Algebra Play — Class 8 Mathematics (Ganita Prakash Part 2)

Test a trick on ten numbers and you have ten pieces of evidence. Carry one letter through the steps and you have a proof.

1. About the Chapter

This is Chapter 6 of Ganita Prakash Part 2 (pages 135–147), the thirteenth chapter of the Class 8 course and the shortest in Part 2. Its opening line sets the terms:

"Over the last two years, we have used algebra to model different situations. We have learned how to solve algebraic equations and find the values of unknown letter-numbers. Let's now have some fun with algebra."

So solving linear equations is assumed knowledge, not the topic. The topic is what algebra is for: investigating tricks and puzzles, explaining why they work, and then inventing new ones.

Five sections, five kinds of puzzle:

What is not in this chapter

Older notes for "Chapter 13" often cover linear equations, the transposition method and word-problem drill. Those are prerequisites, assumed from Grade 7 — a short refresher is kept in the appendix at the end of this page.

Key idea

Algebra's most powerful use is not calculating an answer. It is justifying a statement about infinitely many cases at once.

2. 'Think of a Number' Tricks

The basic trick

Think of a number. Double it. Add four. Divide by two. Subtract the number you first thought of.

The answer is always 2. Try it: 7 → 14 → 18 → 9 → 2. Try 100 → 200 → 204 → 102 → 2. Try 0.5 → 1 → 5 → 2.5 → 2.

Testing is not proving, though. Carry a letter instead:

The x cancels. That single line covers every number that will ever be chosen.

Designing your own

Notice where the 2 came from: it is half the number that was added. Everything else vanished.

• To finish at 3, add 6
• To finish at 5, add 10
• To finish at k, add 2k

The general design rule: arrange the steps so that the coefficient of x is 1 just before you subtract the original number. After that point you are doing arithmetic on a constant and can make the trick as elaborate as you like:

think of x → add 5 → multiply by 3 → subtract 9 → divide by 3 → subtract the original → multiply by 4 → 8

The date trick

Multiply the month by 5, add 6, multiply by 4, add 9, multiply by 5, add the day.

The month has been pushed up to 100M, clear of the two digits a day needs. So subtract 165 and read the last two digits as the day, the rest as the month.

Where does 165 come from? The 6 is later multiplied by 4 and then 5, growing to 120; the 9 is multiplied by 5, growing to 45; and 120 + 45 = 165. Change a step and the constant changes with it — add 3 and 7 instead of 6 and 9 and you subtract 95.

3. Number Pyramids

The rule: each box is the sum of the two boxes directly below it.

      14
    7    7
  3   4   3

Three ways to fill one

Upwards, by adding, when a row is complete. Bottom 5, 4, 5, 0 → 9, 9, 5 → 18, 14 → 32.

Downwards, by subtracting, when you know a box and one of the two feeding it. Given 10 on top, 4 in the middle-left and 1 at bottom-left:

10 − 4 = 6 → 4 − 1 = 3 → 6 − 3 = 3

      10
    4     6
  1    3    3

By algebra, when neither can start. Top 60, bottom 12, ?, 8, middle row empty:

        60                         60
      a     b       →           32    28
   12    c     8              12    20    8

• a + b = 60
• 12 + c = a
• c + 8 = b

Substituting: (12 + c) + (c + 8) = 60, so 20 + 2c = 60 and c = 20, giving a = 32 and b = 28.

Notice that c appears in both middle boxes, which is why the equation had 2c. That doubling reappears immediately as the formula for the top.

The top in terms of the bottom

Three rows, bottom a, b, c:

    a + 2b + c
   a+b     b+c
  a    b     c

Top = (a + b) + (b + c) = a + 2b + c — the middle counts twice.

Four rows, bottom a, b, c, d: top = a + 3b + 3c + d = (a + d) + 3(b + c)

The coefficients 1,1 / 1,2,1 / 1,3,3,1 / 1,4,6,4,1 are the rows of Pascal's triangle — no coincidence, since Pascal's triangle is built by the same add-the-two-above rule. Each coefficient counts the number of upward paths from that bottom box to the top.

The Virahāṅka-Fibonacci pyramid

Write the first n terms of 1, 2, 3, 5, 8, 13, … along the bottom.

          21
       8     13
     3    5     8
   1   2    3     5

Every entry is a Virahāṅka-Fibonacci number. Why: the pyramid rule and the sequence rule are the same rule — add two neighbours. So a run of consecutive terms produces another run of consecutive terms, starting two places further along:

4. Fun with Grids

Calendar magic

Moving one square right adds 1 day; moving one square down adds 7. So a 2 × 2 block with top-left a is

Sum = 4a + 16 = 4(a + 4), so a = (sum ÷ 4) − 4.

Design your own

The 3 × 3 and the cross are the most striking: the neighbours cancel in pairs, since each is as far above the centre as its partner is below.

One caution. The block must fit inside the month, or the +1 and +7 relations break at the edges.

Algebra grids

Shapes stand for numbers; the last column of each row is the sum of the cells to its left. This is simultaneous equations dressed up as pictures.

