Solve this :

$\mathit{Q}\mathbf{.}\mathbf{ } How many perfect cubes are there in the sequence 1^1, 2^2, 3^3, 4^4, ...., {100}^{100} ?$

(A)  33                                                (B)  37
(C)  36                                                (D)  34

Dear Student,

Please find below the solution to the asked query:

Powers are: 1, 2, 3, 4, ... , 100.
 
Multiple of 3 are: 3, 6, 9, ..., 99
 
Here, a = 3 and d = 3
 
Now, a + (n-1)d = 99

Or, 3 + (n-1)3 = 99

Or, 3(n-1) = 96

Or, n-1 = 32

Or, n = 33

Thus there are 33 perfect cube numbers.

Therefore,

Option ( A )                                           ( Ans )


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