Question

Proof that 10 is a solitary number.​

Answer 1

Answer:

Step-by-step explanation:

et σ(n)σ(n) be the sum of divisor function.

For 1010 to have a friendly pair we have to find another solution to

σ(5x)=9xσ(5x)=9x

6 σ(x)=9x (Assuming (5,x)=1)6 σ(x)=9x (Assuming (5,x)=1)

2 σ(x)=3x2 σ(x)=3x

Now Let x=2kpx=2kp where p∈Np∈N ,(p,z)=1(p,z)=1

2(2k+1−1)σ(p)=3⋅2kp2(2k+1−1)σ(p)=3⋅2kp

σ(p)p=3⋅2k−12k+1−1σ(p)p=3⋅2k−12k+1−1

Now as any number has atleast two prime factors one and itself hence ,

σ(p)p>1σ(p)p>1

⟹3⋅2k−1>2