Question-72894

Question Number 72894 by TawaTawa last updated on 04/Nov/19

Answered by mind is power last updated on 04/Nov/19

lets assum n, m are relativ prime

let N = p_{1}^{k 1} . p_{2}^{k 2} \dots . p_{n}^{kn}

σ (N) = \underset{l_{1} = 0}{\sum^{k_{1}}} \underset{l 2 = 0}{\sum^{k 2}} \dots . \underset{\ln = 0}{\sum^{kn}} {\overset{l 1}{p}}_{1} {\overset{l 2}{p}}_{2} \dots {\overset{\ln}{p}}_{n} = \underset{i = 1}{\prod^{n}} (\underset{li = 0}{\sum^{ki}} p_{i}^{li})

σ (N) = \underset{i = 1}{\prod^{n}} \frac{p_{i}^{ki + 1} - 1}{p_{i} - 1}

N_{3}, σ (N_{3}) ⩾ 3 . N_{3}

\Rightarrow \underset{i = 1}{\prod^{n}} \frac{p_{i}^{ki + 1} - 1}{pi - 1} ⩾ 3 . p_{1}^{k 1} \dots . p_{n}^{kn}

\Leftrightarrow (1 + p_{i} \dots . + p_{i}^{k 1}) \dots \dots . . (1 + p_{n} + \dots . + p_{n}^{kn}) ⩾ {3 p}_{1}^{k 1} \dots p_{n}^{kn}

σ (3^{2} 5^{2} 2^{2} = 900) = \frac{26}{2} . \frac{124}{4} . \frac{7}{1} = 13.31.7 = 2821 ⩾ 2700

N_{3} = 900

σ (2^{3} 3^{4} 5^{3}) = 400140 ⩾ 4 . 2^{3} 3^{4} 5^{3}

N_{4} = 81000

Commented by TawaTawa last updated on 04/Nov/19

God bless you sir

Commented by mind is power last updated on 04/Nov/19

y^{'} re welcom