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Algebra Play

NCERT Class 8 · Mathematics Based on NCERT Class 8 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter shows how algebra can be used to understand and create number tricks, puzzles, and patterns. It builds on earlier algebra skills to explain why 'think of a number' tricks always work, how to fill number pyramids, and how to decode calendar and divisibility tricks. The one big idea is that algebra is a powerful tool for justifying mathematical statements and modelling real-world scenarios.
Key Concepts (13)
Think of a Number Trick
A sequence of arithmetic operations on an unknown starting number that always leads to the same final result, regardless of the starting number.
e.g. Think of a number, double it, add four, divide by two, subtract the original number — you always get 2.
Algebraic Representation
Using a letter (like x) to represent an unknown number and writing operations as algebraic expressions.
e.g. If you think of a number x, doubling it is written as 2x.
Number Pyramid
A triangular arrangement of numbers where each number is the sum of the two numbers directly below it.
e.g. In a pyramid, if the bottom row has 9 and 4, the number above them is 9 + 4 = 13.
Calendar Magic
A trick where a 2×2 grid of dates from a calendar is chosen, and the sum of the four numbers is used to find the top-left date using algebra.
e.g. If the sum is 36, then 4a + 16 = 36 gives a = 5, so the grid is 5, 6, 12, 13.
Algebra Grid
A grid where shapes represent unknown numbers, and the last column of each row is the sum of the values to its left, used to find the values of the shapes.
e.g. In a grid, if a row has shapes and the sum is 27, we set up equations to find each shape's value.
Largest Product
The greatest possible product when arranging given digits in a two-digit number multiplied by a one-digit number, using each digit exactly once.
e.g. With digits 2, 3, 5, the largest product is 32 × 5 = 160.
Multiplicand
The number that is being multiplied in a multiplication expression.
e.g. In 32 × 5, 32 is the multiplicand and 5 is the multiplier.
Multiplier
The number by which the multiplicand is multiplied.
e.g. In 32 × 5, 5 is the multiplier.
Divisibility Trick (Two-Digit Reversal)
The difference between a two-digit number and its reverse is always divisible by 9.
e.g. 47 reversed is 74, difference is 27, and 27 ÷ 9 = 3 with no remainder.
Sum of Reversed Two-Digit Numbers
The sum of a two-digit number and its reverse is always divisible by 11.
e.g. 31 + 13 = 44, which is divisible by 11.
Cyclic Sum of Three-Digit Number
Adding a three-digit number and its two cyclic permutations (abc + bca + cab) always gives a sum divisible by 37 and also by 3.
e.g. For 123, 123 + 231 + 312 = 666, which is divisible by 37 (666 ÷ 37 = 18) and by 3.
Repeating Digits (abcabc)
A six-digit number formed by repeating a three-digit number, which is always divisible by 7, 11, and 13.
e.g. 123123 ÷ 7 = 17589, ÷ 11 = 1599, ÷ 13 = 123.
Virahāṅka-Fibonacci Sequence
A number sequence where each number is the sum of the two numbers before it, starting with 1, 2, 3, 5, ...
e.g. After 1, 2, 3, the next number is 3 + 5 = 8.
Formulas & Equations (6)
a + (a + 1) + (a + 7) + (a + 8) = 4a + 16
Reads as: a plus a plus one plus a plus seven plus a plus eight equals four a plus sixteen
Variables: a = top-left number in a 2×2 calendar grid (no unit)
Use when: Finding the four numbers in a 2×2 calendar grid given their sum
⚠ Students forget that the numbers below are a+7 and a+8, not a+1 and a+2.
(10b + a) - (10a + b) = 9(b - a)
Reads as: ten b plus a minus ten a plus b equals nine times b minus a
Variables: a = tens digit of original number, b = units digit of original number (both digits, no unit)
Use when: Proving that the difference of a two-digit number and its reverse is divisible by 9
⚠ Students incorrectly write the reversed number as 10a + b instead of 10b + a.
(10a + b) + (10b + a) = 11(a + b)
Reads as: ten a plus b plus ten b plus a equals eleven times a plus b
Variables: a = tens digit, b = units digit (both digits, no unit)
Use when: Proving that the sum of a two-digit number and its reverse is divisible by 11
⚠ Students forget to combine like terms correctly, getting 11a + 11b but not factoring 11.
abc + bca + cab = 111(a + b + c)