Worked example. Row 1 = ■ ■ ■ = 27 gives 3■ = 27, so ■ = 9. Row 2 = ● ● ■ = 19 gives 2● + 9 = 19, so ● = 5.

Grid A. ■■● = 27 and ●●■ = 21. 2s + c = 27 and s + 2c = 21. Adding: 3(s + c) = 48, so s + c = 16, hence s = 11, c = 5. Third row ●■● = 21.

Grid B. ●◆◆ = 18 and ◆●● = 15. b + 2d = 18 and 2b + d = 15 give d = 7, b = 4. Third row ◆●● = 15.

Rows 2 and 3 of each grid contain the same shapes in a different order and therefore the same total — only the multiset of shapes matters, not their arrangement. That is also the check to make before you start: if two rows hold the same shapes, they give the same equation twice and the puzzle has no unique answer.

5. The Largest Product

Fill three digits into (two-digit number) × (one-digit number), using each once. Which arrangement is largest?

The case 2, 3, 5

Six arrangements, grouped by multiplier:

52 × 3 = 156 beats 53 × 2 = 106 at once. The real contest:

• 32 × 5 = (3 × 10 × 5) + (2 × 5) = 150 + 10
• 52 × 3 = (5 × 10 × 3) + (2 × 3) = 150 + 6

The first terms are identical. So 32 × 5 = 160 wins.

The general rule, proved

Let p < q < r. Grouping and reducing leaves three candidates: rq × p, rp × q, qp × r.

Eliminate rq × p. rq × p = 10rp + qp and rp × q = 10rq + pq share the term pq, and 10rq > 10rp since q > p.

Final comparison.

• qp × r = 10qr + pr
• rp × q = 10rq + pq

Identical first terms again, so it turns on pr against pq — and r > q decides it.

The largest digit is the multiplier; the other two go in decreasing order. Answer: qp × r.

Each margin is exactly what the algebra predicts: for 3, 5, 9 it is 3×9 − 3×5 = 27 − 15 = 12 ✓

The common wrong instinct is to put the two largest digits into the two-digit number — 95 × 3, which gives only 285. The multiplier acts on both digits of the multiplicand, so making the multiplier large matters more.

6. Decoding Divisibility Tricks

Everything in this section rests on one move:

Write the number in expanded form. 'ab' is not a × b — it is 10a + b.

Reverse and subtract

If b > a, the difference is

(10b + a) − (10a + b) = 9b − 9a = 9(b − a)

If a > b it is 9(a − b). Either way the difference is 9 × |a − b| — always a multiple of 9, and the quotient is the difference of the digits.

74 − 47 = 27 = 9 × 3, and 7 − 4 = 3 ✓

Reverse and add

(10a + b) + (10b + a) = 11(a + b)

Always a multiple of 11, and the quotient is the sum of the digits.

The two tricks are twins: subtracting cancels the units and leaves (10 − 1) = 9 lots of the difference; adding reinforces them and leaves (10 + 1) = 11 lots of the sum.

Cycling three digits

abc + bca + cab = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = 111(a + b + c)

Each digit visits the hundreds, tens and units place exactly once — hence 111. And

111 = 3 × 37

so the sum is always divisible by 37 and by 3. Check: 152 + 521 + 215 = 888 = 111 × 8 ✓

Repeating three digits

abcabc = abc × 1000 + abc = abc × 1001, and

1001 = 7 × 11 × 13 — three consecutive primes

So dividing by 7, then 11, then 13 hands back the original number. 486486 ÷ 7 = 69498 ÷ 11 = 6318 ÷ 13 = 486 ✓

7. Worked Examples

Example 1 — The magical ponds

Three ponds each double the flowers dipped in them; the same number k is left at each of three shrines, and the hand ends empty.

So 8x = 7k, and since 7 and 8 share no factor, the smallest solution is k = 8, x = 7.

Check: 7 → 14, leave 8 → 6 → 12, leave 8 → 4 → 8, leave 8 → 0 ✓

The doubling makes early offerings expensive: the k left at shrine 1 would have doubled twice more, costing 4k of the final total, then 2k, then k — and 4k + 2k + k = 7k is the same equation seen from the other side.

Example 2 — Horses and hens

55 heads, 150 legs. With algebra: h + n = 55 and 4h + 2n = 150 give h = 20, n = 35.

Without algebra: if all 55 were hens there would be 110 legs. The 40 extra legs arrive 2 at a time as each hen becomes a horse, so 40 ÷ 2 = 20 horses.

Example 3 — Mother and daughter

Mother is 5× the daughter now; in 6 years she will be 3×.

5d + 6 = 3(d + 6) → 2d = 12 → d = 6, mother 30.

The gap never changes: 30 − 6 = 24, now and always. What falls is the ratio, as the fixed 24 becomes a smaller fraction of a growing age. That gives a second route: if mother = 3 × daughter and the gap is 24, then the daughter is 12 then, so 6 now.

