7.4 Q-7 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.4 — Question 7

Problem Statement

Use the Binomial Theorem to:

Find the remainder when a large power is divided by a number.
Find the last digit (units digit) of a large power.
Test divisibility of a number.

Key Strategy

The core idea 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:

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

All terms except the constant term $1$ will contain $x$ as a factor, making them divisible by $x$ (and hence by the divisor $k$ ).

Part (i): Finding the Remainder

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

Solution:

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

Expanding: $7^{100} = (0 100) + (1 100) \cdot 6 + (2 100) \cdot 6^{2} + \dots + 6^{100}$

$7^{100} = 1 + 6 [(1 100) + (2 100) \cdot 6 + \dots + 6^{99}]$

Every term except $1$ is divisible by $6$ , so: $7^{100} = 6 k + 1 for some integer k$

$Remainder = 1$

Part (ii): Finding the Last Digit

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

Solution:

The last digit depends on the remainder when divided by $10$ .

Write $3^{100} = (3^{2})^{50} = 9^{50}$ .

Now write $9 = 10 - 1$ : $9^{50} = (10 - 1)^{50} = \sum_{r = 0}^{50} (r 50) 1 0^{r} (- 1)^{50 - r}$

$= (- 1)^{50} + (1 50) \cdot 10 \cdot (- 1)^{49} + \dots + 1 0^{50}$

$= 1 + 10 [(1 50) (- 1)^{49} + \dots]$

All terms except $1$ are divisible by $10$ , so: $3^{100} \equiv 1 (mod 10)$

$Last digit = 1$

Part (iii): Testing Divisibility

Example: Show that $1 0^{n} - 1$ is divisible by $9$ for all positive integers $n$ .

Solution:

Write $10 = 9 + 1$ : $1 0^{n} = (9 + 1)^{n} = \sum_{r = 0}^{n} (r n) 9^{r}$

$= 1 + (1 n) \cdot 9 + (2 n) \cdot 9^{2} + \dots + 9^{n}$

$= 1 + 9 [(1 n) + (2 n) \cdot 9 + \dots + 9^{n - 1}]$

Therefore: $1 0^{n} - 1 = 9 [(1 n) + (2 n) \cdot 9 + \dots + 9^{n - 1}]$

This is clearly divisible by $9$ . $■$

Summary of Technique

Goal	Strategy
Find remainder when $a^{n} \div d$	Write $a = d \cdot m + r$ , expand $(d m + r)^{n}$
Find last digit of $a^{n}$	Find remainder when $a^{n} \div 10$
Test divisibility by $d$	Write $a = d k \pm 1$ , show result $\equiv 0 (mod d)$

Exercise 7.4 — Question 7

Problem Statement

Use the Binomial Theorem to:

Find the remainder when a large power is divided by a number.
Find the last digit (units digit) of a large power.
Test divisibility of a number.

Key Strategy

The core idea 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:

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

All terms except the constant term $1$ will contain $x$ as a factor, making them divisible by $x$ (and hence by the divisor $k$ ).

Part (i): Finding the Remainder

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

Solution:

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

Expanding: $7^{100} = (0 100) + (1 100) \cdot 6 + (2 100) \cdot 6^{2} + \dots + 6^{100}$

$7^{100} = 1 + 6 [(1 100) + (2 100) \cdot 6 + \dots + 6^{99}]$

Every term except $1$ is divisible by $6$ , so: $7^{100} = 6 k + 1 for some integer k$

$Remainder = 1$

Part (ii): Finding the Last Digit

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

Solution:

The last digit depends on the remainder when divided by $10$ .

Write $3^{100} = (3^{2})^{50} = 9^{50}$ .

Now write $9 = 10 - 1$ : $9^{50} = (10 - 1)^{50} = \sum_{r = 0}^{50} (r 50) 1 0^{r} (- 1)^{50 - r}$

$= (- 1)^{50} + (1 50) \cdot 10 \cdot (- 1)^{49} + \dots + 1 0^{50}$

$= 1 + 10 [(1 50) (- 1)^{49} + \dots]$

All terms except $1$ are divisible by $10$ , so: $3^{100} \equiv 1 (mod 10)$

$Last digit = 1$

Part (iii): Testing Divisibility

Example: Show that $1 0^{n} - 1$ is divisible by $9$ for all positive integers $n$ .

Solution:

Write $10 = 9 + 1$ : $1 0^{n} = (9 + 1)^{n} = \sum_{r = 0}^{n} (r n) 9^{r}$

$= 1 + (1 n) \cdot 9 + (2 n) \cdot 9^{2} + \dots + 9^{n}$

$= 1 + 9 [(1 n) + (2 n) \cdot 9 + \dots + 9^{n - 1}]$

Therefore: $1 0^{n} - 1 = 9 [(1 n) + (2 n) \cdot 9 + \dots + 9^{n - 1}]$

This is clearly divisible by $9$ . $■$

Summary of Technique

Goal	Strategy
Find remainder when $a^{n} \div d$	Write $a = d \cdot m + r$ , expand $(d m + r)^{n}$
Find last digit of $a^{n}$	Find remainder when $a^{n} \div 10$
Test divisibility by $d$	Write $a = d k \pm 1$ , show result $\equiv 0 (mod d)$