odd numbers sums to even, and the even numbers also sum to even, so the total is even.)
7. Two consecutive terms of the Fibonacci sequence are 377 and 610. What are the next two terms? What were the two preceding terms?
Answer: Next two: 377 + 610 = 987 and 610 + 987 = 1597 → 987, 1597. Preceding two: 610 − 377 = 233 and 377 − 233 = 144 → 144, 233.
8. Determine whether the 25th term of the Fibonacci sequence is even or odd.
Answer: The parity pattern of Fibonacci terms is odd, odd, even, repeating — so every 3rd term is even. Since 25 ÷ 3 leaves remainder 1, the 25th term is odd (in fact the 25th term is 75025).
9. Match Column I with Column II — Column I (numbers of a magic square): (P) −4 to 4, (Q) −1 to 7, (R) 1 to 9, (S) 6 to 14; Column II (magic constant): (i) 15, (ii) 30, (iii) 0, (iv) 9.
Answer: Magic constant = 3 × centre. (P) centre 0 → 0 (iii); (Q) centre 3 → 9 (iv); (R) centre 5 → 15 (i); (S) centre 10 → 30 (ii). Correct option: A — (P)–(iii), (Q)–(iv), (R)–(i), (S)–(ii).
10. To get magic constant 12 you arrange the numbers 0 to 8. Which numbers give magic constant 33? A. 6 to 14, B. 7 to 15, C. 8 to 16, D. 9 to 17.
Answer: Centre number = 33 ÷ 3 = 11, so the 9 consecutive numbers are 7 to 15. Correct option: B. 7 to 15.
11. Which of the following statements are true? (i) 3k + 2 always gives odd numbers; (ii) every odd number can be expressed as 4m + 1 or 4m + 3; (iii) 5n − 1 gives both even and odd numbers; (iv) the sum of two consecutive integers is always odd. Options: A. (i); B. (ii), (iii); C. (i), (ii), (iv); D. (i), (ii), (iii), (iv).
Answer: (i) is false (3 × 1 + 2 = 5 is odd but 3 × 2 + 2 = 8 is even); (ii) is true; (iii) is true (5 × 1 − 1 = 4 even, 5 × 2 − 1 = 9 odd). Since (i) is false, options A, C and D are ruled out, so the correct option is B. (ii), (iii).
12. Solve the following cryptarithms — (i) ONE + ONE + ONE + ONE = TEN, (ii) AB + CB = BBA, (iii) AA + BB = CBC. (Each letter is a digit; the first digit cannot be zero.)
- (i) 4 × ONE = TEN, so O = 1; testing gives ONE = 182 and 4 × 182 = 728 = TEN → O = 1, N = 8, E = 2, T = 7.
- (ii) The sum is three-digit with hundreds digit B, so B = 1; then 9A + 10C = 108 gives A = 2, C = 9 → A = 2, B = 1, C = 9 (21 + 91 = 112).
- (iii) As in Section 6.5, A = 9, B = 2, C = 1 (99 + 22 = 121).
13. A man had 9 cows named A, B, C, D, E, F, G, H, I giving 1, 2, 3, …, 9 litres of milk daily. He wanted to give three cows to each of his three sons so that each son gets the same amount of milk.
(i) Make a magic square with 1 to 9 litres of milk.
| 4 | 9 | 2 |
| 3 | 5 | 7 |
| 8 | 1 | 6 |
(ii) What is the magic constant and what does it indicate?
Answer: The magic constant is 15. It shows that each son gets 15 litres of milk every day.
(iii) How many groupings of cows give the same amount of milk? Name them.
Answer: The three rows, three columns and two diagonals give 8 groups that each total 15 litres: (D, I, B) = (4, 9, 2), (C, E, G) = (3, 5, 7), (H, A, F) = (8, 1, 6), (D, C, H) = (4, 3, 8), (I, E, A) = (9, 5, 1), (B, G, F) = (2, 7, 6), (D, E, F) = (4, 5, 6), (B, E, H) = (2, 5, 8).
