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

Exercise 7.3 — Question 6

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 (units digit) of a large power
Testing divisibility of a large power
Comparing two large numbers

Key Technique

The core idea is to rewrite the base as a sum or difference involving the divisor (or a power of 10 for last-digit problems), then apply the Binomial Theorem:

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

All terms except the last (or first) will contain the divisor as a factor.

Type 1: Finding the Remainder

Method: Write the base as $(divisor + 1)^{n}$ or $(divisor - 1)^{n}$ .

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

$7^{100} = (6 + 1)^{100} = (0 100) 6^{100} + (1 100) 6^{99} + \dots + (99 100) 6^{1} + (100 100) 6^{0}$

Every term except the last contains a factor of $6$ :

$7^{100} = 6 k + 1 for some integer k$

$Remainder = 1$

Type 2: Finding the Last Digit

Method: Write the base as $(10 \pm r)^{n}$ so all terms except the constant term are multiples of 10.

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

Note that $3^{4} = 81$ , so: $3^{100} = (3^{4})^{25} = 8 1^{25} = (80 + 1)^{25}$

Expanding: $(80 + 1)^{25} = (0 25) 8 0^{25} + \dots + (24 25) 8 0^{1} + (25 25) 1^{25}$

All terms except the last are multiples of $80$ (hence multiples of $10$ ): $3^{100} = 10 m + 1 for some integer m$

$Last digit of 3^{100} = 1$

Type 3: Testing Divisibility

Method: Show that the expression equals a multiple of the divisor (plus or minus a remainder).

Example: Show that $6^{n} - 1$ is divisible by $5$ .

$6^{n} = (5 + 1)^{n} = (0 n) 5^{n} + (1 n) 5^{n - 1} + \dots + (n - 1 n) 5 + 1$

$6^{n} - 1 = 5 [(0 n) 5^{n - 1} + \dots + (n - 1 n)] = 5 k$

$6^{n} - 1 is divisible by 5$

Type 4: Comparing Two Large Numbers

Method: Express both numbers as binomial expansions with the same exponent, then compare dominant terms.

Example: Compare $2^{30}$ and $3^{20}$ .

$2^{30} = (2^{3})^{10} = 8^{10} = (1 + 7)^{10}$ $3^{20} = (3^{2})^{10} = 9^{10} = (1 + 8)^{10}$

Since $9 > 8$ , we have $(1 + 8)^{10} > (1 + 7)^{10}$ , therefore:

$3^{20} > 2^{30}$

Exercise 7.3 — Question 6

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 (units digit) of a large power
Testing divisibility of a large power
Comparing two large numbers

Key Technique

The core idea is to rewrite the base as a sum or difference involving the divisor (or a power of 10 for last-digit problems), then apply the Binomial Theorem:

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

All terms except the last (or first) will contain the divisor as a factor.

Type 1: Finding the Remainder

Method: Write the base as $(divisor + 1)^{n}$ or $(divisor - 1)^{n}$ .

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

$7^{100} = (6 + 1)^{100} = (0 100) 6^{100} + (1 100) 6^{99} + \dots + (99 100) 6^{1} + (100 100) 6^{0}$

Every term except the last contains a factor of $6$ :

$7^{100} = 6 k + 1 for some integer k$

$Remainder = 1$

Type 2: Finding the Last Digit

Method: Write the base as $(10 \pm r)^{n}$ so all terms except the constant term are multiples of 10.

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

Note that $3^{4} = 81$ , so: $3^{100} = (3^{4})^{25} = 8 1^{25} = (80 + 1)^{25}$

Expanding: $(80 + 1)^{25} = (0 25) 8 0^{25} + \dots + (24 25) 8 0^{1} + (25 25) 1^{25}$

All terms except the last are multiples of $80$ (hence multiples of $10$ ): $3^{100} = 10 m + 1 for some integer m$

$Last digit of 3^{100} = 1$

Type 3: Testing Divisibility

Method: Show that the expression equals a multiple of the divisor (plus or minus a remainder).

Example: Show that $6^{n} - 1$ is divisible by $5$ .

$6^{n} = (5 + 1)^{n} = (0 n) 5^{n} + (1 n) 5^{n - 1} + \dots + (n - 1 n) 5 + 1$

$6^{n} - 1 = 5 [(0 n) 5^{n - 1} + \dots + (n - 1 n)] = 5 k$

$6^{n} - 1 is divisible by 5$

Type 4: Comparing Two Large Numbers

Method: Express both numbers as binomial expansions with the same exponent, then compare dominant terms.

Example: Compare $2^{30}$ and $3^{20}$ .

$2^{30} = (2^{3})^{10} = 8^{10} = (1 + 7)^{10}$ $3^{20} = (3^{2})^{10} = 9^{10} = (1 + 8)^{10}$

Since $9 > 8$ , we have $(1 + 8)^{10} > (1 + 7)^{10}$ , therefore:

$3^{20} > 2^{30}$