What is the value of positive integer x?

Though E is the correct answer, your analysis of statement 1 is faulty. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is not always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50.

From statement 2 we know that it is perfect square, which isn't sufficient. When we combine, we know it is a perfect square and it's factors add to 31, which gives us two answers as you've rightly mentioned.

True you are correct, I purposefully picked only the perfect squares as to make the solution more concise.I edited my first statement.
Thanks for the point.

VERITAS PREP OFFICIAL SOLUTION

Solution: E.

Statement 2 is clearly not sufficient as plenty of numbers are squares of integers (1, 4, 9, 16, 25, 36, etc.). When testing statement 1, then, it's a good idea to test candidates that will work with statement 2, as well. 25 is a good option: the factors of 25 are 1, 5, and 25 for a sum of 31. x could be 25. The only other number in that list with a chance is 16. And the factors of 16 are 1, 2, 4, 8, and 16. That sum is also 31. Because statements 1 and 2 each allow for two different values of x, the informations is not sufficient and the correct answer is E.

1) SUM OF ALL UNIQUE FACTORS = 31
X= 17x11x2x1 => sum =31
X= 17x13x1 => sum = 31
AD eliminated
2) X=Y^2 , Y = integer
X=1 , Y=1
X=4, Y=2
X=9, Y=3
B eliminated

Combine 1&2

X= Y^2
Y= 17x13x1 => X= (17x13x1)^2
Y= 17x11x2x1 => X= (17x11x2x1)^2
C is eliminated

Ans is E

Concept: to find the SUM of all the unique positive factors of a given value, you can find all the unique combinations of the prime bases of the number and then using common factoring end up with the following Rule/Formula:

If a number - N’s prime factorization is given by:

N = (p)^2 * (q)^1

Then for each unique prime, you add up each Prime from the 0th power to its highest power

And then take the product of each one of the SUMS you find (as illustrated below for the hypothetical value of N above):

SUM of unique positive factors of N = (p^0 + p^1 + p^2) * (q^0 + q^1)


What is the value of integer X = ?

Statement 1: the sum of all the factors of X is 31

Since 31 is a prime number itself, using the above concept to find the SUM of all the unique positive factors, you can infer that there must be only ONE UNIQUE PRIME FACTOR which divides X.

This is because the only 2 positive integers that multiply to 31 are——-> (1) * (31) = 31 ——> and since we include (prime)^0 = 1 in each one of our “factors”, we could never have such a scenario.

For example: if there were 2 prime factors that divided X, such that —- N = (p) (q)

then the SUM of the factors would be:

(p^0 + p^1) * (q^0 + q^1) = 31

We could never have one of the Binomial Factors of the above product be just 1 since any base to the 0th power = 1 already.


Thus If the SUM of all the unique positive factors of X is 31, then we can Infer that X is made up of ONLY 1 Unique Prime Factor.


Case 1: try a value of X composed of only the prime factor 2

If X = (2)^4

Then the sum of all the unique prime factors would be:

(2^0 + 2^1 + 2^2 + 2^3 + 2^4) =

(1 + 2 + 4 + 8 + 16) = 31

X can = 16

Case 2: can we use the prime factor of 3 and get the same result?

3’0 = 1
3’1 = 3
3’2 = 9
3’3 = 27 ———-> too high, will not work


Case 3: X is composed only of the Prime Factor 5

If X = (5)^2

Then the SUM of all the unique prime factors would be:

(5^0 + 5^1 + 5^2) =

(1 + 5 + 25) = 31


X can = 25 as well

Since we have two different values for X, statement 1 not sufficient

*note* X can not = 31 because both 1 and 31 divide into 31 and the sum of the factors would be 32, not 31, in violation of statement 1


S2: X = (Y)^2

All statement 2 tells us is that X must be a perfect square.

Thus X = 16 and X = 25 both work.


Together: since case X can = 16 or 25, the statements will be not sufficient together either.


E

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