7.2 Q-16 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.2 — Question 16

Using the Binomial Theorem, find the remainder when $7^{100}$ is divided by $6$ , and find the last digit of $3^{100}$ .

Key Idea: Write $7 = 6 + 1$ , then expand $(6 + 1)^{100}$ using the Binomial Theorem.

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

Expanding the first few and last terms:

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

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

$7^{100} = 6 \cdot M + 1$

where $M$ is some positive integer.

$∴ Remainder = 1$

Key Idea: Write $3^{100} = (3^{2})^{50} = 9^{50}$ . Now write $9 = 10 - 1$ and expand.

$9^{50} = (10 - 1)^{50} = \sum_{r = 0}^{50} (r 50) 1 0^{50 - r} (- 1)^{r}$

Every term except the last contains a factor of $10$ , so:

$9^{50} = 10 \cdot K + (- 1)^{50} = 10 K + 1$

where $K$ is some positive integer.

The last digit is determined by the units digit of $10 K + 1$ , which is $1$ .

$∴ Last digit of 3^{100} = 1$

Goal	Strategy
Find remainder when $a^{n} \div m$	Write $a = m \cdot k \pm 1$ , expand, all terms except last divisible by $m$
Find last digit of $a^{n}$	Write $a^{n}$ in terms of a power of $10 \pm 1$ , last term gives units digit