7.2 Q-20 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.2 — Question 20

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

Strategy: 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:

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

$= 6^{100} + 100 \cdot 6^{99} + \dots + 100 \cdot 6 + 1$

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

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

$Remainder = 1$

Strategy: Write $3^{100} = (3^{4})^{25} = 8 1^{25}$ . Now write $81 = 80 + 1$ and expand.

$8 1^{25} = (80 + 1)^{25} = \sum_{r = 0}^{25} (r 25) 8 0^{25 - r} \cdot 1^{r}$

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

Every term except the last contains a factor of $80$ (and hence a factor of $10$ ), so:

$8 1^{25} = 10 \cdot (integer) + 1$

The units digit of $3^{100}$ is therefore $1$ .

Goal	Strategy
Find remainder when $a^{n} \div k$	Write $a = k \cdot m \pm 1$ , expand, all terms except the constant are divisible by $k$
Find last digit of $a^{n}$	Find remainder when $a^{n}$ is divided by $10$ using the same technique