7.4 Q-9 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.4 — Question 9

This question applies the Binomial Theorem to solve problems involving:

Finding the remainder when a large power is divided by a number
Finding the last digit of a large power
Testing divisibility of a number
Comparing two large numbers

Key Technique: Rewriting the Base

The core strategy is to rewrite the base as $(1 + k m)$ or $(k m - 1)$ where $k m$ is a multiple of the divisor, then expand using the Binomial Theorem:

$(1 + k m)^{n} = \sum_{r = 0}^{n} (r n) (k m)^{r} = 1 + n (k m) + (2 n) (k m)^{2} + \dots$

All terms except the first ( $r = 0$ ) are divisible by $k m$ , hence by $m$ .

Finding the Remainder

Method: Write the base as $(1 + d \cdot q)$ where $d$ is the divisor.

Example: Find the remainder when $7^{100}$ is divided by $6$ .

$7^{100} = (6 + 1)^{100} = \sum_{r = 0}^{100} (r 100) 6^{r} \cdot 1^{100 - r}$

$= 1 + (1 100) \cdot 6 + (2 100) \cdot 6^{2} + \dots + 6^{100}$

Every term with $r \geq 1$ contains a factor of $6$ , so they are all divisible by $6$ .

$∴ 7^{100} = 6 k + 1 for some integer k$

Remainder $= 1$

Finding the Last Digit

The last digit of $N$ equals the remainder when $N$ is divided by $10$ .

Example: Find the last digit of $3^{100}$ .

$3^{100} = (3^{2})^{50} = 9^{50} = (10 - 1)^{50}$

$= \sum_{r = 0}^{50} (r 50) 1 0^{r} \cdot (- 1)^{50 - r}$

All terms with $r \geq 1$ are divisible by $10$ . The $r = 0$ term is:

$(0 50) \cdot 1 0^{0} \cdot (- 1)^{50} = 1$

$∴ 3^{100} = 10 k + 1$

Last digit $= 1$

Testing Divisibility

Example: Show that $6^{n} - 1$ is divisible by $5$ for all positive integers $n$ .

$6^{n} = (5 + 1)^{n} = \sum_{r = 0}^{n} (r n) 5^{r} = 1 + 5 n + (2 n) 5^{2} + \dots$

$6^{n} - 1 = 5 n + (2 n) 5^{2} + \dots = 5 (n + (2 n) 5 + \dots)$

Since every term has a factor of $5$ , $6^{n} - 1$ is divisible by $5$ . $■$

Comparing Two Large Numbers

Example: Which is greater: $e^{π}$ or $π^{e}$ ? (Conceptual approach using binomial approximation)

For FBISE purposes, comparison problems typically involve showing $a^{b} > b^{a}$ by writing one in terms of the other plus a small increment and expanding.

General approach: Write $a = b (1 + x)$ where $x > 0$ , then:

$a^{b} = b^{b} (1 + x)^{b} \approx b^{b} (1 + b x + \frac{b ( b - 1 )}{2} x^{2} + \dots)$

Compare with $b^{a} = b^{b (1 + x)} = b^{b} \cdot b^{b x}$ and use the expansion to determine which is larger.

Summary of Techniques

Goal	Strategy
Remainder of $a^{n} \div d$	Write $a = (d \cdot q \pm 1)$ , expand, isolate constant term
Last digit of $a^{n}$	Find remainder when $a^{n} \div 10$
Divisibility by $d$	Show $a^{n} - c = d \cdot (integer)$
Compare large numbers	Express one as $(1 + x)$ times the other, expand

Exercise 7.4 — Question 9

This question applies the Binomial Theorem to solve problems involving:

Finding the remainder when a large power is divided by a number
Finding the last digit of a large power
Testing divisibility of a number
Comparing two large numbers

Key Technique: Rewriting the Base

The core strategy is to rewrite the base as $(1 + k m)$ or $(k m - 1)$ where $k m$ is a multiple of the divisor, then expand using the Binomial Theorem:

$(1 + k m)^{n} = \sum_{r = 0}^{n} (r n) (k m)^{r} = 1 + n (k m) + (2 n) (k m)^{2} + \dots$

All terms except the first ( $r = 0$ ) are divisible by $k m$ , hence by $m$ .

Finding the Remainder

Method: Write the base as $(1 + d \cdot q)$ where $d$ is the divisor.

Example: Find the remainder when $7^{100}$ is divided by $6$ .

$7^{100} = (6 + 1)^{100} = \sum_{r = 0}^{100} (r 100) 6^{r} \cdot 1^{100 - r}$

$= 1 + (1 100) \cdot 6 + (2 100) \cdot 6^{2} + \dots + 6^{100}$

Every term with $r \geq 1$ contains a factor of $6$ , so they are all divisible by $6$ .

$∴ 7^{100} = 6 k + 1 for some integer k$

Remainder $= 1$

Finding the Last Digit

The last digit of $N$ equals the remainder when $N$ is divided by $10$ .

Example: Find the last digit of $3^{100}$ .

$3^{100} = (3^{2})^{50} = 9^{50} = (10 - 1)^{50}$

$= \sum_{r = 0}^{50} (r 50) 1 0^{r} \cdot (- 1)^{50 - r}$

All terms with $r \geq 1$ are divisible by $10$ . The $r = 0$ term is:

$(0 50) \cdot 1 0^{0} \cdot (- 1)^{50} = 1$

$∴ 3^{100} = 10 k + 1$

Last digit $= 1$

Testing Divisibility

Example: Show that $6^{n} - 1$ is divisible by $5$ for all positive integers $n$ .

$6^{n} = (5 + 1)^{n} = \sum_{r = 0}^{n} (r n) 5^{r} = 1 + 5 n + (2 n) 5^{2} + \dots$

$6^{n} - 1 = 5 n + (2 n) 5^{2} + \dots = 5 (n + (2 n) 5 + \dots)$

Since every term has a factor of $5$ , $6^{n} - 1$ is divisible by $5$ . $■$

Comparing Two Large Numbers

Example: Which is greater: $e^{π}$ or $π^{e}$ ? (Conceptual approach using binomial approximation)

For FBISE purposes, comparison problems typically involve showing $a^{b} > b^{a}$ by writing one in terms of the other plus a small increment and expanding.

General approach: Write $a = b (1 + x)$ where $x > 0$ , then:

$a^{b} = b^{b} (1 + x)^{b} \approx b^{b} (1 + b x + \frac{b ( b - 1 )}{2} x^{2} + \dots)$

Compare with $b^{a} = b^{b (1 + x)} = b^{b} \cdot b^{b x}$ and use the expansion to determine which is larger.

Summary of Techniques

Goal	Strategy
Remainder of $a^{n} \div d$	Write $a = (d \cdot q \pm 1)$ , expand, isolate constant term
Last digit of $a^{n}$	Find remainder when $a^{n} \div 10$
Divisibility by $d$	Show $a^{n} - c = d \cdot (integer)$
Compare large numbers	Express one as $(1 + x)$ times the other, expand