Example 4 — The cowherds

Naina has twice Gauri's cows; giving 3 away would even them up.

2g − 3 = g + 3 → g = 6, Naina 12.

Handing over 3 moves one up by 3 and the other down by 3, so it closes a gap of 6 — which must therefore have been the gap. Getting the direction of a transfer right is where marks are usually lost.

Example 5 — The dosa cart

Rent ₹5000/day (fixed), ₹10 per dosa (variable). Profit = np − (5000 + 10n).

• 100 dosas, ₹2000 profit: cost 6000, revenue needed 8000, so p = ₹80
• ₹50 a dosa, ₹2000 profit: 40n = 7000, so n = 175

Each ₹50 dosa contributes ₹40 to rent-plus-profit. Break-even alone needs 5000 ÷ 40 = 125 dosas; every dosa after that is profit.

Example 6 — The odd-number fractions

1/3, (1+3)/(5+7), (1+3+5)/(7+9+11) → 1/3, 4/12, 9/27 — all equal 1/3.

The sum of the first n odd numbers is n². So

• numerator = n²
• denominator = (2n)² − n² = 3n²

giving n²/3n² = 1/3 exactly, for every n.

Example 7 — Karim and the Genie

Coins double each round; the genie takes 8. One round: x ↦ 2x − 8.

8x − 48 = 8 (the third doubling left exactly 8) → x = 7

7 → 14, pay 8 → 6 → 12, pay 8 → 4 → 8, pay 8 → 0

To empty the purse in exactly k rounds the genie sets c = 2ᵏx/(2ᵏ − 1). With k = 3 and x = 7 that is 8 — precisely what it charged. The offer sounded trifling only because 8 was just above 7.

8. Common Mistakes

1. Testing instead of proving. Ten examples are ten examples. One letter settles every case.
2. Reading 'ab' as a × b. In digit puzzles it means 10a + b. Write that first, always.
3. Only ever adding upwards in a pyramid. The rule runs backwards as a subtraction, and when neither direction can start, name the unknowns.
4. Forgetting the middle counts twice. The top of a three-row pyramid is a + 2b + c, not a + b + c.
5. Putting the two big digits in the multiplicand. 95 × 3 = 285 loses badly to 53 × 9 = 477.
6. Getting a transfer's direction wrong. Three cows given away close a gap of six, not three.
7. Memorising 165 instead of deriving it. Change the added numbers and the constant changes.
8. Ignoring the fixed cost in a profit question. Rent does not scale with sales; separate it out.

9. Real-World Applications

• Checksums — Aadhaar numbers, ISBNs and bank account numbers carry a check digit so that a single mistyped or transposed digit is caught. Transposition is caught by exactly the 9 and 11 arguments in this chapter.
• Mental arithmetic — knowing 1001 = 7 × 11 × 13 and 111 = 3 × 37 turns awkward divisions into instant ones.
• Puzzle design — every card trick with a guaranteed outcome is built the same way: arrange the steps so the unknown cancels.
• Break-even analysis — the dosa cart is real costing. Fixed cost, variable cost, contribution per unit.
• Compound growth with a fixed charge — Karim's tree is a loan in disguise. Whether it grows or collapses turns on the comparison c < x, which is the arithmetic behind every EMI schedule.

10. Conclusion

The chapter's own summary is three lines:

• Algebra is very useful in modelling and understanding numerical scenarios, which is why it occurs in almost all of mathematics, science and beyond.
• Algebra is an indispensable tool in justifying mathematical statements.
• It was applied here to 'think of a number' tricks, number pyramids, grids, forming numbers to maximise products, divisibility tricks and various other problems.

The second line is the one that matters. Everywhere else, algebra finds a number; here it settles a claim about every number at once — and it settles a dozen apparently unrelated puzzles with the same handful of moves: write the expanded form, name the unknown, and see what cancels.

The chapter also keeps asking you to invent: change the trick, design your own calendar block, build a puzzle for a friend. That is not decoration. Solving a puzzle shows you followed a method; designing one shows you understood why it works.

Appendix — Linear Equations: The Assumed Prerequisite

The chapter opens by saying equation-solving is already known from Grade 7. This is the refresher.

The balance principle

An equation stays true if you do the same thing to both sides — add, subtract, multiply or divide (never by zero).

3x + 5 = 20 → subtract 5 → 3x = 15 → divide by 3 → x = 5

Transposition

Moving a term across the equals sign flips its operation. It is the balance principle written more briefly.

5x − 7 = 3x + 9 → 5x − 3x = 9 + 7 → 2x = 16 → x = 8

Standard steps

1. Clear brackets
2. Clear fractions by multiplying through by the LCM of the denominators
3. Collect the variable terms on one side, constants on the other
4. Simplify and divide by the coefficient
5. Verify by substituting into the original equation

Translating a word problem

State clearly what the letter means, then convert each phrase in turn. "A number is doubled and 7 added; the result is 19" becomes 2x + 7 = 19, so x = 6.

Verification must be against the original wording, not against your own equation — that is how a mis-translation is caught.