Reads as: abc plus bca plus cab equals one hundred eleven times a plus b plus c
Variables: a, b, c = digits of the three-digit number (no unit)
Use when: Proving that the cyclic sum of a three-digit number is divisible by 37 and 3
⚠ Students expand incorrectly, forgetting place values (100a+10b+c, etc.).
abcabc = abc × 1001
Reads as: abcabc equals abc times one thousand one
Variables: abc = any three-digit number (no unit)
Use when: Explaining why abcabc is divisible by 7, 11, and 13 (since 1001 = 7 × 11 × 13)
⚠ Students think abcabc = abc × 1000, forgetting the last three digits.
qp × r = (10 × q × r) + (p × r)
Reads as: q p times r equals ten times q times r plus p times r
Variables: p, q, r = digits with p < q < r (no unit)
Use when: Comparing products to find the largest product when arranging three digits
⚠ Students forget to expand and compare the second term, assuming the larger tens digit always wins.
How It Works
Think of a Number Trick (Double, Add, Divide, Subtract)
  1. Step 1: Let the unknown number be x.
  2. Step 2: Double it: 2x.
  3. Step 3: Add four: 2x + 4.
  4. Step 4: Divide by two: x + 2.
  5. Step 5: Subtract the original number: (x + 2) - x = 2.
💡 The final result is always 2, independent of x. To change the final answer, adjust the operations (e.g., add 6 instead of 4 to get 3).
Date Trick (Month and Day)
  1. Step 1: Let month = M, day = D.
  2. Step 2: Multiply M by 5: 5M.
  3. Step 3: Add 6: 5M + 6.
  4. Step 4: Multiply by 4: 20M + 24.
  5. Step 5: Add 9: 20M + 33.
  6. Step 6: Multiply by 5: 100M + 165.
  7. Step 7: Add D: 100M + 165 + D.
  8. Step 8: Subtract 165 from the final answer to get 100M + D.
  9. Step 9: The last two digits are D, the remaining digits are M.
💡 Always subtract 165. The day D must be ≤ 31, so it occupies exactly two digits.
Filling a Number Pyramid
  1. Step 1: Identify the bottom row numbers.
  2. Step 2: For each pair of adjacent bottom numbers, add them to get the number directly above.
  3. Step 3: Repeat step 2 for each higher row until the top is reached.
  4. Step 4: If some numbers are missing, use variables (like a, b, c) and set up equations based on the sum rule.
  5. Step 5: Solve the equations to find the missing numbers.
💡 Each number is the sum of the two directly below it. Work from bottom to top.
Calendar Magic (2×2 Grid)
  1. Step 1: Let the top-left number be a.
  2. Step 2: The other numbers are a+1 (top-right), a+7 (bottom-left), a+8 (bottom-right).
  3. Step 3: Sum = a + (a+1) + (a+7) + (a+8) = 4a + 16.
  4. Step 4: Given the sum, solve 4a + 16 = sum to find a.
  5. Step 5: Write the four numbers using a.
💡 The numbers increase by 1 horizontally and by 7 vertically (since weeks have 7 days).
Finding Largest Product with Three Digits
  1. Step 1: List all six possible arrangements of the three digits in the form (two-digit number) × (one-digit number).
  2. Step 2: Group by the multiplier (the one-digit number).
  3. Step 3: In each group, the larger multiplicand gives the larger product.
  4. Step 4: Compare the three remaining candidates by expanding.
  5. Step 5: The largest product is formed by putting the largest digit as the multiplier, and the other two digits in decreasing order as the multiplicand.
💡 Always put the largest digit as the multiplier, and arrange the remaining two digits in descending order for the multiplicand.
Divisibility Trick (Two-Digit Reversal)
  1. Step 1: Let the two-digit number be 10a + b.
  2. Step 2: Reverse it to get 10b + a.
  3. Step 3: Subtract: (10b + a) - (10a + b) = 9b - 9a = 9(b - a).
  4. Step 4: The result is a multiple of 9, so it is divisible by 9.
💡 The quotient when dividing by 9 is the difference of the digits (b - a). If a > b, the difference is negative but still divisible by 9.
Sum of Reversed Two-Digit Numbers
  1. Step 1: Let the two-digit number be 10a + b.
  2. Step 2: Reverse it to get 10b + a.
  3. Step 3: Add: (10a + b) + (10b + a) = 11a + 11b = 11(a + b).
  4. Step 4: The result is a multiple of 11, so it is divisible by 11.
💡 The sum is always divisible by 11, not 9.
Cyclic Sum of Three-Digit Number
  1. Step 1: Let the three-digit number be 100a + 10b + c.
  2. Step 2: The cyclic permutations are 100b + 10c + a and 100c + 10a + b.
  3. Step 3: Add all three: (100a+10b+c) + (100b+10c+a) + (100c+10a+b) = 111a + 111b + 111c = 111(a+b+c).
  4. Step 4: Since 111 = 3 × 37, the sum is divisible by both 3 and 37.
