Any number of the form p^{a}q^{b}r^{c} will have (a + 1) (b + 1)(c + 1) factors, where p, q, r are prime. (This is a very important idea)
For any number N of the form p^{a}q^{b}r^{c}, the sum of the factors will be (1 + p^{1} + p^{2} + p^{3} + …+ p^{a}) (1 + q^{1} + q^{2} + q^{3} + …+ q^{b}) (1 + r^{1} + r^{2} + r^{3} + …+ r^{c}).
Sum of factors of number N is 124. 124 can be factorized as 2^{2} * 31. It can be written as 4 * 31, or 2 * 62 or 1 * 124.
2 cannot be written as (1 + p^{1} + p^{2}…p^{a}) for any value of p.
4 can be written as (1 + 3)
So, we need to see if 31 can be written in that form.
The interesting bit here is that 31 can be written in two different ways.
31 = (1 + 2^{1} + 2^{2} + 2^{3} + 2^{4})
31 = ( 1 + 5 + 5^{2})
Or, the number N can be 3 * 2^{4} or 3 * 5^{2}. Or N can be 48 or 75.
Correct Answer: More than one such number exists.
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