Latest TCS Aptitude Question SOLUTION: 29.the number 2^32 -1 has several divisors greater than 1 and less than 100.the sum of these divisors is clue:2^2n+1 is a prime for n=1,2,3,4

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29.the number 2^32 -1 has several divisors greater than 1 and less than 100.the sum of these divisors is clue:2^2n+1 is a prime for n=1,2,3,4 a)172 b)125 c)91 d)176

Asked In TCS (8 years ago)

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avijit sharma  (8 years ago)

To solve this qn we have use this fact-- (2^2n)+1 = prime for n=1,2,3,4 & the formula a2-b2 = (a+b)(a-b)
Now
(2^32 )-1 = (2^16 -1) (2^16+1) = (2^8 -1) (2^8+1)(2^16+1) = (2^4 -1) (2^4+1) (2^8+1)(2^16+1)
Repeting the formula again & again we get:
(2^32 )-1 = (2 -1) (2 +1)(2^2 +1) (2^4+1) (2^8+1)(2^16+1)
= 1 x 3 x 5 x 17 x (factor larger than 100) x (factor larger than 100)
So factors smaller than 100 are 3,5,17,15,51,85
So sum = 3+5+17+15+51+85 = 176

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aditya vasudeva  (8 years ago)

Consider the case of 2^5 = 32 .. 2x2x2x2x2x2 = 32 .... divisor of 32 will be 2, 4, 8 , 16 , 32 .. similarly for 2^6 = 64 divisor will be 2,4,8,16,32,64 ... now for 2^31 divisors between 1 and 100 will be 2,4,8,16,32,64 der sum will be 126 which is most nearby to answer 125.... so answer is 125

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avijit sharma  (8 years ago)

Aditya ...It is not asking 2^31...It is asking (2^32)-1

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Parth Shah  (8 years ago)

Hey AVIJIT can you please elobrate how you came up with 15,51 and 85?

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aditya vasudeva  (8 years ago)

Thanx avijit

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shilpi  (8 years ago)

bit confusing .... not getting the sol

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