Question

is the number 2³×3²¹ perfectly divisible by 6? is it perfectly divisible by 8 ?​

Answer 1

Answer:

yes this number is not perfectly divisible by 6      

Step-by-step explanation:

but it is divisible by 8

2^3 [3^7]^3/ 2*3= 2^2 * 3^20 divisibility by six

2^3 [3^7]^3/ 2^3= 3^21 divisibility by 8

Answer 2

Step-by-step explanation:

Given :-

2^3 × 3^21

To find :-

Is 2^3 × 3^21 perfectly divisible by 6? is it perfectly divisible by 8 ?

Solution :-

Given number = 2^3 × 3^21

It is divisible by 6

Reason :-

It can be written as

(2×2^2 )×(3×3^20)

Since a^m × a^n = a^(m+n)

It can be arranged as

=> (2×3)×2^2×3^20

=> 6×2^2×3^20

=> It is the multiple of 6

=> It is divisible by 6.

and

Given number = 2^3 × 3^21

It is divisible by 8

Reason :-

2^3×3^21

=> (2×2×2)×3^21

=> 8×3^21

=> It is the multiple of 8

=> It is divisible by 8.

Answer:-

Given number is perfectly divisible by both 6 and 8.

Used formulae :-

• a^m × a^n = a^(m+n)

Points to know :-

Divisibility by 6:-

A number is divisible by both 2 and 3 is divisible by 6.

Divisibility by 2:-

A number has one of the digits 0 or 2,4,6,8 on its ones place is divisible by 2

Divisibility by 3:-

The sum of all the digits in the given number is divisible by 3 then the number is divisible by 3

Divisibility by 8:-

A number with more than three digits is divisible by 8,if the last three digits from units place are zeroes or the number formed by the last three digits from units place is divisible by 8.