Question

A 2 digit number has exactly 3 factors excluding 1 and the number itself. What is the sum of all the values of the number

Answer 1

First make prime factorization of an integer n=ap∗bq∗crn=ap∗bq∗cr, where aa, bb, and cc are prime factors of nn and pp, qq, and rr are their powers.

The number of factors of nn will be expressed by the formula (p+1)(q+1)(r+1)(p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: 450=21∗32∗52450=21∗32∗52

Total number of factors of 450 including 1 and 450 itself is (1+1)∗(2+1)∗(2+1)=2∗3∗3=18(1+1)∗(2+1)∗(2+1)=2∗3∗3=18factors.

Back to the question:

Since 5 is a prime number, it cannot be the product of two integers greater than 1, which implies that a number having 5 factors must be of a form of (prime)^4 --> the number of factors = (4 + 1).

There are only 2 two-digit numbers which can be written this way: 2^4 = 16 and 3^4 = 81.


16 has factor - 2,4,8
81 has factor-3,927