Question

How many perfect cubes are there in the sequences 1^1, 2^2,3^3,4^4,5^5,.....100^100?

Answer 1

Given: The sequences 1^1, 2^2,3^3,4^4,5^5,.....100^100

To find: Perfect cubes in the sequences?

Solution:

• Now we have given the sequence where powers are:

            1, 2, 3, 4, ... , 100.

•  Now the multiple of 3 in powers are: 3, 6, 9, ..., 99

•  So let a (first term) be 3 and d (common difference) be 3

•  Now the formula is a + (n-1)d = last term

• So putting the value in it:

            3 + ( n - 1 ) 3 = 99

             3 x ( n - 1 ) = 96

            n - 1 = 32

             n = 33

Answer:

         So there are 33 perfect cube numbers.

Answer 2

Answer:

Step-by-step explanation:

let n = 3x, where x is a positive integer.

n^n can be written as (3x)^(3x) = (((3x)^x)^3)

So, n^n, will be a perfect cube for all multiples of 3.

Below 100, the following are multples of 3.

n = {3, 6, 9, ..., 99}, where n=3x for x = {1,2,3,...,33}

So, these are total of 33 numbers.

In addition, we have some exceptions.

1. 1^1 = 1^3, so it s perfect cube.
2. 8^8 = 8^6 * 8 * 8 = (8^2)^3 * 2^3 * 2^3 =((8^2) *2 *2)^3
3. 64^64=64^63 * 64 = ((64^21)^3) * 4^3 = ((64^21)*4)^3

So, 1^1, 8^8, and 64^64 are also perfect cubes.

So, in total we have 33 + 3 = 36 perfect cubes.