Question

What is the remainder when 2^223 is divided by 9?

Answer 1

$2^{223} \div 9 = 2^{6(37) + 1} \div 9$

$2^{223} \div 9 = [ (2^6)^{37} \times 2^1} ] \div 9$

$2^{223} \div 9 = [ (64)^{37} \times 2^1} ] \div 9$

$2^{223} \div 9 = (64 \div 9)^{37} \times (2 \div 9)^1}$

Find the remainder:

$(64 \div 9)^{37} = \text {Reminder = } 1^{37} = 1$

$(2 \div 9) = \text {Reminder = } 2$

$2^{223} \div 9 = \text {Reminder = } (1)(2) = 2$

Answer: The remainder is 2

Answer 2

The answer is 49.55 and it is forming after 28 groups that let it divisible by 2*2 when divided by 223.

It will give the remainder as 49.55 and apply the rule cited above.

The remainder value must be analyzed well and thus it gives best solution when divided by 223.