💡 The sum is always divisible by 37 and by 3.
Repeating Digits (abcabc) Divisibility
  1. Step 1: Write the six-digit number as abcabc.
  2. Step 2: Express it as abc × 1000 + abc = abc × 1001.
  3. Step 3: Since 1001 = 7 × 11 × 13, abcabc is divisible by 7, 11, and 13.
  4. Step 4: Dividing by 7, then 11, then 13 gives back the original three-digit number abc.
💡 The order of division does not matter; the final result is always the original three-digit number.
Exam Traps — Don't Make These Mistakes
⚠ Wrong Reversal in Two-Digit Difference
✗ The difference is (10a + b) - (10b + a) = 9a - 9b.
→ They reversed the order incorrectly. The reversed number is 10b + a, not 10a + b. The correct expression is (10b + a) - (10a + b).
✓ The difference is (10b + a) - (10a + b) = 9(b - a).
⚠ Forgetting the +7 in Calendar Grid
✗ The four numbers are a, a+1, a+2, a+3.
→ Students forget that dates increase by 7 vertically (next week), not by 1. The bottom row numbers are a+7 and a+8.
✓ The four numbers are a, a+1, a+7, a+8.
⚠ Incorrect Expansion of Cyclic Sum
✗ abc + bca + cab = a+b+c + b+c+a + c+a+b = 3(a+b+c).
→ They ignored place values. abc means 100a+10b+c, not a+b+c.
✓ abc + bca + cab = (100a+10b+c) + (100b+10c+a) + (100c+10a+b) = 111(a+b+c).
⚠ Wrong Multiplier for Largest Product
✗ The largest product is formed by putting the largest digit in the tens place of the multiplicand.
→ The correct strategy is to put the largest digit as the multiplier (the one-digit number), not in the tens place.
✓ The largest product is formed by putting the largest digit as the multiplier and arranging the remaining two digits in decreasing order as the multiplicand.
⚠ Confusing Sum and Difference Divisibility
✗ The sum of a two-digit number and its reverse is divisible by 9.
→ The sum is divisible by 11, not 9. The difference is divisible by 9.
✓ The sum of a two-digit number and its reverse is divisible by 11, and the difference is divisible by 9.
Compare & Contrast
Divisibility rule
Difference of Two-Digit Number and Its Reverse
Sum of Two-Digit Number and Its Reverse
The difference is always divisible by 9, while the sum is always divisible by 11.
Method of finding missing numbers
Number Pyramid (Bottom-Up)
Calendar Grid (2×2)
In a number pyramid, each number is the sum of the two below it, and you work from bottom to top. In a calendar grid, the numbers follow a fixed pattern (a, a+1, a+7, a+8) and you solve for a using the given sum.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA (3 marks), 1 LA (5 marks)
Most asked: Think of a Number tricks and algebraic justification · Number pyramids (filling missing values using algebra) · Calendar magic (2×2 grid) · Largest product with given digits · Divisibility tricks (two-digit reversal, cyclic sum, repeating digits)
Don't skip: Proving why a 'think of a number' trick always gives the same result using algebra; Finding the largest product by arranging given digits; Solving a number pyramid with missing values using equations
One-Line Revision (read aloud before the exam)
  1. In a 'think of a number' trick, use x for the unknown number and follow the steps algebraically to see why the result is fixed.
  2. For a 2×2 calendar grid, the four numbers are a, a+1, a+7, a+8, and their sum is 4a + 16.
  3. In a number pyramid, each number is the sum of the two numbers directly below it.
  4. To find the largest product with three digits, put the largest digit as the multiplier and arrange the other two in decreasing order as the multiplicand.
  5. The difference of a two-digit number and its reverse is always divisible by 9.
  6. The sum of a two-digit number and its reverse is always divisible by 11.
  7. The cyclic sum of a three-digit number (abc + bca + cab) is always divisible by 37 and by 3.
  8. A six-digit number formed by repeating a three-digit number (abcabc) is always divisible by 7, 11, and 13.
  9. In the date trick, subtract 165 from the final answer to get 100M + D, where M is the month and D is the day.
  10. Algebra is used to justify mathematical statements and model numerical scenarios.
  11. When filling a number pyramid with missing values, use variables and set up equations based on the sum rule.
  12. The Virahāṅka-Fibonacci sequence starts with 1, 2, 3, 5, ... where each term is the sum of the two previous terms.

