Exercise 7.2 — Question 14

Problem Statement

Use the Binomial Theorem to show that for any positive integer $n$ :

$3^{n} = 1 + n \cdot 2 + (2 n) \cdot 4 + (3 n) \cdot 8 + \dots$

and hence find the remainder when $3^{100}$ is divided by $4$ , and determine the last digit of $3^{100}$ .

Key Concepts Needed

Binomial Theorem

For any positive integer $n$ :

$(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$

The general term is:

$T_{r + 1} = (r n) a^{n - r} b^{r}$

Solution

Step 1: Express $3^{100}$ in a useful form

Write $3 = 1 + 2$ , so:

$3^{100} = (1 + 2)^{100}$

Apply the Binomial Theorem with $a = 1$ , $b = 2$ , $n = 100$ :

$3^{100} = (0 100) (1)^{100} (2)^{0} + (1 100) (1)^{99} (2)^{1} + (2 100) (1)^{98} (2)^{2} + \dots$

$3^{100} = 1 + 100 \cdot 2 + (2 100) \cdot 4 + (3 100) \cdot 8 + \dots$

$3^{100} = 1 + 200 + (2 100) \cdot 4 + (3 100) \cdot 8 + \dots$

Step 2: Find the remainder when $3^{100}$ is divided by $4$

From the expansion:

$3^{100} = 1 + 200 + (2 100) \cdot 4 + (3 100) \cdot 8 + \dots$

Every term from the third term onwards contains a factor of $4$ , so they are all divisible by $4$ .

Group the expansion:

$3^{100} = 1 + 200 + 4 [(2 100) + (3 100) \cdot 2 + \dots]$

Now:

$1 + 200 = 201$
$201 = 4 \times 50 + 1$

So:

$3^{100} = 4 \times 50 + 1 + 4 [(2 100) + \dots] = 4 k + 1$

where $k$ is some positive integer.

$Remainder when 3^{100} is divided by 4 = 1$

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

The last digit of a number is its remainder when divided by $10$ .

Observe the pattern of last digits of powers of $3$ :

Power	Value	Last Digit
$3^{1}$	$3$	3
$3^{2}$	$9$	9
$3^{3}$	$27$	7
$3^{4}$	$81$	1
$3^{5}$	$243$	3

The last digits repeat with a cycle of 4: $3, 9, 7, 1, 3, 9, 7, 1, \dots$

Since $100 = 4 \times 25$ , we have $100 \equiv 0 (mod 4)$ , which corresponds to the 4th position in the cycle.

$Last digit of 3^{100} = 1$

Verification using Binomial Theorem:

Write $3^{100} = (4 - 1)^{100}$ :

$(4 - 1)^{100} = (0 100) 4^{100} - (1 100) 4^{99} + \dots + (99 100) 4 - 1$

All terms except the last contain a factor of $4$ (and hence $10$ does not divide them cleanly), but we need divisibility by $10$ . Instead, write $3^{4} = 81 = 80 + 1 = 8 \times 10 + 1$ :

$3^{100} = (3^{4})^{25} = (80 + 1)^{25}$

Expanding:

$= (0 25) 8 0^{25} + (1 25) 8 0^{24} + \dots + (24 25) 80 + 1$

Every term except the last contains a factor of $80$ , hence a factor of $10$ . Therefore:

$3^{100} \equiv 1 (mod 10)$

The last digit of $3^{100}$ is $1$ .

Summary of Results

Question	Answer
Remainder when $3^{100} \div 4$	$1$
Last digit of $3^{100}$	$1$

Exercise 7.2 — Question 14

Problem Statement

Use the Binomial Theorem to show that for any positive integer $n$ :

$3^{n} = 1 + n \cdot 2 + (2 n) \cdot 4 + (3 n) \cdot 8 + \dots$

and hence find the remainder when $3^{100}$ is divided by $4$ , and determine the last digit of $3^{100}$ .

Key Concepts Needed

Binomial Theorem

For any positive integer $n$ :

$(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$

The general term is:

$T_{r + 1} = (r n) a^{n - r} b^{r}$

Solution

Step 1: Express $3^{100}$ in a useful form

Write $3 = 1 + 2$ , so:

$3^{100} = (1 + 2)^{100}$

Apply the Binomial Theorem with $a = 1$ , $b = 2$ , $n = 100$ :

$3^{100} = (0 100) (1)^{100} (2)^{0} + (1 100) (1)^{99} (2)^{1} + (2 100) (1)^{98} (2)^{2} + \dots$

$3^{100} = 1 + 100 \cdot 2 + (2 100) \cdot 4 + (3 100) \cdot 8 + \dots$

$3^{100} = 1 + 200 + (2 100) \cdot 4 + (3 100) \cdot 8 + \dots$

Step 2: Find the remainder when $3^{100}$ is divided by $4$

From the expansion:

$3^{100} = 1 + 200 + (2 100) \cdot 4 + (3 100) \cdot 8 + \dots$

Every term from the third term onwards contains a factor of $4$ , so they are all divisible by $4$ .

Group the expansion:

$3^{100} = 1 + 200 + 4 [(2 100) + (3 100) \cdot 2 + \dots]$

Now:

$1 + 200 = 201$
$201 = 4 \times 50 + 1$

So:

$3^{100} = 4 \times 50 + 1 + 4 [(2 100) + \dots] = 4 k + 1$

where $k$ is some positive integer.

$Remainder when 3^{100} is divided by 4 = 1$

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

The last digit of a number is its remainder when divided by $10$ .

Observe the pattern of last digits of powers of $3$ :

Power	Value	Last Digit
$3^{1}$	$3$	3
$3^{2}$	$9$	9
$3^{3}$	$27$	7
$3^{4}$	$81$	1
$3^{5}$	$243$	3

The last digits repeat with a cycle of 4: $3, 9, 7, 1, 3, 9, 7, 1, \dots$

Since $100 = 4 \times 25$ , we have $100 \equiv 0 (mod 4)$ , which corresponds to the 4th position in the cycle.

$Last digit of 3^{100} = 1$

Verification using Binomial Theorem:

Write $3^{100} = (4 - 1)^{100}$ :

$(4 - 1)^{100} = (0 100) 4^{100} - (1 100) 4^{99} + \dots + (99 100) 4 - 1$

All terms except the last contain a factor of $4$ (and hence $10$ does not divide them cleanly), but we need divisibility by $10$ . Instead, write $3^{4} = 81 = 80 + 1 = 8 \times 10 + 1$ :

$3^{100} = (3^{4})^{25} = (80 + 1)^{25}$

Expanding:

$= (0 25) 8 0^{25} + (1 25) 8 0^{24} + \dots + (24 25) 80 + 1$

Every term except the last contains a factor of $80$ , hence a factor of $10$ . Therefore:

$3^{100} \equiv 1 (mod 10)$

The last digit of $3^{100}$ is $1$ .

Summary of Results

Question	Answer
Remainder when $3^{100} \div 4$	$1$
Last digit of $3^{100}$	$1$

7.2 Q-14

Exercise 7.2 — Question 14

Problem Statement

Key Concepts Needed

Binomial Theorem

Solution

Step 1: Express $3^{100}$ in a useful form

Step 2: Find the remainder when $3^{100}$ is divided by $4$

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

Summary of Results

7.2 Q-14

Exercise 7.2 — Question 14

Problem Statement

Key Concepts Needed

Binomial Theorem

Solution

Step 1: Express $3^{100}$ in a useful form

Step 2: Find the remainder when $3^{100}$ is divided by $4$

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

Summary of Results