Let's analyze the problem step-by-step. Given:
- nnn is a whole number.
- When nnn is divided by 4, the remainder is 3.
This means we can write:
n=4k+3n=4k+3n=4k+3
for some integer kkk.
Step 1: Express 2n2n2n
Multiply both sides of the equation by 2:
2n=2(4k+3)=8k+62n=2(4k+3)=8k+62n=2(4k+3)=8k+6
Step 2: Find the remainder when 2n2n2n is divided by 4
Divide 2n2n2n by 4:
2n=8k+6=4(2k)+62n=8k+6=4(2k)+62n=8k+6=4(2k)+6
When dividing by 4, the term 4(2k)4(2k)4(2k) is divisible by 4 and leaves no remainder. So, the remainder depends on 6 modulo 4:
6÷4=1 remainder 26\div 4=1\text{ remainder }26÷4=1 remainder 2
Answer:
The remainder when 2n2n2n is divided by 4 is 2.