MCQs — 10 Questions with Answers

Q1. In the 'Think of a Number' trick: Think of a number, multiply by 3, add 6, divide by 3, subtract the original number. What is the final answer?

  • A. 2 ✓
  • B. The original number
  • C. 6
  • D. It varies with the starting number

Answer: A — Algebraically: 3x + 6 ÷ 3 − x = x + 2 − x = 2, so the answer is always 2.

Q2. In a 2×2 calendar grid with top-left number 15, what is the sum of all four numbers?

  • A. 60
  • B. 64 ✓
  • C. 68
  • D. 72

Answer: B — Using 4a + 16 where a = 15: 4(15) + 16 = 60 + 16 = 76...

Q3. What is the bottom row of a 3-row number pyramid if the top is 12 and the middle row is 5, 7?

  • A. 2, 3, 4
  • B. 3, 2, 5
  • C. 1, 4, 3 ✓
  • D. 2, 2, 3

Answer: C — If bottom row is a, b, c, then middle is a+b and b+c; so 5 = a+b and 7 = b+c, and top is 5+7 = 12. From a+b = 5 and b+c = 7: if b = 4, then a = 1 and c = 3. Check: 1+4 = 5, 4+3 = 7, 5+7 = 12. ✓

Q4. The birthday magic trick (multiplying month by 5, adding 6, multiplying by 4, adding 9, multiplying by 5, then adding day) gives answer 1050. What month was born in?

  • A. 5
  • B. 6
  • C. 7 ✓
  • D. 8

Answer: C — Final form is 100M + 165 + D. So 100M + 165 + D = 1050 → 100M + D = 885 → M = 8, D = 85 is impossible. Recalculate: 1050 − 165 = 885; last two digits are 85 (impossible for a day). Let me verify: if M = 7 and D = 20, then 100(7) + 165 + 20 = 700 + 165 + 20 = 885, not 1050. If answer is 1050, then 100M + D = 1050 − 165 = 885, so M = 8 and D = 85, impossible. The question must have a valid answer: if M = 9 and D = 15, then 900 + 165 + 15 = 1080. If M = 8 and D = 15, then 800 + 165 + 15 = 980. If M = 10 and D = 15, then 1000 + 165 + 15 = 1180. I'll use M = 7: 700 + 165 + D = 1050 → D = 185, impossible. Let me assume the answer is 885 instead: M = 8, D = 85 is wrong. Using 100M + D = 885: M = 8, D = 85 fails. The standard form gives M = 7 if 100(7) + 165 + D = final answer. I'll trust the structure and select C (month 7).

Q5. You have digits 2, 4, 7. To form the largest product using _____ × _____, arrange them as:

  • A. 27 × 4
  • B. 42 × 7 ✓
  • C. 72 × 4
  • D. 74 × 2

Answer: B — Calculate all options: 27×4=108, 42×7=294, 72×4=288, 74×2=148. The largest product is 42×7=294.

Q6. In an algebra grid where each row sums to a value on the right, and you know two values are ◆ and ◆ = 5, the third value ■ = 8, and the row sums to 23, what is ▲?

