Hi Bunuel,

In 2) it says 2√x^2 is a prime number. Does that not mean that x is a 1 since -2 cannot be a prime number. I think answer is B

if all the numbers were 1/5 , the median would have been (1/5 + 1/5)/2= 1/5.
the reciprocal of 1/5 is 5 a prime number. so we can have all 1/2 or all 1/5 . in both the cases the numbers are not less than 1/5.Please correct me where I am going wrong. option A is sufficient.

Cannot a set be say {1, 2, 3, 4, 5, 7} --> the product of the smallest and largest = 1*7 = 7 = prime.
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Factors (at least on GMAT) are positive divisors.
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How do you know when considering (2) that a, b, and c are non-negative?
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How do you know when considering (2) that a, b, and c are non-negative?[/quote]

As you rightly pointed out - The factorial of a negative number is undefined.
We have a set [a!,b!,c!]. Now if c! is 2. Then that leaves a,b with 0,1.[/quote]

My question still remains. How do you know that when considering (2)??? You cannot use info from (1), when considering (2).
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Hi There Bunel,

Was looking at this solution, the question says that the X and Y are consecutive perfect squares. One could interpret it to mean that they are perfect squares that follow each other ie 4, 9, 16, 25 etc, that is the perfect squares or consecutive primes, which should be different in meaning from consecutive integers. How come the solution assumes that they are consecutive integers? while solving it, I was able to tell that the numbers are prime, but thought the answer cannot be ascertained because it could be any primes, any of the sets that follow each other, so I chose E instead. Please clarify.

[quote="Bunuel"]8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Answer: B.

Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.



Hi Bunel,

For your solution in 1, if all the terms are 1/2, wouldn't the median be 1/2+1/2 divided by 2? since it says median, not mean. Same applies to 1/7, Such that the reciprocal in both cases cannot be prime.

Just to clarify: all solutions here are correct 100%.

As it's explained in the solution, the product of any three integers in the set to be negative, we can have TWO cases:
a. the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}.
b. the set consists of more than 3 terms, then the set can only have negative numbers.
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In reference to Q1: In second statement although √x² can have value x=1 and x=-1, yet the 2√x² will be prime only when it's value is positive; There are no negative primes. Therefore x can take just value=1.
In my view Option B is correct answer.

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