NCERT Solutions Ganita Prakash (Part 1) Chapter 2 –282.2 Exponential Notation and Operations · 2.3 The Other Side of Powers — In-text Questions

Book page 27 Updated on2026-09-05

Q1.

How many 5-digit passwords are possible?

Answer

10⁵ = 1,00,000.

Each of the 5 slots has 10 choices (0 to 9)
10 × 10 × 10 × 10 × 10 = 10⁵
= 1,00,000 passwords

The list runs 00000, 00001, 00002, …, 00010, 00011, …, 00100, …, 00999, …, 30456, …, 99998, 99999.

Why it happens: writing every number from 0 to 99,999 with exactly five digits — padding with leading zeros — matches passwords to numbers one for one. There are 99,999 – 0 + 1 = 1,00,000 such numbers, which confirms the count independently of the multiplication principle.

Q2.

Estu says, “Next time, I will buy a lock that has 6 slots with the letters A to Z. I feel it is safer.” How many passwords are possible with such a lock?

Answer

26⁶ = 30,89,15,776 — about 30.9 crore, or 3.09 × 10⁸.

26 × 26 × 26 × 26 × 26 × 26 = 26⁶
26² = 676
26³ = 17,576
26⁶ = (26³)² = 17,576 × 17,576
= 30,89,15,776

Estu is right. Compared with the 1,00,000 of the 5-digit lock, this is more than 3000 times as many passwords.

Why it happens: two things improved at once. The base rose from 10 to 26 (more symbols per slot) and the exponent rose from 5 to 6 (one more slot). Of the two, raising the exponent is usually the stronger move: 26⁷ would be 26 times bigger again, whereas widening the alphabet by one letter to 27⁶ multiplies the count by only about 1.26.

Tip: (26³)² is much less work than six multiplications — the power-of-a-power rule paying its way.

Q3.

Think about how many combinations are possible in different contexts. Some examples are — (i) Pincodes of places in India — The Pincode of Vidisha in Madhya Pradesh is 464001. The Pincode of Zemabawk in Mizoram is 796017. (ii) Mobile numbers. (iii) Vehicle registration numbers. Try to find out how these numbers or codes are allotted/generated.

Answer

Work out the raw count first, then see how the real system spends it.

Code	Structure	Possible codes
Pincode	6 digits, 0–9	10⁶ = 10,00,000
Mobile number	10 digits, but the first must be 6, 7, 8 or 9	4 × 10⁹ = 4,00,00,00,000
Vehicle number	2 letters + 2 digits + 2 letters + 4 digits	26² × 10² × 26² × 10⁴ ≈ 4.57 × 10¹¹

The real systems are not random. In a Pincode the first digit names one of the postal zones (4 = central India, so Vidisha; 7 = eastern India, so Zemabawk), the first two digits fix the sub-zone or state circle, the third fixes the sorting district, and the last three name the individual post office. Vehicle numbers start with the state code (MP, MZ, TN …) and then the RTO district number. Mobile numbers begin with an operator-and-circle block.

Why it happens: a code that carries meaning is easier to route and to check, but meaning costs capacity — once the first digit is fixed by geography, only 10⁵ Pincodes remain for that zone. Every real coding system settles somewhere between "pure counting" (maximum codes) and "readable structure" (fewer codes, fewer mistakes).

Q4.

What is 2¹⁰⁰ ÷ 2²⁵ in powers of 2?

Answer

2⁷⁵.

2¹⁰⁰ ÷ 2²⁵ = 2^{100 – 25} = 2⁷⁵

Why subtraction is the right operation: writing it out, there are 100 twos on top and 25 twos below. Each two below cancels one above, and 25 cancellations leave 100 – 25 = 75 twos.

In general: n^a ÷ n^b = n^a–b, where n ≠ 0 and a > b

Why it happens: division undoes multiplication, so where the product rule added exponents the quotient rule must subtract them. The chapter reaches this rule by halving a line of length 2⁴ units: 2⁴ ÷ 2³ = 2¹. The condition n ≠ 0 is essential, since you cannot cancel zeros.

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