7.4 Q-13 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.4 — Question 13

This question applies the Binomial Theorem to solve problems involving:

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

Key Strategy

The core idea is to rewrite the base as $(1 + m)$ or $(k m \pm 1)$ so that the Binomial expansion isolates the remainder or last digit.

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

Technique 1 — Finding the Remainder

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

Solution:

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

All terms with $k \geq 1$ contain a factor of $6$ , so: $7^{100} = 6 M + (0 100) \cdot 1 = 6 M + 1$

Remainder = 1

Technique 2 — Finding the Last Digit

The last digit of a number equals its remainder when divided by 10.

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

Solution:

Write $3^{100} = (3^{4})^{25} = 8 1^{25}$ .

Now write $81 = 80 + 1$ : $8 1^{25} = (80 + 1)^{25} = \sum_{k = 0}^{25} (k 25) 8 0^{k}$

All terms with $k \geq 1$ are divisible by $80$ , hence by $10$ . The last term is $1$ .

$8 1^{25} = 10 M + 1$

Last digit = 1

Technique 3 — Testing Divisibility

Example: Show that $6^{n} - 5 n - 1$ is divisible by $25$ for all positive integers $n$ .

Solution:

Write $6 = 5 + 1$ : $6^{n} = (5 + 1)^{n} = (0 n) + (1 n) 5 + (2 n) 5^{2} + \dots + 5^{n}$ $6^{n} = 1 + 5 n + (2 n) 5^{2} + \dots + 5^{n}$

Subtracting $5 n + 1$ : $6^{n} - 5 n - 1 = (2 n) 5^{2} + (3 n) 5^{3} + \dots + 5^{n}$

Every term contains $5^{2} = 25$ as a factor, so $6^{n} - 5 n - 1$ is divisible by 25. $■$

Technique 4 — Comparing Two Large Numbers

Example: Which is greater: $2^{30}$ or $3^{20}$ ?

Solution:

Rewrite with the same exponent: $2^{30} = (2^{3})^{10} = 8^{10}, 3^{20} = (3^{2})^{10} = 9^{10}$

Since $9 > 8$ and both exponents are equal and positive: $9^{10} > 8^{10} ⟹ 3^{20} > 2^{30}$

Summary Table

Goal	Strategy
Remainder of $a^{n} \div m$	Write $a = m \cdot q \pm 1$ , expand, isolate constant term
Last digit of $a^{n}$	Find remainder when $a^{n}$ is divided by $10$
Divisibility by $m^{2}$	Write $a = m + 1$ , expand, subtract lower terms
Compare $a^{p}$ vs $b^{q}$	Rewrite as equal exponents, compare bases

Exercise 7.4 — Question 13

This question applies the Binomial Theorem to solve problems involving:

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

Key Strategy

The core idea is to rewrite the base as $(1 + m)$ or $(k m \pm 1)$ so that the Binomial expansion isolates the remainder or last digit.

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

Technique 1 — Finding the Remainder

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

Solution:

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

All terms with $k \geq 1$ contain a factor of $6$ , so: $7^{100} = 6 M + (0 100) \cdot 1 = 6 M + 1$

Remainder = 1

Technique 2 — Finding the Last Digit

The last digit of a number equals its remainder when divided by 10.

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

Solution:

Write $3^{100} = (3^{4})^{25} = 8 1^{25}$ .

Now write $81 = 80 + 1$ : $8 1^{25} = (80 + 1)^{25} = \sum_{k = 0}^{25} (k 25) 8 0^{k}$

All terms with $k \geq 1$ are divisible by $80$ , hence by $10$ . The last term is $1$ .

$8 1^{25} = 10 M + 1$

Last digit = 1

Technique 3 — Testing Divisibility

Example: Show that $6^{n} - 5 n - 1$ is divisible by $25$ for all positive integers $n$ .

Solution:

Write $6 = 5 + 1$ : $6^{n} = (5 + 1)^{n} = (0 n) + (1 n) 5 + (2 n) 5^{2} + \dots + 5^{n}$ $6^{n} = 1 + 5 n + (2 n) 5^{2} + \dots + 5^{n}$

Subtracting $5 n + 1$ : $6^{n} - 5 n - 1 = (2 n) 5^{2} + (3 n) 5^{3} + \dots + 5^{n}$

Every term contains $5^{2} = 25$ as a factor, so $6^{n} - 5 n - 1$ is divisible by 25. $■$

Technique 4 — Comparing Two Large Numbers

Example: Which is greater: $2^{30}$ or $3^{20}$ ?

Solution:

Rewrite with the same exponent: $2^{30} = (2^{3})^{10} = 8^{10}, 3^{20} = (3^{2})^{10} = 9^{10}$

Since $9 > 8$ and both exponents are equal and positive: $9^{10} > 8^{10} ⟹ 3^{20} > 2^{30}$

Summary Table

Goal	Strategy
Remainder of $a^{n} \div m$	Write $a = m \cdot q \pm 1$ , expand, isolate constant term
Last digit of $a^{n}$	Find remainder when $a^{n}$ is divided by $10$
Divisibility by $m^{2}$	Write $a = m + 1$ , expand, subtract lower terms
Compare $a^{p}$ vs $b^{q}$	Rewrite as equal exponents, compare bases