14 marksGanita Prakash Cl-8 Part 2, Section 6.5, pages 142-143
Fill the digits 2, 3 and 5 into a two-digit number times a one-digit number, using each digit once. What is the largest product possible? Work through it systematically rather than by trial.
Hint. There are only six arrangements — group them by which digit is the single multiplier.
Count the arrangements. The single digit can be chosen in 3 ways, and the remaining two digits can be ordered in 2 ways, giving 6 arrangements:
23 × 5, 25 × 3, 32 × 5, 35 × 2, 52 × 3, 53 × 2
Group them by the multiplier.
| Multiplier | Options |
|---|
| 2 | 35 × 2, 53 × 2 |
| 3 | 25 × 3, 52 × 3 |
| 5 | 23 × 5, 32 × 5 |
Cut each pair down to one. Within a pair the multiplier is the same, so the larger two-digit number wins. That leaves three candidates:
53 × 2, 52 × 3, 32 × 5
Compare the survivors. 52 × 3 = 156 beats 53 × 2 = 106 straight away, so the contest is between 52 × 3 and 32 × 5. Expand both:
- 32 × 5 = (3 × 10 × 5) + (2 × 5) = 150 + 10
- 52 × 3 = (5 × 10 × 3) + (2 × 3) = 150 + 6
The first terms are identical — both are 10 × 3 × 5 — so everything turns on the second term, and 2 × 5 = 10 beats 2 × 3 = 6.
Therefore 32 × 5 = 160 is the largest.
The full table, for confirmation.
| Product | Value |
|---|
| 32 × 5 | 160 |
| 52 × 3 | 156 |
| 23 × 5 | 115 |
| 53 × 2 | 106 |
| 25 × 3 | 75 |
| 35 × 2 | 70 |
The pattern. The largest digit became the multiplier (5), and the other two were arranged in decreasing order to form the two-digit number (32). Whether that always happens is settled in the next question.
✦ The largest product is 32 × 5 = 160. The systematic route is to group the six arrangements by multiplier, keep the larger multiplicand in each pair, and then compare 53 × 2, 52 × 3 and 32 × 5 — where the last two share the term 10 × 3 × 5 and differ only in 2 × 5 against 2 × 3.
25 marksGanita Prakash Cl-8 Part 2, Section 6.5, pages 143-144
Prove using algebra that for any three digits p < q < r, the largest product of the form (two-digit number) × (one-digit number) is obtained by using r as the multiplier and arranging q and p in decreasing order as the multiplicand.
Hint. Group the six products by multiplier exactly as in the numerical case, and compare the two survivors term by term.
Set up. Let the three digits be p, q, r with p < q < r. Writing 'qp' for the two-digit number with tens digit q and units digit p, the six products grouped by multiplier are
| Multiplier | Options |
|---|
| p | qr × p, rq × p |
| q | pr × q, rp × q |
| r | pq × r, qp × r |
Step 1 — reduce each pair. In each pair the multiplier is the same, so the arrangement with the larger tens digit gives the larger two-digit number and hence the larger product. Since p < q < r, the survivors are
rq × p, rp × q, qp × r
Step 2 — eliminate one. Compare rq × p with rp × q. Both two-digit numbers have tens digit r, and rq > rp exactly when q > p — but the multipliers differ too, so expand:
- rq × p = (10r + q)p = 10rp + qp
- rp × q = (10r + p)q = 10rq + pq
The second terms qp and pq are the same product. The first terms are 10rp and 10rq, and since q > p we get 10rq > 10rp. So rp × q is larger, and rq × p is eliminated.
Step 3 — the final comparison. Compare qp × r with rp × q:
- qp × r = (10q + p)r = (10 × q × r) + (p × r)
- rp × q = (10r + p)q = (10 × r × q) + (p × q)
The first terms are identical — both equal 10qr. So the comparison reduces entirely to the second terms:
p × r against p × q
Since r > q and p > 0, we have pr > pq, so qp × r is the larger.
Conclusion. The largest of all six is qp × r: the largest digit r is the multiplier, and the remaining two digits are placed in decreasing order (q in the tens place, p in the units) to form the multiplicand.
Check against the worked case. With p = 2, q = 3, r = 5 the rule gives qp × r = 32 × 5 = 160, which matches the table exactly ✓
Why the rule makes sense. The tens digit of the multiplicand is multiplied by 10 and then by the multiplier, so the two big digits q and r should be the ones that meet in that term — one as the tens digit and the other as the multiplier. Which of them takes which role is then decided by the small leftover term, and that is where p × r beats p × q.
✦ For p < q < r the largest product is qp × r — the largest digit as the multiplier, the other two in decreasing order as the two-digit number. The proof reduces the six arrangements to three, eliminates rq × p because 10rq > 10rp, and then compares qp × r with rp × q, which share the term 10qr and differ only in p × r against p × q, where r > q decides it.
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.
