7.3 Q-8 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.3 — Question 8

This question applies the Binomial Theorem to 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 large power

To apply the Binomial Theorem to remainder/divisibility problems, rewrite the base as: $a^{n} = (1 + k m)^{n} or a^{n} = (k m - 1)^{n}$ where $k m$ is a multiple of the divisor. Then expand using: $(1 + x)^{n} = 1 + (1 n) x + (2 n) x^{2} + \dots + x^{n}$

All terms except the first (or last) will be divisible by the divisor, isolating the remainder.

Worked Example 1 — Remainder

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

Write $7 = 6 + 1$ , so: $7^{100} = (6 + 1)^{100} = (0 100) 6^{100} + (1 100) 6^{99} + \dots + (99 100) 6 + 1$

Every term except the last ( $1$ ) contains a factor of $6$ , so: $7^{100} = 6 k + 1 for some integer k$

$Remainder = 1$

Worked Example 2 — Last Digit

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

The last digit of $3^{n}$ cycles with period 4: $3, 9, 7, 1, 3, 9, 7, 1, \dots$

Using Binomial Theorem: write $3^{100} = (10 - 7)^{100}$ ... or more directly: $3^{100} = (3^{4})^{25} = 8 1^{25}$

Write $81 = 80 + 1$ : $8 1^{25} = (1 + 80)^{25} = 1 + (1 25) (80) + (2 25) (80)^{2} + \dots$

Every term except $1$ is divisible by $10$ , so the last digit of $3^{100}$ is $1$ .

Worked Example 3 — Divisibility

Show that $4^{50} - 1$ is divisible by $3$ .

Write $4 = 1 + 3$ : $4^{50} = (1 + 3)^{50} = 1 + (1 50) (3) + (2 50) (3)^{2} + \dots$

$4^{50} - 1 = (1 50) (3) + (2 50) (9) + \dots = 3 [(1 50) + (2 50) (3) + \dots]$

Since every term has a factor of $3$ , $4^{50} - 1$ is divisible by $3$ . $■$

Summary of Technique

Goal	Rewrite base as	Result
Remainder when dividing by $d$	$a = 1 + d \cdot m$	All terms except $1$ divisible by $d$
Last digit	$a^{n} = (\dots + 1)^{k}$ with $10 ∣$ other terms	Last digit $= 1$
Divisibility by $d$	$a = 1 + d \cdot m$	$a^{n} - 1$ divisible by $d$

Exercise 7.3 — Question 8

This question applies the Binomial Theorem to 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 large power

Key Strategy

All terms except the first (or last) will be divisible by the divisor, isolating the remainder.

Worked Example 1 — Remainder

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

Write $7 = 6 + 1$ , so: $7^{100} = (6 + 1)^{100} = (0 100) 6^{100} + (1 100) 6^{99} + \dots + (99 100) 6 + 1$

Every term except the last ( $1$ ) contains a factor of $6$ , so: $7^{100} = 6 k + 1 for some integer k$

$Remainder = 1$

Worked Example 2 — Last Digit

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

The last digit of $3^{n}$ cycles with period 4: $3, 9, 7, 1, 3, 9, 7, 1, \dots$

Using Binomial Theorem: write $3^{100} = (10 - 7)^{100}$ ... or more directly: $3^{100} = (3^{4})^{25} = 8 1^{25}$

Write $81 = 80 + 1$ : $8 1^{25} = (1 + 80)^{25} = 1 + (1 25) (80) + (2 25) (80)^{2} + \dots$

Every term except $1$ is divisible by $10$ , so the last digit of $3^{100}$ is $1$ .

Worked Example 3 — Divisibility

Show that $4^{50} - 1$ is divisible by $3$ .

Write $4 = 1 + 3$ : $4^{50} = (1 + 3)^{50} = 1 + (1 50) (3) + (2 50) (3)^{2} + \dots$

$4^{50} - 1 = (1 50) (3) + (2 50) (9) + \dots = 3 [(1 50) + (2 50) (3) + \dots]$

Since every term has a factor of $3$ , $4^{50} - 1$ is divisible by $3$ . $■$

Summary of Technique

Goal	Rewrite base as	Result
Remainder when dividing by $d$	$a = 1 + d \cdot m$	All terms except $1$ divisible by $d$
Last digit	$a^{n} = (\dots + 1)^{k}$ with $10 ∣$ other terms	Last digit $= 1$
Divisibility by $d$	$a = 1 + d \cdot m$	$a^{n} - 1$ divisible by $d$