Question

9.
Let N be a natural number having 4 factors, If sum of all factors excluding Nis 31. Then the number of
possible values of N is?
​

Answer 1

Answer:

3.1

Step-by-step explanation:

let the numbers be n ,2n,3n and 4n

than, its sum is 31

so, it means

n+2n+3n+4n =31 (sum of 4 natural number )

10n=31

n =31/10

n=3.1

Answer 2

Answer:

4

Step-by-step explanation:

A number has four factor is two cases: it's a prime number raised to the third power, so N=n3N=n3.

So in this case the factors are n3, n2, nn and 1. Their sum is 31, so n2+n+1=31, n=5 and n3=125 or n=-6, but the natural numbers are positive so this is not valid. ( and 6 is not prime )

The other valid case is the one in which the number N=a∗bN=a∗b, where a and b are prime.

In this case the factors are 1,a,b,a∗b, their sum is a+b+1=31 or a+b=30, where a and B are prime.

This sum is valid if the numbers a,b are 7,23 or 13,17 or 11,19. (please note that 1 and 29 is not a valid option)

So N can be 7∗23=161 or 13∗17=221 or 11∗19=209.

Total values 4