If $$n = {2^{p - 1}}\left( {{2^p} - 1} \right),$$ where $${{2^p} - 1}$$ is a prime, then the sum of the divisors of $$n$$ is equal to
A.
$$n$$
B.
$$2n$$
C.
$$pn$$
D.
$$p^n$$
Answer :
$$2n$$
Solution :
If $$N = {p_1}^{{\alpha _1}}{p_2}^{{\alpha _2}}$$ then the sum of the divisors
of $$N$$ is
$$\left( {\frac{{{p_1}^{{\alpha _1} + 1} - 1}}{{{p_1} - 1}}} \right)\left( {\frac{{{p_2}^{{\alpha _2} + 1} - 1}}{{{p_2} - 1}}} \right)$$
Releted MCQ Question on Algebra >> Permutation and Combination
Releted Question 1
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$ and $$^n{C_{r + 1}} = 126,$$ then $$r$$ is:
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are
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