Question 91
If $$6^{15} + 6^{16} + 6^{17} + 6^{18}$$ is divided by 24, find the remainder of finite series.
Solution
$$6^{15} + 6^{16} + 6^{17} + 6^{18}$$
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.
