Class 8 › Mathematics › Chapter
Algebra Play
Using letter-numbers to explain 'Think of a Number' tricks, number pyramids, calendar and shape grids, the largest-product problem, divisibility tricks and word puzzles.
1. 'Think of a Number' Tricks
Core idea: let the chosen number be a letter-number $x$ and apply each step. If $x$ cancels out, the result is the same for every starting number.
Example: the answer is always 2
$x \to 2x \to 2x+4 \to \dfrac{2x+4}{2} = x+2 \to x+2-x = 2$.
To get a different final value (say $3$ or $5$), change the number added in step 3 (add $6$ or $10$).
Date trick (month $M$, day $D$)
Steps: $\times 5$, $+6$, $\times 4$, $+9$, $\times 5$, $+D$ give $5M \to 5M+6 \to 20M+24 \to 20M+33 \to 100M+165 \to 100M+165+D$.
Decode: subtract $165$ from the answer to get $100M + D$. Since $D \le 31$ takes two digits, the last two digits are the day and the digits before them are the month.
Examples: $291 - 165 = 126 \Rightarrow$ 26 Jan. $1390 - 165 = 1225 \Rightarrow$ 25 Dec.
- If you change the steps, the number you subtract at the end changes too (it is the constant term of the final expression).
- To invent a trick: design steps so the variable cancels, or so the answer encodes the secret number in a readable form.
2. Number Pyramids
Rule: each number is the sum of the two numbers directly below it.
Top of a 3-row pyramid in terms of the bottom row.
- Filling from the top: use subtraction. For top $10$ with $4$ and $?$ below and bottom-left $1$: right $= 10-4 = 6$, next bottom $= 4-1 = 3$, last bottom $= 6-3 = 3$.
- Unknown in the middle: use letter-numbers and form equations. For bottom $12, c, 8$ and top $60$: $a = 12+c$, $b = c+8$, $a+b = 60 \Rightarrow 20+2c = 60 \Rightarrow c = 20$, giving row $32, 28$.
- 2 rows: top $= a + b$.
- 3 rows: top $= a + 2b + c$ (middle counted twice).
- 4 rows: top $= a + 3b + 3c + d$ (coefficients $1, 3, 3, 1$). In general the coefficients follow Pascal's triangle, so the top can be found without building the pyramid.
- Example: bottom $4, 13, 8$ gives top $4 + 26 + 8 = 38$.
- Virahāṅka-Fibonacci numbers ($1, 2, 3, 5, \dots$, each is the sum of the previous two) in the bottom row give a pyramid where the patterns can be studied using these rules.
3. Fun with Grids
Calendar magic
$2 \times 2$ grid
Let the top-left be $a$. The grid is $a,\ a+1,\ a+7,\ a+8$ (a week is $7$ days). Sum $= 4a + 16$.
Decode
Find the grid from the sum
Sum $36$: $4a+16 = 36 \Rightarrow a = 5$, grid $5, 6, 12, 13$. Sum $40 \Rightarrow a = 6$.
- Own tricks: choose a different shape or size (a $3 \times 3$ block, an L-shape, a row of 3). Write every cell in terms of one letter, find the sum, then solve.
- Algebra grids (shapes as numbers): in each row the last column is the sum of the others. Use a row with a single unknown shape first, then substitute. Example: three squares $= 27 \Rightarrow$ square $= 9$; two circles + square $= 19 \Rightarrow 2c + 9 = 19 \Rightarrow$ circle $= 5$.
4. The Largest Product
Problem: place three digits in a box pattern $\square\square \times \square$ (each digit once) to get the largest product.
Result
For digits $p \lt q \lt r$, the largest product is $\overline{qp} \times r$
Reasoning
1. There are $6$ arrangements. Group them by multiplier; in each pair, the larger tens digit in the two-digit number wins. This leaves $\overline{rq}\times p,\ \overline{rp}\times q,\ \overline{qp}\times r$.
2. $\overline{rp}\times q \gt \overline{rq}\times p$ because $q \gt p$.
3. Expand: $\overline{qp}\times r = 10qr + pr$ and $\overline{rp}\times q = 10rq + pq$. First terms equal; since $r \gt q$, $pr \gt pq$, so $\overline{qp}\times r$ is larger.
- Rule: use the largest digit as the multiplier, and arrange the other two in decreasing order as the two-digit number.
- Examples: digits $2, 3, 5$: $32 \times 5 = 160$. Digits $1, 3, 7$: $31 \times 7 = 217$. Digits $3, 5, 9$: $53 \times 9 = 477$.