(iv) Determine the total milk obtained each day.
Answer: Total = 1 + 2 + 3 + … + 9 = 45 litres (three sons × 15 litres = 45 litres).
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. The magic constant of a 3 × 3 magic square is how many times the middle number?
(a) 2 times (b) 3 times (c) 4 times (d) 5 times
Answer: (b) 3 times
2. The magic constant of a magic square made from 1 to 9 is —
(a) 12 (b) 15 (c) 18 (d) 45
Answer: (b) 15
3. In the Virahanka–Fibonacci pattern, the number after 13 is —
(a) 18 (b) 21 (c) 26 (d) 34
Answer: (b) 21
4. The 50th even number is —
(a) 50 (b) 99 (c) 100 (d) 102
Answer: (c) 100
5. The pattern of even numbers is —
(a) 2n − 1 (b) 2n (c) 2n + 1 (d) n + 2
Answer: (b) 2n
6. The magic constant of the 4 × 4 Chautisa Yantra is —
(a) 15 (b) 30 (c) 34 (d) 45
Answer: (c) 34
7. The sum of any two odd numbers is —
(a) always even (b) always odd (c) sometimes even (d) zero
Answer: (a) always even
8. The middle number of the Lo Shu magic square is —
(a) 1 (b) 4 (c) 5 (d) 9
Answer: (c) 5
9. The numbers of the pattern 8n + 1 are always —
(a) even (b) odd (c) multiples of 8 (d) prime
Answer: (b) odd
10. In the Fibonacci sequence, each term (after the first two) is the —
(a) double of one term (b) sum of the previous two terms (c) product of two terms (d) square of one term
Answer: (b) sum of the previous two terms
Fill in the Blanks
- The pattern of odd numbers is ______. (2n − 1)
- The magic constant of the general form of a 3 × 3 magic square is ______. (3n)
- The famous Chinese magic square is called the ______. (Lo Shu magic square)
- In numbers in disguise, a letter can take any value from 0 to ______. (9)
- The sum of two consecutive natural numbers is always ______. (odd)
True or False
- The sum of an even number of odd numbers is even. — True
- The sum of three odd numbers is even. — False (it is odd)
- In a magic square every row, column and diagonal has the same sum. — True
- The Fibonacci sequence begins 1, 1, 2, 3, 5, … — True
- An even number cannot be split into two equal pairs. — False (it can be split into pairs)
Short Answer Questions
1. What is a magic constant?
Answer: The magic constant is the equal sum obtained from each row, each column and both main diagonals of a magic square.
2. Explain the difference between even and odd numbers with examples.
Answer: Numbers that can be divided into two equal pairs are even (e.g. 2, 4, 6); numbers that leave one over when made into pairs are odd (e.g. 1, 3, 5).
3. What is the difference between the Virahanka and Fibonacci patterns?
Answer: The Virahanka pattern begins 1, 2, 3, 5, 8, … while the Fibonacci pattern begins 1, 1, 2, 3, 5, … In both, each term is the sum of the previous two terms.
4. What is an alphametic (cryptarithm) problem?
Answer: A number puzzle in which the digits of numbers are written as letters and combined with operations such as addition, and the digit represented by each letter has to be found.
Key Terms
| Term | Meaning |
|---|---|
| Even number | A number that can be divided into two equal pairs |
| Odd number | A number that leaves one over when made into pairs |
| Pattern | A sequence of numbers formed by a rule |
| Magic square | A square whose rows, columns and diagonals have the same sum |
| Magic constant | The common sum of a magic square |
| Generalisation | Expressing a rule in a general form |
| Consecutive numbers | Numbers that follow one after another |
| Sequence | An ordered list of terms following a rule |
| Negative number | A number less than zero |
| Cryptarithm / Alphametic | A puzzle where letters stand for the digits of numbers |