7.4 Q-6 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.4 — Question 6

This question applies the Binomial Theorem to practical problems involving:

Expanding and simplifying binomial expressions
Finding remainders when large powers are divided by a number
Finding the last digit (units digit) of large powers
Testing divisibility

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

The Binomial Theorem states:

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

For problems involving remainders and last digits, rewrite the base so that one part is a multiple of the divisor:

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

Finding the Remainder

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

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

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

All terms with $r \geq 1$ contain a factor of $6$ , so:

$7^{100} = (0 100) \cdot 1 + 6 k = 1 + 6 k$

Remainder = 1

Finding the Last Digit

Method: The last digit of $a^{n}$ depends only on the last digit of $a$ . Use the Binomial Theorem by writing $a = (multiple of 10) + d$ where $d$ is the units digit.

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

$3^{100} = (10 - 7)^{100} — not ideal; use pattern instead$

Better: $3^{100} = (3^{4})^{25} = 8 1^{25}$

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

All terms with $r \geq 1$ are multiples of $80$ (hence multiples of $10$ ), so:

$8 1^{25} \equiv 1 (mod 10)$

Last digit = 1

Testing Divisibility

Method: Show that $a^{n} - c$ is divisible by $d$ by expanding $a^{n}$ using the Binomial Theorem and showing all terms are multiples of $d$ except a constant that equals $c$ .

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

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

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

$5^{2 n} - 1 = 24 n + (2 n) 576 + \dots = 8 (3 n + \dots)$

Every term is divisible by $8$ , so $5^{2 n} - 1$ is divisible by $8$ . $■$

Exercise 7.4 — Question 6

This question applies the Binomial Theorem to practical problems involving:

Expanding and simplifying binomial expressions
Finding remainders when large powers are divided by a number
Finding the last digit (units digit) of large powers
Testing divisibility

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

The Binomial Theorem states:

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

For problems involving remainders and last digits, rewrite the base so that one part is a multiple of the divisor:

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

Finding the Remainder

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

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

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

All terms with $r \geq 1$ contain a factor of $6$ , so:

$7^{100} = (0 100) \cdot 1 + 6 k = 1 + 6 k$

Remainder = 1

Finding the Last Digit

Method: The last digit of $a^{n}$ depends only on the last digit of $a$ . Use the Binomial Theorem by writing $a = (multiple of 10) + d$ where $d$ is the units digit.

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

$3^{100} = (10 - 7)^{100} — not ideal; use pattern instead$

Better: $3^{100} = (3^{4})^{25} = 8 1^{25}$

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

All terms with $r \geq 1$ are multiples of $80$ (hence multiples of $10$ ), so:

$8 1^{25} \equiv 1 (mod 10)$

Last digit = 1

Testing Divisibility

Method: Show that $a^{n} - c$ is divisible by $d$ by expanding $a^{n}$ using the Binomial Theorem and showing all terms are multiples of $d$ except a constant that equals $c$ .

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

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

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

$5^{2 n} - 1 = 24 n + (2 n) 576 + \dots = 8 (3 n + \dots)$

Every term is divisible by $8$ , so $5^{2 n} - 1$ is divisible by $8$ . $■$