Question

Let $1 , \alpha_1, \alpha_2........\alpha_k$ are divisors of number $N = 2^{n-1}(2^n-1)$ where $2^n-1$ is prime number and $1 \ \textless \ \alpha_1\ \textless \ \alpha_2 \ \textless \ .......\ \textless \ \alpha_k$.

1) Then find the value of k.
2) Find the value of $1 + \dfrac{1}{\alpha_1 } + \dfrac{1}{\alpha_2}.......\dfrac{1}{\alpha_k}$

Answer 1

Answer:

Hello ,

here, N = 2^(n-1)(2^n -1)

Given : (2^n -1) is a prime number.

If we put 2^n -1=7

then we get

n=3

so, N = 4 × 7 = 28

since , factors of N=28 are

{ 1, 2 , 4 , 7 ,14 , 28 }

i.e {1, alp1 , alp2 , alp 3,...alp k}

1) Thus , alpk=28

so, k = 5 ans...

2) 1 + 1/alp1 + 1/alp2 +..... + 1/ alp k

= 1 + 1/2 + 1/4 + 1/7 + 1/14 + 1/28

= 2 ans...

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Answer 2

Step-by-step explanation:

Hello ,

here, N = 2^(n-1)(2^n -1)

Given : (2^n -1) is a prime number.

If we put 2^n -1=7

then we get

n=3

so, N = 4 × 7 = 28

since , factors of N=28 are

{ 1, 2 , 4 , 7 ,14 , 28 }

i.e {1, alp1 , alp2 , alp 3,...alp k}

1) Thus , alpk=28

so, k = 5 ans...

2) 1 + 1/alp1 + 1/alp2 +..... + 1/ alp k

= 1 + 1/2 + 1/4 + 1/7 + 1/14 + 1/28

= 2 ans...

Thank you! !

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