7.4 Q-12 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.4 — Question 12

This question applies the Binomial Theorem to problems involving remainders, last digits, and divisibility of large powers.

Key Technique: Rewriting the Base

To apply the Binomial Theorem to a large power $a^{n}$ , rewrite the base $a$ as $(1 + k m)$ or $(k m \pm 1)$ where $m$ is the divisor (or a convenient number), then expand:

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

All terms except the constant term $1$ contain $m$ as a factor, making divisibility and remainder analysis straightforward.

Finding the Remainder

Method: Write $a = (d \cdot q + r_{0})$ so that $a^{n} = (1 + d \cdot q)^{n}$ or similar, expand, and identify the remainder.

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

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

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

Every term except $1$ is divisible by $6$ , so:

$7^{100} = 6 k + 1 \Rightarrow Remainder = 1$

Finding the Last Digit

The last digit of $a^{n}$ equals the remainder when $a^{n}$ is divided by $10$ .

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

$3^{100} = (10 - 7)^{100}$

Alternatively, note $3^{4} = 81$ , so $3^{100} = (3^{4})^{25} = 8 1^{25}$ .

$8 1^{25} = (80 + 1)^{25} = \sum_{r = 0}^{25} (r 25) 8 0^{r}$

Every term except $1^{25} = 1$ is divisible by $10$ , so the last digit of $3^{100}$ is $1$ .

Testing Divisibility

Example: Show that $5^{2 n} - 1$ is divisible by $24$ for all positive integers $n$ .

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

$5^{2 n} - 1 = (1 n) 24 + (2 n) 2 4^{2} + \dots = 24 [n + (2 n) 24 + \dots]$

Since every term contains factor $24$ , we conclude $24 ∣ (5^{2 n} - 1)$ . $■$

Summary of Steps

Goal	Strategy
Remainder of $a^{n} \div d$	Write $a = (d \cdot q \pm 1)$ , expand, all terms except $\pm 1$ divisible by $d$
Last digit of $a^{n}$	Find remainder of $a^{n} \div 10$ using same method
Divisibility by $d$	Show $a^{n} - c = d \cdot (integer)$ via expansion

Exercise 7.4 — Question 12

This question applies the Binomial Theorem to problems involving remainders, last digits, and divisibility of large powers.

Key Technique: Rewriting the Base

To apply the Binomial Theorem to a large power $a^{n}$ , rewrite the base $a$ as $(1 + k m)$ or $(k m \pm 1)$ where $m$ is the divisor (or a convenient number), then expand:

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

All terms except the constant term $1$ contain $m$ as a factor, making divisibility and remainder analysis straightforward.

Finding the Remainder

Method: Write $a = (d \cdot q + r_{0})$ so that $a^{n} = (1 + d \cdot q)^{n}$ or similar, expand, and identify the remainder.

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

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

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

Every term except $1$ is divisible by $6$ , so:

$7^{100} = 6 k + 1 \Rightarrow Remainder = 1$

Finding the Last Digit

The last digit of $a^{n}$ equals the remainder when $a^{n}$ is divided by $10$ .

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

$3^{100} = (10 - 7)^{100}$

Alternatively, note $3^{4} = 81$ , so $3^{100} = (3^{4})^{25} = 8 1^{25}$ .

$8 1^{25} = (80 + 1)^{25} = \sum_{r = 0}^{25} (r 25) 8 0^{r}$

Every term except $1^{25} = 1$ is divisible by $10$ , so the last digit of $3^{100}$ is $1$ .

Testing Divisibility

Example: Show that $5^{2 n} - 1$ is divisible by $24$ for all positive integers $n$ .

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

$5^{2 n} - 1 = (1 n) 24 + (2 n) 2 4^{2} + \dots = 24 [n + (2 n) 24 + \dots]$

Since every term contains factor $24$ , we conclude $24 ∣ (5^{2 n} - 1)$ . $■$

Summary of Steps

Goal	Strategy
Remainder of $a^{n} \div d$	Write $a = (d \cdot q \pm 1)$ , expand, all terms except $\pm 1$ divisible by $d$
Last digit of $a^{n}$	Find remainder of $a^{n} \div 10$ using same method
Divisibility by $d$	Show $a^{n} - c = d \cdot (integer)$ via expansion