Question

Is the number 2² × 3²¹ perfectly divisible by 6? Is it perfectly divisible by 8?

Please Explain this question with Answer.​

Answer 1

Answer:

6, yes. 8, no.

Step-by-step explanation:

6 can be re-written as 2*3. So on dividing, we get,

 $\frac{2^{2}*3^{21} }{2*3}$

= $\frac{2^{2-1} * 3^{21-1} }{1}$

= $2^{1} * 3^{20}$

Hence, $2^{2} * 3^{21}$ is perfectly divisible by 6.

8 can be re-written as 2*2*2. So on dividing, we get,

 $\frac{2^{2} *3^{21} }{2*2*2}$

= $\frac{2^{2-2}*3^{21} }{2}$

= $\frac{1*3^{21} }{2}$           [∵$a^{0} = 1$]

Therefore,  $2^{2} * 3^{21}$ is not perfectly divisible by 8.

I hope you find the solution helpful :)