NCERT Solutions Ganita Prakash (Part 2) Chapter 6 – 1436.5 The Largest Product — In-text Questions

Book page 142 Updated on2026-09-05

Q1.

Fill the digits 2, 3, and 5 in ⬜⬜ × ⬜, using each digit once. What is the largest product possible?

Answer

The largest is 32 × 5 = 160.

Arrangement	Product	Arrangement	Product
23 × 5	115	35 × 2	70
25 × 3	75	52 × 3	156
32 × 5	160	53 × 2	106

Why it happens: there are only six arrangements, because the multiplier can be any of the three digits and the remaining two can go either way round. The winner puts the largest digit (5) as the multiplier and the other two in decreasing order (32). The reason is place value: the tens digit of the multiplicand gets multiplied by 10 and by the multiplier, so it should not be the biggest digit — the biggest digit does more work standing outside, where it multiplies everything.

Q2.

How do we find the largest product among these six options?

Answer

Compare in pairs instead of testing all six one by one.

Group by multiplier: (35 × 2, 53 × 2) (25 × 3, 52 × 3) (23 × 5, 32 × 5)
In each pair the multiplier is the same, so the bigger multiplicand wins:
53 × 2, 52 × 3, 32 × 5 survive.

52 × 3 = 156 > 106 = 53 × 2, so 53 × 2 is out.
Now expand the last two:
32 × 5 = (3 × 10 × 5) + (2 × 5) = 150 + 10
52 × 3 = (5 × 10 × 3) + (2 × 3) = 150 + 6
The first parts are equal; the second part is larger for 32 × 5.
Largest = 32 × 5 = 160

Why it happens: splitting each product by place value is what makes the comparison easy. Both 32 × 5 and 52 × 3 contain the term 3 × 5 × 10 — the same two digits multiplied and scaled by ten. So they differ only in the smaller term, 2 × 5 against 2 × 3. Comparing two things that share most of their structure is far quicker, and far more convincing, than computing every product.

Q3.

In this case, we used the largest digit as the multiplier. The other two digits were arranged in decreasing order to form the multiplicand. Will this always be the case? Let us find out using algebra.

Answer

Yes — it is always the case. Take any three digits p, q, r with p < q < r. There are six arrangements; group them by multiplier.

qr × p, rq × p | pr × q, rp × q | pq × r, qp × r

In each pair the multiplier is the same, and the multiplicand with the larger tens digit is the larger number. So only three survive:

rq × p, rp × q, qp × r

Step 1 — rp × q beats rq × p. Write both out by place value:

rq × p = (10r + q) × p = 10rp + qp
rp × q = (10r + p) × q = 10rq + pq
The second terms are equal (qp = pq). Since q > p, we get 10rq > 10rp.
So rp × q > rq × p.

Step 2 — qp × r beats rp × q.

qp × r = (10q + p) × r = 10qr + pr
rp × q = (10r + p) × q = 10rq + pq
Again the first terms are equal (10qr = 10rq). Since r > q, we get pr > pq.
So qp × r > rp × q.

The largest product is therefore qp × r = (10q + p) × r — the largest digit outside as the multiplier, the other two in decreasing order.

Why it happens: in every product one pair of digits gets the ×10 boost and one pair does not. Writing (10q + p) × r as 10qr + pr shows that the two largest digits q and r are the ones multiplied together and scaled by 10 — the best possible use of the boost. Any other arrangement wastes the ×10 on a pair that includes the smallest digit p. Testing examples could never establish this for all digit sets; the letter-numbers do it in two lines.

Check it yourself: with p = 2, q = 3, r = 5 the rule gives (10 × 3 + 2) × 5 = 32 × 5 = 160 — exactly the answer found by listing all six. ✓

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