7.4 Applications of Binomial Theorem: Cyclicity, Remainders, and Divisibility | Mathematical Induction And Binomial Theorem | Mathematics 11

This section covers powerful applications of the Binomial Theorem to solve problems involving:

The last digit (unit digit) of large powers
Remainders when large powers are divided by a number
Divisibility proofs
Comparing large numbers without full computation

1. Cyclicity of Unit Digits

The unit digit of $a^{n}$ follows a repeating cycle as $n$ increases. This cycle is called the cyclicity of $a$ .

Cycles for Common Bases

Base	Cycle of Unit Digits	Cycle Length
2	2, 4, 8, 6	4
3	3, 9, 7, 1	4
4	4, 6	2
7	7, 9, 3, 1	4
8	8, 4, 2, 6	4
9	9, 1	2
0, 1, 5, 6	constant	1

Method

Identify the cycle length $c$ for the base.
Compute $r = n mod c$ .
If $r = 0$ , the unit digit is the last in the cycle; otherwise it is the $r$ -th element.

Example

Find the unit digit of $7^{102}$ .

Cycle of 7: $7, 9, 3, 1$ (length 4). $102 = 4 \times 25 + 2 ⟹ r = 2$ Unit digit of $7^{102}$ = unit digit of $7^{2} = 9$ .

2. Finding Remainders Using the Binomial Theorem

For $(a + b)^{n}$ divided by $a$ : $(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$ Every term except the last term $b^{n}$ contains at least one factor of $a$ . Therefore: $(a + b)^{n} \equiv b^{n} (mod a)$

Strategy: Rewrite the Base

To find the remainder of $x^{n} \div d$ , rewrite $x$ as $(k d \pm 1)$ or $(k d \pm r)$ so the expansion simplifies.

Example 1

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

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

All terms except the last contain a factor of 6, so: $7^{100} \equiv 1^{100} = 1 (mod 6)$

Remainder $= 1$ .

Example 2

Find the remainder when $5^{103}$ is divided by 13.

Note $5^{2} = 25 = 2 (13) - 1$ , so: $5^{103} = 5 \cdot (5^{2})^{51} = 5 \cdot (2 \cdot 13 - 1)^{51}$

By the Binomial Theorem: $(2 \cdot 13 - 1)^{51} \equiv (- 1)^{51} = - 1 (mod 13)$

Therefore: $5^{103} \equiv 5 \times (- 1) = - 5 \equiv 8 (mod 13)$

Remainder $= 8$ .

3. Finding Last Two Digits

To find the last two digits of $x^{n}$ , we need $x^{n} (mod 100)$ .

Strategy

Write $x = 10 k \pm r$ and expand. Terms with $(10 k)^{2}$ or higher are multiples of 100 and do not affect the last two digits. Only the last two terms of the expansion matter.

Example

Find the last two digits of $3^{100}$ .

$3^{100} = (3^{4})^{25} = 8 1^{25} = (80 + 1)^{25}$

Expanding: $(80 + 1)^{25} = 1 + 25 (80) + (2 25) (80)^{2} + \dots$

Since $(80)^{2} = 6400$ is a multiple of 100, only the first two terms matter: $\equiv 1 + 2000 \equiv 2001 \equiv 01 (mod 100)$

Last two digits: 01.

4. Divisibility Proofs

Proving $a^{n} - b^{n}$ is Divisible by $(a - b)$

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

Every term except $b^{n}$ contains $(a - b)$ as a factor. Therefore: $a^{n} - b^{n} = (a - b) \cdot Q$ for some integer $Q$ , proving $(a - b) ∣ (a^{n} - b^{n})$ .

Example: Prove $9^{n + 1} - 8 n - 9$ is divisible by 64

Write $9^{n + 1} = 9 \cdot 9^{n} = 9 (1 + 8)^{n}$ .

Expand $(1 + 8)^{n}$ : $(1 + 8)^{n} = 1 + 8 n + (2 n) 8^{2} + (3 n) 8^{3} + \dots$

Multiply by 9: $9^{n + 1} = 9 + 72 n + 9 (2 n) (64) + \dots$

Now subtract $8 n + 9$ : $9^{n + 1} - 8 n - 9 = 72 n - 8 n + 9 (2 n) (64) + \dots = 64 n + 64 \cdot 9 (2 n) + \dots$

Every term contains 64 as a factor, so $64 ∣ (9^{n + 1} - 8 n - 9)$ . $■$

5. Comparing Large Numbers Using the Binomial Theorem

The Binomial Theorem allows us to compare or bound large expressions without computing them exactly.

Key Idea

For $x > 0$ : $(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2} x^{2} + \dots > 1 + n x$

If $1 + n x > K$ , then $(1 + x)^{n} > K$ .

Example

Show that $(1.1)^{10000} > 1000$ .

