Question

A series consists of some numbers such that the summation of the divisors of any number of that series is 1 less than twice of that number. For example the divisors of 4 are 1,2 and 4 and the sum is 7. If the numbers of this sequence are arranged in ascending order then what is the 9th term?

Answer 1

Answer:

The 9th term is 17179869184

Step-by-step explanation:

Here we have that;

The divisors of 1 = 1 which is 1 less than 1*2

The sum of the divisors of 2 = 1 + 2  = 3 which is 1 less than 2*2

The sum of the divisors of 4 = 1 + 2 + 4 which is 1 less than 4*2

The sum of the divisors of 8 = 1 + 2 + 4 + 8 which is 1 less than 8*2

The sum of the divisors of 32 = 1 + 2 + 4 + 8 + 16 + 32 which is 1 less than 32*2

The sum of the divisors of 256 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 which is 1 less than

The next numbers are therefore,

32 × 256 = 8192

256 × 8192 = 2097152 and the 9th term is therefore;

8192 × 2097152 = 17179869184.