Question 91

If $$6^{15} + 6^{16} + 6^{17} + 6^{18}$$ is divided by 24, find the remainder of finite series.

Solution

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$$6^{15} + 6^{16} + 6^{17} + 6^{18}$$

Let's take $$6^{15}$$ common from the whole.

We get $$6^{15}$$(1+$$6^{1}$$+$$6^{2}$$+$$6^{3}$$).

$$6^{15}$$ can be written as $$2^{15}$$*$$3^{15}$$.

24 can be written as $$2^{3}*3$$.

So, $$6^{15}$$ is divisible by 24. Hence, the remainder is zero.

All options are wrong in this P.Y.Q.

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