How many numbers of the form a × b, where a and b are prime numbers

Question

How many numbers of the form a × b, where a and b are prime numbers, have sum of factors, excluding the factor as the number itself, equal to 31?

(A) One

(B) Two

(C) Three

(D) Four

(E) Six

sanjitscorps18 · Answer

The sum of the factors = (a^2 - 1)(b^2 - 1)/(a-1)(b-1)

Now 31 = sum of factors, excluding the factor as the number itself

=> 31 = (a^2 - 1)(b^2 - 1)/(a-1)(b-1) - ab

=> 31 = (a+1)(b+1) - ab

=> 31 = a + b + 1

=> a + b = 30

Now a and b are prime numbers, hence possible values are

(a, b) = (13, 17) or (11, 19) or (7, 23)

(29,1) cannot ben an option as 1 is not a prime number

IMHO Option C

QuantMadeEasy · Answer

Step 1: Understanding the question 
When a and b are the prime numbers, factors of a x b are 1, a, b and a x b. 
a x b is not be included

Sum of the factors = 1 + a + b = 31
hence, a+b = 30
Prime numbers whose sum is 30 are (7,23), (11,19), (13,17)

C is correct

bo12345 · Answer

Bunuel , shouldn't the answer be 6 considering numbers can interchange?

Bunuel · Answer

No, we are not asked about the ordered pairs, so (7,23) is the same as (23,7).

RiyaJ0032 · Answer

that is the definition of an "unordered pair"

where the interchange won't matter

but in ordered pair, the interchange does matter

for eg, if a = 7, b = 23
I can form another pair where a= 23, and b = 7

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