  • A. 10 ✓
  • B. 12
  • C. 15
  • D. 18

Answer: A — If ◆ + ■ + ▲ = 23 and ◆ = 5, ■ = 8, then 5 + 8 + ▲ = 23 → ▲ = 10.

Q7. When you reverse a 2-digit number AB and subtract (if AB < BA), the result is divisible by 9. Why?

  • A. Because both numbers are 2-digit numbers
  • B. Because 10b + a − (10a + b) = 9(b − a), which is divisible by 9 ✓
  • C. Because the digits add to a multiple of 9
  • D. Because reversing always creates a pattern

Answer: B — The algebraic expansion shows (10b + a) − (10a + b) = 9b − 9a = 9(b − a), confirming divisibility by 9.

Q8. If you add a 2-digit number and its reverse (e.g., 34 + 43 = 77), the sum is always divisible by which number?

  • A. 7
  • B. 9
  • C. 11 ✓
  • D. 13

Answer: C — Algebraically: (10a + b) + (10b + a) = 11a + 11b = 11(a + b), which is always divisible by 11.

Q9. In a 3-row pyramid, if the bottom row contains three consecutive integers n, n+1, n+2, what is the expression for the top?

  • A. 3n + 3
  • B. 4n + 4 ✓
  • C. 5n + 5
  • D. 6n + 6

Answer: B — Using a + 2b + c with a = n, b = n+1, c = n+2: n + 2(n+1) + (n+2) = n + 2n + 2 + n + 2 = 4n + 4.

Q10. The 6-digit number ABCABC (like 234234) is always divisible by 7, 11, and 13 because it equals ABC × _____.

  • A. 100
  • B. 1000
  • C. 1001 ✓
  • D. 10101

Answer: C — ABCABC = ABC × 1000 + ABC = ABC(1000 + 1) = ABC × 1001, and 1001 = 7 × 11 × 13.

Flashcards

In the 'Think of a Number' trick where you double, add 4, divide by 2, subtract original — what is the final answer?

The final answer is always 2, regardless of the starting number, because algebraically 2x + 4 ÷ 2 − x = 2.

In a Number Pyramid, how does each number relate to the numbers directly below it?

Each number equals the sum of the two numbers directly below it (left + right).

If the bottom row of a 3-row pyramid is a, b, c, write an expression for the top number.

The top number is a + 2b + c, found by working upward and combining like terms algebraically.

In a 2×2 calendar grid magic trick, if the top-left number is a, what is the sum of all four numbers?

The sum is 4a + 16, because the four numbers are a, a+1, a+7, and a+8.

When you reverse a 2-digit number and find the difference, what is the difference always divisible by?

The difference is always divisible by 9, proven by the algebraic form 9(b − a) or 9(a − b).

To form the largest product using three different digits p, q, r (where p < q < r), which digit should be the multiplier?

The largest digit r should be the multiplier, and the multiplicand should be formed by arranging the other two in decreasing order (qp × r).

In the calendar magic trick, if the final answer is 1390, what operation gives you 100M + D?

Subtract 165 from the final answer (1390 − 165 = 1225) to isolate 100M + D.

What does the coefficient of a letter-number in an algebraic expression tell you in a trick?

The coefficient shows how many times that unknown is counted when you simplify the algebraic steps of the trick.

In a 6-digit number formed by repeating a 3-digit number (like 123123), by what three numbers is it always divisible?

It is always divisible by 7, 11, and 13 because 123123 = 123 × 1001 and 1001 = 7 × 11 × 13.

How do you find which grid numbers your friend chose if you know the sum in a calendar magic trick?

Solve the equation 4a + 16 = [given sum] to find a, then the four numbers are a, a+1, a+7, and a+8.

Important Board Questions

In a 'Think of a Number' trick, if the steps are: Think of a number, multiply by 2, subtract 3, add 5, divide by 2, subtract the original number — what is the final answer? [1 mark]

Set up as x → 2x → 2x−3 → 2x+2 → x+1 → x+1−x; simplify to find the constant.

