Mathematics Q 2

Let 1 , a 1 , a 2 , , a k 1,a_1,a_2, \ldots , a_k be all the positive divisors of N = 2 74207280 ( 2 74207281 1 ) N = { 2 }^{ 74207280 }( { 2 }^{ 74207281 }-1 ) arranged in ascending order. And let T T denote the number of ways to express N N as the product of two coprime factors.

Compute K + ( i = 1 K 1 a i ) + T \displaystyle K + \left(\sum_{i=1}^K \dfrac1{a_i} \right) + T .


The answer is 148414564.

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1 solution

N = 2 n 1 ( 2 n 1 ) N = 2 n 1 ( p ) W h e r e n = 74207281 a n d p = 2 74207281 1 i s a P r i m e N u m b e r D i v i s i o r s O f N a r e { 1 , 2 , 2 2 , 2 3 , 2 4 , 2 5 , 2 6 , . . . . . . , 2 n 1 , p , 2 p , 4 p , 8 p , . . . . . , 2 n 1 p } N o w k = 2 n 1 w h i c h i s T o t a l N u m b e r o f D i v i s o r s i = 1 k 1 a i = 1 2 + 1 2 2 + 1 2 3 + . . . . . 1 2 n 1 + 1 p ( 1 2 + 1 2 2 + 1 2 3 + . . . . . 1 2 n 1 ) = 1 + 1 2 + 1 2 2 + 1 2 3 + . . . . . 1 2 n 1 + 1 p ( 1 2 + 1 2 2 + 1 2 3 + . . . . . 1 2 n 1 ) 1 T a k i n g L . C . M . = ( 2 0 + 2 1 + 2 2 + . . . . . 2 n 1 ) ( p 0 + p 1 ) p 2 n 1 1 = 1 T = 2 r 1 w h e r e r i s t h e N u m b e r O f P r i m e N u m b e r s H e r e r = 2 T = 2 N o o f w a y s t o e x p r e s s N a s a p r o d u c t o f T w o C o p r i m e F a c t o r s N o w K + i = 1 k 1 a i + T = 2 ( n ) 1 + 1 + 2 = 2 ( 74207281 ) + 2 = 148414564 N={ 2 }^{ n-1 }\left( { 2 }^{ n }-1 \right) \\ N={ 2 }^{ n-1 }(p)\quad \\ Where\\ \\ n=74207281\\ and\quad p={ 2 }^{ 74207281 }-1\quad is\quad a\quad Prime\quad Number\\ \\ Divisiors\quad Of\quad N\quad are\quad \\ \left\{ 1,2,{ 2 }^{ 2 }{ ,2 }^{ 3 },{ 2 }^{ 4 },{ 2 }^{ 5 },{ 2 }^{ 6 },......,{ 2 }^{ n-1 },p,2p,4p,8p,.....,{ 2 }^{ n-1 }p \right\} \\ \\ Now\quad k=2n-1\quad which\quad is\quad Total\quad Number\quad of\quad Divisors\\ \\ \sum _{ i=1 }^{ k }{ \frac { 1 }{ { a }_{ i } } } =\frac { 1 }{ 2 } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 3 } } +.....\frac { 1 }{ { 2 }^{ n-1 } } +\frac { 1 }{ p } \left( \frac { 1 }{ 2 } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 3 } } +.....\frac { 1 }{ { 2 }^{ n-1 } } \right) \\ =1+\frac { 1 }{ 2 } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 3 } } +.....\frac { 1 }{ { 2 }^{ n-1 } } +\frac { 1 }{ p } \left( \frac { 1 }{ 2 } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 3 } } +.....\frac { 1 }{ { 2 }^{ n-1 } } \right) -1\\ Taking\quad L.C.M.\\ =\frac { \left( { 2 }^{ 0 }{ +2 }^{ 1 }+{ 2 }^{ 2 }+.....{ 2 }^{ n-1 } \right) \left( { p }^{ 0 }+{ p }^{ 1 } \right) }{ p{ 2 }^{ n-1 } } -1\\ \\ =1\\ T={ 2 }^{ r-1 }\quad where\quad r\quad is\quad the\quad Number\quad Of\quad Prime\quad Numbers\\ Here\quad r=2\\ T=2\quad No\quad of\quad ways\quad to\quad express\quad N\quad as\quad a\quad product\quad of\quad Two\quad Co-prime\quad Factors\\ \\ Now\quad K+\sum _{ i=1 }^{ k }{ \frac { 1 }{ { a }_{ i } } } +T=2\left( n \right) -1+1+2\\ =2(74207281)+2=148414564\\

Moderator note:

How do you know that p = 2 74207281 1 p={ 2 }^{ 74207281 }-1 is a prime number?

It is The Largest Known Prime Number Till Date.

Rishabh Deep Singh - 5 years ago

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