7.2 Q-12 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.2 — Question 12

This question applies the Binomial Theorem to solve problems involving remainders, last digits, and divisibility of large powers.

Key Technique: Rewriting a Base

To apply the Binomial Theorem to a large power like $a^{n}$ , rewrite the base as a sum or difference involving a convenient number:

$a^{n} = (b + r)^{n} or a^{n} = (b - r)^{n}$

Then expand using the Binomial Theorem:

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

The general term is:

$T_{r + 1} = (r n) a^{n - r} b^{r}$

Finding the Remainder

To find the remainder when a large power is divided by a number $d$ :

Write the base as $(d + k)$ or $(d - k)$ for a small integer $k$ .
Expand using the Binomial Theorem.
All terms except the last (constant) term will be divisible by $d$ .
The remainder is determined by the last term.

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

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

Every term except the last ( $r = 100$ ) contains a factor of $6$ , so:

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

$∴ Remainder = 1$

Finding the Last Digit

The last (units) digit of $a^{n}$ equals the remainder when $a^{n}$ is divided by $10$ .

Write the base as $(10 + k)$ or $(10 - k)$ .
Expand; all terms except the last are divisible by $10$ .
The last digit comes from the constant term.

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

$3^{100} = (3^{2})^{50} = 9^{50} = (10 - 1)^{50}$

$= \sum_{r = 0}^{50} (r 50) 1 0^{50 - r} (- 1)^{r}$

All terms with $r < 50$ contain a factor of $10$ , so:

$9^{50} = 10 m + (- 1)^{50} = 10 m + 1$

$∴ Last digit of 3^{100} = 1$

Testing Divisibility

To show $a^{n} - 1$ (or similar expressions) is divisible by some number $d$ :

Write $a = (1 + (a - 1))$ or a similar form.
Expand and show all terms are divisible by $d$ .

Middle Terms

For the expansion of $(a + b)^{n}$ :

If $n$ is even: one middle term, $T_{\frac{n}{2} + 1}$
If $n$ is odd: two middle terms, $T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$

Exercise 7.2 — Question 12

This question applies the Binomial Theorem to solve problems involving remainders, last digits, and divisibility of large powers.

Key Technique: Rewriting a Base

To apply the Binomial Theorem to a large power like $a^{n}$ , rewrite the base as a sum or difference involving a convenient number:

$a^{n} = (b + r)^{n} or a^{n} = (b - r)^{n}$

Then expand using the Binomial Theorem:

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

The general term is:

$T_{r + 1} = (r n) a^{n - r} b^{r}$

Finding the Remainder

To find the remainder when a large power is divided by a number $d$ :

Write the base as $(d + k)$ or $(d - k)$ for a small integer $k$ .
Expand using the Binomial Theorem.
All terms except the last (constant) term will be divisible by $d$ .
The remainder is determined by the last term.

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

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

Every term except the last ( $r = 100$ ) contains a factor of $6$ , so:

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

$∴ Remainder = 1$

Finding the Last Digit

The last (units) digit of $a^{n}$ equals the remainder when $a^{n}$ is divided by $10$ .

Write the base as $(10 + k)$ or $(10 - k)$ .
Expand; all terms except the last are divisible by $10$ .
The last digit comes from the constant term.

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

$3^{100} = (3^{2})^{50} = 9^{50} = (10 - 1)^{50}$

$= \sum_{r = 0}^{50} (r 50) 1 0^{50 - r} (- 1)^{r}$

All terms with $r < 50$ contain a factor of $10$ , so:

$9^{50} = 10 m + (- 1)^{50} = 10 m + 1$

$∴ Last digit of 3^{100} = 1$

Testing Divisibility

To show $a^{n} - 1$ (or similar expressions) is divisible by some number $d$ :

Write $a = (1 + (a - 1))$ or a similar form.
Expand and show all terms are divisible by $d$ .

Middle Terms

For the expansion of $(a + b)^{n}$ :

If $n$ is even: one middle term, $T_{\frac{n}{2} + 1}$
If $n$ is odd: two middle terms, $T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$