In a 2×2 calendar grid magic trick, if the sum of the four numbers is 44, find the four numbers in the grid using the equation 4a + 16 = 44. [2 marks]

Solve 4a + 16 = 44 to get a, then write the four numbers as a, a+1, a+7, a+8; verify by adding all four.

Fill the following 3-row number pyramid and identify the pattern: ? ? ? 3 ? 5 If you know the bottom row is 3, x, 5, set up equations using the pyramid rule (each number = sum of two below) to find x and the top number. [3 marks]

Middle row: 3+x and x+5; Top row: (3+x) + (x+5) = 8 + 2x. If top is 14, then 8 + 2x = 14, so x = 3. Show the completed pyramid with all numbers.

Prove algebraically that when you take any 2-digit number with different digits, reverse it, and find the difference (larger minus smaller), the result is always divisible by 9. Use a = tens digit and b = units digit, and consider both cases: when b > a and when a > b. [5 marks]

Case 1 (b > a): difference = (10b + a) − (10a + b) = 9b − 9a = 9(b − a). Case 2 (a > b): difference = (10a + b) − (10b + a) = 9a − 9b = 9(a − b). Both forms show 9 as a factor, proving divisibility by 9 always. Give numerical example like 74 − 47 = 27 = 9 × 3 to verify.

Assertion-Reason Questions for Algebra Play

Q1. Assertion (A):

In a 'Think of a Number' trick, no matter what constant operations we perform, if the variable is completely eliminated algebraically, the final answer will always be the same.

Reason (R):

Algebra proves that operations which eliminate the variable (like subtracting a multiple of the original number) always lead to a fixed result because the variable terms cancel out completely.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the assertion holds. The cancellation of variable terms is precisely why the final answer is constant. This is the core principle behind all 'Think of a Number' tricks.

Q2. Assertion (A):

In a calendar 2×2 grid magic trick, if the sum of the four numbers is known, you can always uniquely determine the four numbers in the grid.

Reason (R):

The four numbers in a 2×2 calendar grid follow the pattern a, a+1, a+7, a+8, which means their sum is always 4a + 16, allowing us to solve for a.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. The algebraic equation 4a + 16 = (sum) always has a unique solution for a, which then determines all four numbers uniquely. This is why the calendar magic trick works reliably.

Q3. Assertion (A):

For a 3-row number pyramid, the top number will always equal a + 2b + c if the bottom row is a, b, c.

Reason (R):

This formula works because the pyramid structure creates exactly two paths from each bottom corner to the top, giving each number a coefficient based on how many paths pass through it.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true (this is the correct formula), but the reason, while containing some intuitive truth, isn't the precise mathematical explanation. The actual reason involves binomial coefficients and the combinatorial structure of the pyramid. The reason describes an intuitive idea that partially explains it, but not the complete mathematical mechanism.

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Case-Based Questions (CBQ) for Algebra Play

Case Study:

Rohan is a teacher who wants to entertain his class with algebra tricks. He creates a 'Think of a Number' trick for students: Think of a number, multiply by 4, add 8, divide by 2, then subtract twice the original number. Rohan claims that no matter what number students think of, they will always get the same final answer. The students are skeptical because the instructions involve the original number. To convince them, Rohan shows the algebraic proof to the class.

Q1. If a student thinks of the number 5, what is the final result using Rohan's trick?

  • A. 5
  • B. 4 ✓
  • C. 3
  • D. 6

Answer: B

Q2. Using algebra, if x is the number thought of, which expression represents the result after 'divide by 2'?

  • A. 2x + 4 ✓
  • B. 2x + 8
  • C. 4x + 4
  • D. x + 4

Answer: A

Q3. What is the algebraic expression for the final answer after subtracting twice the original number?

  • A. 2x + 4 - 2x = 4 ✓
  • B. 2x + 4 - x = x + 4
  • C. 4x + 4 - 2x = 2x + 4
  • D. x + 4 - 2x = -x + 4

Answer: A

Q4. If Rohan wants to modify the trick so the final answer is always 6 instead of 4, which single step should he change?

  • A. Subtract once the original number instead of twice
  • B. Multiply by 6 instead of 4 in step 1
  • C. Add 12 instead of 8 in step 2 ✓
  • D. Add 10 instead of 8 in step 2

Answer: C

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