5. Decoding Divisibility Tricks
Place value: a two-digit number $ab = 10a + b$; a three-digit number $abc = 100a + 10b + c$.
| Trick | Algebra | Result |
|---|---|---|
| Reverse a 2-digit number, find the difference | $(10b + a) - (10a + b) = 9(b - a)$ | Divisible by $9$; quotient is $b - a$ (the difference of the digits) |
| Reverse, find the sum | $(10a + b) + (10b + a) = 11(a + b)$ | Divisible by $11$; quotient is $a + b$ |
| Add the 3 cyclic numbers $abc, bca, cab$ | $111(a + b + c) = 37 \times 3(a + b + c)$ | Divisible by $37$ and by $3$ (also by $111$) |
| Repeat a 3-digit number: $abcabc$ | $1000\,abc + abc = 1001 \times abc$ | $1001 = 7 \times 11 \times 13$, so dividing by $7$, $11$, $13$ returns $abc$ |
- If $a \gt b$ in the first trick, the difference is $9(a - b)$: still a multiple of $9$.
- Example: $47 \to 74$; $74 - 47 = 27 = 9 \times 3$ and $4 + 7 = 11$ gives $11 \times 11 = 121$ for the sum trick.
6. Word Problems & Puzzles
Method
1. Name the unknown with a letter-number. 2. Translate each statement into an expression or equation. 3. Solve, then check against the story.
| Problem | Setup | Answer |
|---|---|---|
| Horses & hens: $55$ heads, $150$ legs | All hens $\Rightarrow 110$ legs; extra $40$ legs, each horse adds $2$; or $4h + 2(55-h) = 150$ | $20$ horses, $35$ hens |
| Mother is $5\times$ daughter; in $6$ years $3\times$ | $5d + 6 = 3(d + 6)$ | Daughter is $6$ |
| Gauri & Naina's cows (Naina has twice; after giving $3$, equal) | $2g - 3 = g + 3$ | Gauri $6$, Naina $12$ |
| Dosa cart: rent $₹5000$, cost $₹10$/dosa, profit $₹2000$ | (i) $100p = 5000 + 1000 + 2000$. (ii) $50n - 10n - 5000 = 2000$ | (i) $₹80$ each. (ii) $175$ dosas |
| Flowers doubled at $3$ ponds, equal $s$ left at each shrine | Start $x$: $8x - 6s = s \Rightarrow 8x = 7s$ | Smallest: $x = 7$, $s = 8$ per shrine |
| Karim & the genie: double, pay $8$, three times, left with $8$ | $8x - 56 = 8$ | Started with $8$ coins |
- Karim (ii): with $x$ coins and cost $c$ per round, after $3$ rounds he has $8x - 7c$; he gains only if $c \lt x$. (iii) The genie takes everything if $8x - 7c = 0$, i.e. $c = \tfrac{8x}{7}$.
- Sum of odd numbers: $\dfrac{1}{3},\ \dfrac{1+3}{5+7},\ \dfrac{1+3+5}{7+9+11}$ all equal $\dfrac13$. The first $n$ odd numbers sum to $n^2$ and the next $n$ odd numbers sum to $3n^2$, so the ratio is always $\tfrac13$.
7. Key Results and Quick Revision
| Topic | Key expression |
|---|---|
| Date trick | Answer $- 165 = 100M + D$ |
| 3-row pyramid | Top $= a + 2b + c$ |
| 4-row pyramid | Top $= a + 3b + 3c + d$ |
| Calendar $2 \times 2$ | Sum $= 4a + 16$ |
| Largest product | $\overline{qp} \times r$ with $p \lt q \lt r$ |
| Reverse difference / sum | $9(b - a)$ / $11(a + b)$ |
| Cyclic 3-digit sum | $111(a + b + c) = 37 \times 3(a+b+c)$ |
| Repeated 3-digit | $abcabc = 1001 \times abc$ |
Common mistakes:
• Writing $ab$ as $a \times b$ instead of $10a + b$.
• Forgetting that the middle number is counted twice in a 3-row pyramid.
• Taking calendar neighbours as $a + 1$ below instead of $a + 7$.
• Not simplifying fully, so the letter-number does not cancel.
• Not checking that answers fit the story (whole animals, ages, days).
• Assuming a trick works for all numbers without proving it algebraically.
• Writing $ab$ as $a \times b$ instead of $10a + b$.
• Forgetting that the middle number is counted twice in a 3-row pyramid.
• Taking calendar neighbours as $a + 1$ below instead of $a + 7$.
• Not simplifying fully, so the letter-number does not cancel.
• Not checking that answers fit the story (whole animals, ages, days).
• Assuming a trick works for all numbers without proving it algebraically.