NCERT Solutions Ganita Prakash (Part 1) Chapter 3 .3 Advantages of a Base-n System — In-text Questions

Book page 673 Updated on2026-09-05

Q1.

Find the following products — (i) ∩ × tadpole (ii) coil × lotus (iii) lotus × lotus (iv) finger × kneeling man

Answer

Add the powers of 10 each time.

	Product	In powers of 10	Answer
(i)	∩ × tadpole	10¹ × 10⁵ = 10¹⁺⁵	10⁶ — the kneeling man
(ii)	coil × lotus	10² × 10³ = 10²⁺³	10⁵ — the tadpole
(iii)	lotus × lotus	10³ × 10³ = 10³⁺³	10⁶ — the kneeling man
(iv)	finger × kneeling man	10⁴ × 10⁶ = 10⁴⁺⁶	10¹⁰

Thus the product of any two landmark numbers is another landmark number.

Why it happens: part (iv) is worth pausing over. 10¹⁰ is a landmark number all right, but the Egyptian symbol list stops at 10⁷ (the sun). So the answer exists as a number but cannot be written down in Egyptian numerals. This is precisely the shortcoming set out on page 69: a system that gives each landmark its own symbol needs an unending supply of symbols.

Tip: 10¹⁰ is one thousand crore, or 1 followed by ten zeroes. In our system it is no trouble at all — the same ten digits handle it.

Q2.

Does this property hold true in the base-5 system that we created? Does this hold for any number system with a base?

Answer

Yes in the base-5 system, and yes in every base-n system.

In base 5: ⬡ × □ = 25 × 5 = 125 = ○   (5² × 5¹ = 5³)
              ○ × □ = 125 × 5 = 625 = ∿   (5³ × 5¹ = 5⁴)
              ⬡ × ⬡ = 25 × 25 = 625 = ∿   (5² × 5² = 5⁴)

In general: n^a × n^b = n^a+b

Why it happens: n^a is n written down as a factor a times, and n^b is n written down b times. Multiply them and you have n written down a + b times — which is n^a+b, again a power of n, and therefore again a landmark number. Nothing about 10 or 5 was used; the argument is about counting factors, so it holds for every base.

Check it yourself: now compare V × L = CCL in the Roman system. Roman landmarks are 1, 5, 10, 50, 100, 500, 1000 — the ratios are 5, 2, 5, 2, 5, 2, so they are not the powers of any one number, and the property fails. That single failure is the whole reason Roman multiplication is hard.

Q3.

What can we conclude about the product of a number and ∩ (10), in the Egyptian system?

Answer

Every symbol in the numeral is replaced by the next symbol up, and the number of each kind stays the same.

The book shows why with two examples.

(i) coil coil × ∩
coil coil is the same as coil + coil, so
(coil + coil) × ∩ = (coil × ∩) + (coil × ∩) (distributive law)
= lotus + lotus = lotus lotus
In numbers: (100 + 100) × 10 = 1000 + 1000 = 2000 ✓

Why it happens: the distributive law lets you multiply each part separately, and each part is a landmark, so each part simply steps up one place in the symbol list. The counts are untouched — two coils become two lotus, two arches become two coils. In Hindu numerals this same statement reads: “to multiply by 10, write a 0 at the end.” 121 becomes 1210, and the digits 1, 2, 1 have not changed at all, they have only moved one place left.

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