Math Talk — Multiplying Landmark Numbers in a Base System — A Story of Numbers

1. 12 marksGanita Prakash Cl-8 Part 1, Math Talk, page 66

   In the Egyptian system, what is any landmark number multiplied by the symbol for 10? Find the products for the landmarks 10, 100, 1000 and 10000.

   Hint. Write each landmark as a power of 10 first, then use the fact that multiplying by 10 adds one to the exponent.

   Every Egyptian landmark number is a power of 10, so multiplying by 10 raises the power by one — which is exactly the next landmark up.

   10¹ × 10 = 10² 10² × 10 = 10³ 10³ × 10 = 10⁴ 10⁴ × 10 = 10⁵

   General result: any landmark number multiplied by 10 gives the next landmark number.

   The reason is worth stating carefully. Ten copies of a landmark are, by the very construction of the system, grouped together to form the next landmark — that is how the sequence 1, 10, 100, 1000, … was built in the first place. So multiplying by 10 is not a calculation at all in this system; it is simply moving one step up the ladder of landmarks.

   In the Egyptian numeral this means you replace every symbol by the next symbol up, and the job is done.

   ✦ 10¹×10 = 10², 10²×10 = 10³, 10³×10 = 10⁴, 10⁴×10 = 10⁵ — any landmark multiplied by 10 gives the next landmark.

2. 22 marksGanita Prakash Cl-8 Part 1, Math Talk, page 66

   In the Egyptian system, what is any landmark number multiplied by the symbol for 100 (that is, 10²)? Find the products for the landmarks 10, 100, 1000 and 10000.

   Hint. Multiplying by 10² means multiplying by 10 twice, so the power goes up by two.

   Multiplying by 10² = 100 means multiplying by 10 twice over, so the exponent rises by two each time:

   10¹ × 10² = 10³ 10² × 10² = 10⁴ 10³ × 10² = 10⁵ 10⁴ × 10² = 10⁶

   General result: the product of any two landmark numbers is again a landmark number.

   This follows from the law for multiplying powers of the same base, 10ᵃ × 10ᵇ = 10ᵃ⁺ᵇ. Since every landmark is a power of 10, adding the exponents can only ever produce another power of 10, and therefore another landmark.

   This property is the real payoff of choosing powers of a single number as landmarks. In the Roman system the landmarks are 1, 5, 10, 50, 100, 500, 1000 — and V × L = 250, which is not a landmark at all, so the answer has to be laboriously regrouped. Here no regrouping is ever needed at this stage.

   ✦ 10¹×10² = 10³, 10²×10² = 10⁴, 10³×10² = 10⁵, 10⁴×10² = 10⁶ — the product of any two landmark numbers is again a landmark number.

3. 32 marksGanita Prakash Cl-8 Part 1, Math Talk, page 67

   Does the property that the product of two landmark numbers is again a landmark number hold true in the base-5 system we created? Does it hold for any number system with a base?

   Hint. Write the landmarks as powers of the base and multiply two of them.

   Yes to both.

   In the base-5 system. The landmarks are 5⁰, 5¹, 5², 5³, … Multiplying any two of them gives 5ᵃ × 5ᵇ = 5ᵃ⁺ᵇ, which is again a power of 5, and therefore again a landmark. For instance 25 × 125 = 5² × 5³ = 5⁵ = 3125, which is indeed the sixth landmark of the system.

   In any base-n system. The landmarks are n⁰, n¹, n², n³, … and the same law gives nᵃ × nᵇ = nᵃ⁺ᵇ, again a power of n, hence again a landmark.

   So this is not a special feature of base 10 or base 5 — it is a consequence of the definition of a base system, since insisting that every landmark be a power of one fixed number is exactly what forces products to stay inside the family.

   This is the structural reason the chapter gives for why base systems are efficient not just for writing numbers but for computing with them, and it is the property the Roman system lacks.

   ✦ Yes, and yes — in any base-n system nᵃ × nᵇ = nᵃ⁺ᵇ, which is again a power of n and hence again a landmark.

4. 41 markGanita Prakash Cl-8 Part 1, Math Talk, page 67

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

   Hint. Apply the distributive property to each landmark making up the number.

   We get the next landmark number for each part — that is, every symbol in the numeral moves up one step.

   Here is why. Any number in this system is a sum of landmark numbers. Multiplying that whole sum by 10 distributes across the sum: (a + b + c) × 10 = (a × 10) + (b × 10) + (c × 10).

   And we already know each individual landmark multiplied by 10 gives the next landmark. So every landmark in the number is simply replaced by the one above it.

   For example, 324 = 100 + 100 + 100 + 10 + 10 + 1 + 1 + 1 + 1. Multiplying by 10 turns each 100 into a 1000, each 10 into a 100 and each 1 into a 10, giving 1000 + 1000 + 1000 + 100 + 100 + 10 + 10 + 10 + 10 = 3240 ✓

   So multiplying by the base is done purely by symbol substitution, with no arithmetic at all — which in the Hindu system shows up as the familiar rule of simply appending a 0.

   ✦ Every symbol in the numeral is replaced by the symbol for the next landmark up, so the whole number moves one step up the ladder.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp103.pdf). This chapter is a history of number *systems* — tally marks, Roman, Egyptian, base-5, Mesopotamian, Mayan, Chinese rod and Hindu numerals — not a chapter on rational numbers. Questions appear as numbered 'Figure it Out' blocks plus 'Math Talk'/'Try This' prompts in the running text; every numeric answer here is checked against the book's own printed answer key at the end of the chapter. Four sub-parts whose questions exist only as printed Egyptian/base-5 glyph images (the addition drills on pages 65 and the products on page 68) are deliberately omitted rather than guessed at, since the operands cannot be recovered from the text.. Questions are referenced from the NCERT textbook for identification.