$4^{61} + 4^{62} + 4^{63} + 4^{64}$ is divisible by :
(A) 3        (B) 10      (C) 11      (D) 13

Given :

The given expression is $4^{61} +4^{62} + 4^{63} + 4^{64} $

To do :

We have to find the given expression $4^{61} +4^{62} + 4^{63} + 4^{64} $ is divisible by which of the given options.

Solution :

$4^1 = 4$

$4^2 = 16$

$4^3 = 64$

$4^4 = 256$

From the above we can infer that,

$4^{odd}$ ends with the digit 4.

$4^{even} $ ends with the digit 6.

So, $4^{61} = ......4$

$4^{62} = ......6$

$4^{63} = ........4$

$4^{64} = ........6$

$4^{61} +4^{62} + 4^{63} + 4^{64}  = .....4 + .......6 + ........4 + .......6 = ........0$

Therefore, sum of $4^{61} +4^{62} + 4^{63} + 4^{64} $ is ends with the digit 0.

Any number ends with the digit 0 is divisible by 10.

Therefore, $4^{61} +4^{62} + 4^{63} + 4^{64} $ is divisible by 10.

 Option B is correct.

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Updated on: 10-Oct-2022

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