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

Exercise 7.4 — Question 11

Key Concepts Covered

This question applies the Binomial Theorem to:

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

Concept 1: Finding the Remainder Using Binomial Theorem

To find the remainder when $(a + b)^{n}$ is divided by $a$ , rewrite the base so one part is a multiple of the divisor.

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

Every term except the last ( $r = n$ ) contains at least one factor of $a$ , so: $(a + b)^{n} \equiv b^{n} (mod a)$

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

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

All terms with $r < 100$ are divisible by $6$ . The last term is $(100 100) \cdot 6^{0} \cdot 1^{100} = 1$ .

$∴ 7^{100} \equiv 1 (mod 6)$

Remainder = 1

Concept 2: Finding the Last Digit Using Cyclicity

The units digit of powers of any integer follows a repeating cycle (cyclicity).

Steps:

List the units digits of successive powers until the pattern repeats.
Note the cycle length $c$ .
Compute $n \div c$ and find the remainder $k$ .
The units digit of $a^{n}$ equals the units digit of $a^{k}$ (if $k = 0$ , use $a^{c}$ ).

Example: Find the last digit of $7^{102}$ .

Power	Units digit
$7^{1}$	7
$7^{2}$	9
$7^{3}$	3
$7^{4}$	1
$7^{5}$	7 (repeats)

Cycle length $= 4$ . Now $102 \div 4 = 25$ remainder $2$ .

So the units digit of $7^{102}$ = units digit of $7^{2} = 9$ .

Concept 3: Divisibility Testing

To show $N$ is divisible by $d$ , express $N = (d \cdot k \pm 1)^{n}$ and expand. All terms except $\pm 1$ will be multiples of $d$ , so $N \equiv \pm 1 (mod d)$ , or adjust accordingly.

Example: Show $1 0^{n} - 1$ is divisible by $9$ .

$1 0^{n} = (9 + 1)^{n} = \sum_{r = 0}^{n} (r n) 9^{n - r} \cdot 1^{r}$

All terms except the last are divisible by $9$ , and the last term is $1$ . So $1 0^{n} \equiv 1 (mod 9)$ , meaning $1 0^{n} - 1 \equiv 0 (mod 9)$ . ✓

Exercise 7.4 — Question 11

Key Concepts Covered

This question applies the Binomial Theorem to:

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

Concept 1: Finding the Remainder Using Binomial Theorem

To find the remainder when $(a + b)^{n}$ is divided by $a$ , rewrite the base so one part is a multiple of the divisor.

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

Every term except the last ( $r = n$ ) contains at least one factor of $a$ , so: $(a + b)^{n} \equiv b^{n} (mod a)$

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

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

All terms with $r < 100$ are divisible by $6$ . The last term is $(100 100) \cdot 6^{0} \cdot 1^{100} = 1$ .

$∴ 7^{100} \equiv 1 (mod 6)$

Remainder = 1

Concept 2: Finding the Last Digit Using Cyclicity

The units digit of powers of any integer follows a repeating cycle (cyclicity).

Steps:

List the units digits of successive powers until the pattern repeats.
Note the cycle length $c$ .
Compute $n \div c$ and find the remainder $k$ .
The units digit of $a^{n}$ equals the units digit of $a^{k}$ (if $k = 0$ , use $a^{c}$ ).

Example: Find the last digit of $7^{102}$ .

Power	Units digit
$7^{1}$	7
$7^{2}$	9
$7^{3}$	3
$7^{4}$	1
$7^{5}$	7 (repeats)

Cycle length $= 4$ . Now $102 \div 4 = 25$ remainder $2$ .

So the units digit of $7^{102}$ = units digit of $7^{2} = 9$ .

Concept 3: Divisibility Testing

To show $N$ is divisible by $d$ , express $N = (d \cdot k \pm 1)^{n}$ and expand. All terms except $\pm 1$ will be multiples of $d$ , so $N \equiv \pm 1 (mod d)$ , or adjust accordingly.

Example: Show $1 0^{n} - 1$ is divisible by $9$ .

$1 0^{n} = (9 + 1)^{n} = \sum_{r = 0}^{n} (r n) 9^{n - r} \cdot 1^{r}$

All terms except the last are divisible by $9$ , and the last term is $1$ . So $1 0^{n} \equiv 1 (mod 9)$ , meaning $1 0^{n} - 1 \equiv 0 (mod 9)$ . ✓