Exercise 7.2 — Question 11

Overview

This question applies the Binomial Theorem to practical problems involving:

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

These are key applications of the Binomial Theorem covered under SLOs M-11-A-39 and M-11-A-40.

Key Technique: Rewriting as $(1 + x)^{n}$ or $(a + b)^{n}$

The core idea is to rewrite the base so that one part is a multiple of the divisor:

$a^{n} = (1 + (a - 1))^{n} or a^{n} = (k \cdot m + r)^{n}$

Then expand using the Binomial Theorem:

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

All terms except the last (constant) term will be divisible by the chosen modulus.

Finding the Remainder

Method: Write the base as $(multiple of divisor + 1)$ or $(multiple of divisor + r)$ , expand, and identify the remainder.

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

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

Every term except the last ( $r = 100$ ) contains a factor of $6$ :

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

$7^{100} = 6 (integer) + 1$

$∴ Remainder = 1$

Finding the Last Digit

Method: The last digit of $a^{n}$ equals the last digit of $(a^{n} mod 10)$ . Write the base as a multiple of $10$ plus its units digit, then expand.

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

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

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

All terms with $r < 50$ contain a factor of $10$ , so they contribute $0$ to the last digit:

$9^{50} = 10 (integer) + (- 1)^{50} = 10 (integer) + 1$

$∴ Last digit of 3^{100} = 1$

Testing Divisibility

Method: Show that $a^{n} - k$ is divisible by $m$ by writing $a^{n} = (m \cdot q + r)^{n}$ and expanding.

Example: Show that $5^{2 n} - 1$ is divisible by $8$ for all positive integers $n$ .

$5^{2 n} = 2 5^{n} = (24 + 1)^{n} = \sum_{r = 0}^{n} (r n) 2 4^{n - r} \cdot 1^{r}$

$= (0 n) 2 4^{n} + (1 n) 2 4^{n - 1} + \dots + (n - 1 n) 24 + 1$

Every term except the last contains $24 = 8 \times 3$ , so:

$5^{2 n} = 8 (integer) + 1$

$5^{2 n} - 1 = 8 (integer)$

$∴ 8 ∣ (5^{2 n} - 1) ✓$

General Term Reminder

$T_{r + 1} = (r n) a^{n - r} b^{r}, r = 0, 1, 2, \dots, n$

Exercise 7.2 — Question 11

Overview

This question applies the Binomial Theorem to practical problems involving:

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

These are key applications of the Binomial Theorem covered under SLOs M-11-A-39 and M-11-A-40.

Key Technique: Rewriting as $(1 + x)^{n}$ or $(a + b)^{n}$

The core idea is to rewrite the base so that one part is a multiple of the divisor:

$a^{n} = (1 + (a - 1))^{n} or a^{n} = (k \cdot m + r)^{n}$

Then expand using the Binomial Theorem:

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

All terms except the last (constant) term will be divisible by the chosen modulus.

Finding the Remainder

Method: Write the base as $(multiple of divisor + 1)$ or $(multiple of divisor + r)$ , expand, and identify the remainder.

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

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

Every term except the last ( $r = 100$ ) contains a factor of $6$ :

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

$7^{100} = 6 (integer) + 1$

$∴ Remainder = 1$

Finding the Last Digit

Method: The last digit of $a^{n}$ equals the last digit of $(a^{n} mod 10)$ . Write the base as a multiple of $10$ plus its units digit, then expand.

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

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

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

All terms with $r < 50$ contain a factor of $10$ , so they contribute $0$ to the last digit:

$9^{50} = 10 (integer) + (- 1)^{50} = 10 (integer) + 1$

$∴ Last digit of 3^{100} = 1$

Testing Divisibility

Method: Show that $a^{n} - k$ is divisible by $m$ by writing $a^{n} = (m \cdot q + r)^{n}$ and expanding.

Example: Show that $5^{2 n} - 1$ is divisible by $8$ for all positive integers $n$ .

$5^{2 n} = 2 5^{n} = (24 + 1)^{n} = \sum_{r = 0}^{n} (r n) 2 4^{n - r} \cdot 1^{r}$

$= (0 n) 2 4^{n} + (1 n) 2 4^{n - 1} + \dots + (n - 1 n) 24 + 1$

Every term except the last contains $24 = 8 \times 3$ , so:

$5^{2 n} = 8 (integer) + 1$

$5^{2 n} - 1 = 8 (integer)$

$∴ 8 ∣ (5^{2 n} - 1) ✓$

General Term Reminder

$T_{r + 1} = (r n) a^{n - r} b^{r}, r = 0, 1, 2, \dots, n$

7.2 Q-11

Exercise 7.2 — Question 11

Overview

Key Technique: Rewriting as $(1 + x)^{n}$ or $(a + b)^{n}$

Finding the Remainder

Finding the Last Digit

Testing Divisibility

General Term Reminder

7.2 Q-11

Exercise 7.2 — Question 11

Overview

Key Technique: Rewriting as $(1 + x)^{n}$ or $(a + b)^{n}$

Finding the Remainder

Finding the Last Digit

Testing Divisibility

General Term Reminder