$(1.1)^{10000} = (1 + 0.1)^{10000}$

The second term alone is: $(1 10000) (0.1)^{1} = 10000 \times 0.1 = 1000$

Since all terms are positive: $(1.1)^{10000} = 1 + 1000 + (positive terms) > 1001 > 1000 ✓$

Summary Table

Application	Key Technique
Last digit of $a^{n}$	Cyclicity: find $n mod c$
Remainder of $x^{n} \div d$	Write $x = k d \pm r$ , expand, last term gives remainder
Last two digits	Write $x = 10 k \pm r$ , terms with $(10 k)^{2}$ vanish mod 100
Divisibility	Substitute $a = (a - b) + b$ or $x = (k d + 1)$ , factor out
Comparing large numbers	Use $1 + n x$ as a lower bound for $(1 + x)^{n}$

This section covers powerful applications of the Binomial Theorem to solve problems involving:

The last digit (unit digit) of large powers
Remainders when large powers are divided by a number
Divisibility proofs
Comparing large numbers without full computation

1. Cyclicity of Unit Digits

The unit digit of $a^{n}$ follows a repeating cycle as $n$ increases. This cycle is called the cyclicity of $a$ .

Cycles for Common Bases

Base	Cycle of Unit Digits	Cycle Length
2	2, 4, 8, 6	4
3	3, 9, 7, 1	4
4	4, 6	2
7	7, 9, 3, 1	4
8	8, 4, 2, 6	4
9	9, 1	2
0, 1, 5, 6	constant	1

Method

Identify the cycle length $c$ for the base.
Compute $r = n mod c$ .
If $r = 0$ , the unit digit is the last in the cycle; otherwise it is the $r$ -th element.

Example

Find the unit digit of $7^{102}$ .

Cycle of 7: $7, 9, 3, 1$ (length 4). $102 = 4 \times 25 + 2 ⟹ r = 2$ Unit digit of $7^{102}$ = unit digit of $7^{2} = 9$ .

2. Finding Remainders Using the Binomial Theorem

Key Principle

Strategy: Rewrite the Base

To find the remainder of $x^{n} \div d$ , rewrite $x$ as $(k d \pm 1)$ or $(k d \pm r)$ so the expansion simplifies.

Example 1

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

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

All terms except the last contain a factor of 6, so: $7^{100} \equiv 1^{100} = 1 (mod 6)$

Remainder $= 1$ .

Example 2

Find the remainder when $5^{103}$ is divided by 13.

Note $5^{2} = 25 = 2 (13) - 1$ , so: $5^{103} = 5 \cdot (5^{2})^{51} = 5 \cdot (2 \cdot 13 - 1)^{51}$

By the Binomial Theorem: $(2 \cdot 13 - 1)^{51} \equiv (- 1)^{51} = - 1 (mod 13)$

Therefore: $5^{103} \equiv 5 \times (- 1) = - 5 \equiv 8 (mod 13)$

Remainder $= 8$ .

3. Finding Last Two Digits

To find the last two digits of $x^{n}$ , we need $x^{n} (mod 100)$ .

Strategy

Write $x = 10 k \pm r$ and expand. Terms with $(10 k)^{2}$ or higher are multiples of 100 and do not affect the last two digits. Only the last two terms of the expansion matter.

Example

Find the last two digits of $3^{100}$ .

$3^{100} = (3^{4})^{25} = 8 1^{25} = (80 + 1)^{25}$

Expanding: $(80 + 1)^{25} = 1 + 25 (80) + (2 25) (80)^{2} + \dots$

Since $(80)^{2} = 6400$ is a multiple of 100, only the first two terms matter: $\equiv 1 + 2000 \equiv 2001 \equiv 01 (mod 100)$

Last two digits: 01.

4. Divisibility Proofs

Proving $a^{n} - b^{n}$ is Divisible by $(a - b)$

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

Every term except $b^{n}$ contains $(a - b)$ as a factor. Therefore: $a^{n} - b^{n} = (a - b) \cdot Q$ for some integer $Q$ , proving $(a - b) ∣ (a^{n} - b^{n})$ .

Example: Prove $9^{n + 1} - 8 n - 9$ is divisible by 64

Write $9^{n + 1} = 9 \cdot 9^{n} = 9 (1 + 8)^{n}$ .

Expand $(1 + 8)^{n}$ : $(1 + 8)^{n} = 1 + 8 n + (2 n) 8^{2} + (3 n) 8^{3} + \dots$

Multiply by 9: $9^{n + 1} = 9 + 72 n + 9 (2 n) (64) + \dots$

Now subtract $8 n + 9$ : $9^{n + 1} - 8 n - 9 = 72 n - 8 n + 9 (2 n) (64) + \dots = 64 n + 64 \cdot 9 (2 n) + \dots$

Every term contains 64 as a factor, so $64 ∣ (9^{n + 1} - 8 n - 9)$ . $■$

The second term alone is: $(1 10000) (0.1)^{1} = 10000 \times 0.1 = 1000$

Since all terms are positive: $(1.1)^{10000} = 1 + 1000 + (positive terms) > 1001 > 1000 ✓$

Summary Table

Application	Key Technique
Last digit of $a^{n}$	Cyclicity: find $n mod c$
Remainder of $x^{n} \div d$	Write $x = k d \pm r$ , expand, last term gives remainder
Last two digits	Write $x = 10 k \pm r$ , terms with $(10 k)^{2}$ vanish mod 100
Divisibility	Substitute $a = (a - b) + b$ or $x = (k d + 1)$ , factor out
Comparing large numbers	Use $1 + n x$ as a lower bound for $(1 + x)^{n}$