13 marksGanita Prakash Cl-8 Part 2, Section 6.6, pages 144-145
Mukta's trick: choose a two-digit number with different digits, reverse the digits, find the difference of the two numbers, and divide by 9 — there is never a remainder. Explain with algebra why this always works, taking the case b > a.
Hint. Write the number as 10a + b rather than as the two symbols side by side.
Write the numbers in expanded form. A two-digit number with tens digit a and units digit b is worth
10a + b
and reversing the digits gives
10b + a
This is the step that makes the trick transparent: the digits are positions, and only expanded form exposes the 10s.
Take b > a, so the reversed number is the larger and the difference is
(10b + a) − (10a + b)
= 10b + a − 10a − b
= (10b − b) + (a − 10a)
= 9b − 9a
= 9(b − a)
Since b − a is a whole number, the difference is 9 times a whole number, so dividing by 9 leaves no remainder — every time, for every such number.
The chapter's example. Start with 47: reversed it is 74, and 74 − 47 = 27. Here a = 4, b = 7, so 9(b − a) = 9 × 3 = 27 ✓ and 27 ÷ 9 = 3 exactly.
Two more, to be sure.
- 15 → 51, difference 36 = 9 × 4, and b − a = 5 − 1 = 4 ✓
- 29 → 92, difference 63 = 9 × 7, and b − a = 9 − 2 = 7 ✓
Why the digits must differ. If a = b the difference is 0. That is still divisible by 9, but the trick loses its point — which is why the instruction says to choose a number with different digits.
✦ Writing the number as 10a + b and its reverse as 10b + a, the difference is (10b + a) − (10a + b) = 9(b − a), which is a multiple of 9. So dividing by 9 can never leave a remainder.
22 marksGanita Prakash Cl-8 Part 2, Section 6.6, page 145
In Mukta's reverse-and-subtract trick, what happens if a > b instead of b > a?
Hint. The subtraction runs the other way — do the same algebra and see what changes.
Now the original number is the larger. With a > b, the number 10a + b exceeds its reverse 10b + a, so the difference is taken the other way round:
(10a + b) − (10b + a)
= 10a + b − 10b − a
= (10a − a) + (b − 10b)
= 9a − 9b
= 9(a − b)
Again a multiple of 9, so dividing by 9 still leaves no remainder.
The two cases together. Whichever digit is larger, the difference is
9 × |a − b|
— nine times the (positive) difference of the digits. The trick therefore works for every two-digit number, and the person performing it never has to know which digit was bigger.
Example. Start with 74: reversed it is 47, and 74 − 47 = 27 = 9 × |7 − 4| ✓ — the same 27 the chapter got from starting with 47, since only the difference of the digits matters, not their order.
A neat consequence. Because the result depends only on |a − b|, the trick can produce only the nine values 0, 9, 18, 27, 36, 45, 54, 63, 72. Every two-digit number in existence lands on one of those.
✦ If a > b the difference is 9(a − b), again a multiple of 9. Combining both cases, the difference is always 9 × |a − b|, so the trick works for every two-digit number regardless of which digit is larger.
Solutions written by the tuition.in editorial team and checked against NCERT Ganita Prakash Grade 8 Part 2 (hegp206.pdf), Chapter 6 'Algebra Play', pages 135-147. HAND-WRITTEN throughout. Part 2 books carry NO printed answer key, so every answer was derived from first principles and independently recomputed in Python. FIGURES READ OFF HIGH-DPI RENDERS AND RE-SOLVED: the three p.138 pyramids (bottom rows 6,2 / 3,4,3 / 5,4,5,0 giving tops 8, 14 and 32); the p.138 top-down pyramid 10 / 4,6 / 1,3,3; the p.139 letter-number pyramid 60 / 32,28 / 12,20,8; the three p.139 four-row pyramids, whose bottom rows solve to 4,9,6,1 (top 50), 5,14,7,2 (top 70) and 3,5,5,2 (top 35); the p.140 Figure it Out bottom rows 4,13,8 / 7,11,3 / 10,14,25 (tops 38, 32, 63) and 8,19,21,13 / 7,18,19,6 / 9,7,5,11 (tops 141, 124, 56); and the two p.142 algebra grids, which solve to square 11 and circle 5 with a third-row total of 21, and circle 4 and diamond 7 with a third-row total of 15 (the fourth row of that grid is printed blank for the student to design). The Virahanka-Fibonacci result was proved and then checked by computation for n = 2 to 6: every entry of the pyramid is a Virahanka-Fibonacci number, row k from the bottom is the consecutive run V(2k-1) to V(n+k-1), and the top is V(2n-1) - so a 29-row pyramid tops out at V57 = 591286729879. All six arrangements were enumerated for each Largest Product question: 32 x 5 = 160, 31 x 7 = 217 and 53 x 9 = 477, each beating its runner-up by exactly the margin the algebra predicts (4, 4 and 12). The date trick decodes as 100M + 165 + D, giving 4 November, 29 February and 31 January for 1269, 394 and 296. The puzzle answers were each verified by substitution: 7 flowers with 8 per shrine, 20 horses and 35 hens, a daughter of 6 and mother of 30, Gauri 6 cows and Naina 12, a dosa price of Rs 80 or a target of 175 dosas, the fraction sequence equal to 1/3 because n^2/(4n^2 - n^2) = 1/3, and Karim starting with 7 coins with the genie's general ruinous charge being c = 2^k x / (2^k - 1).. Questions are referenced from the NCERT textbook for identification.
