With respect to this question, are we making an assumption that n doesn't have any negative factors? The question stem is silent on the negative factors of n.

Factors (at least on GMAT) are positive divisors.
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Hi Bunuel ,
From S2 we get c=2 and we know a<b<c and that a,b,c are non negetive int. So, it's quite evident theat a,b have to be 0,1 ? Hence, sufficient? Am i missing something here ?

How do you know when considering (2) that a, b, and c are non-negative?
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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.

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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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).[/quote]

@#$%! Dammm!!!! Sorry. In my head the factorials were a part of the original statement. Need to be more careful

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.

Hi Bunel,

in the solution for statement 1, I need some clarity. The question says that the product of "any" 3 integers is negative. Does that not mean that any three integers selected and multiplied will "certainly" be negative? Bearing in mind that we were not told the number in the set, it just says the number is more than 2. So, any three integers selected could turn out to be positive, positive, positive. Therefore, in my opinion, the only way this statement will always be true is if all the items are negative, such that "any 3" will surely mean negative. Based on this, I'ld say 1 is sufficient. Please let me know where I got this wrong.

I don't have any issues with statement 2 not being sufficient.

[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.

Please read the highlighted part.

Notice that since x and y are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers.

x and y are consecutive perfect squares, so x and y could be:
\(x = 1\) and \(y = 4\) --> \(\sqrt{x}=1\) and \(\sqrt{y}=2\), consecutive integers;
\(x = 4\) and \(y = 9\) --> \(\sqrt{x}=2\) and \(\sqrt{y}=3\), consecutive integers;
\(x = 9\) and \(y = 16\) --> \(\sqrt{x}=3\) and \(\sqrt{y}=4\), consecutive integers;
...
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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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1/2+1/2 divided by 2 is 1/2. It's reciprocal is 2, which is a prime.
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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.

Posted from my mobile device

You are wrong. This is addressed several times on previous pages. I'll try again: if x = -1, then \(2\sqrt{x^2}=2=prime\). Also, \(\sqrt{x^2}\) cannot be negative for any value of x.
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Oh I can see the mistake now. Thanks.

Posted from my mobile device

Hi Bunuel,

Why Can't a be 1 in this case? This will imply that they are not consecutive integers.

There is no condition defined for a in these 2 statements.

The median of 1, 1 and 2 is also 1.

From (1) we goth that b = 1, and if a = 1, then the condition given in the stem which says that a < b < c won't be satisfied.

Hope it's clear.
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If \(0 \lt x \lt y\) and \(x\) and \(y\) are consecutive perfect squares, what is the remainder when \(y\) is divided by \(x\)?

Notice that since \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers.

(1) Both \(x\) and \(y\) have 3 positive factors. This statement implies that \(x=(prime_1)^2\) and \(y=(prime_2)^2\). From above we have that \(\sqrt{x}=prime_1\) and \(\sqrt{y}=prime_2\) are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, \(x=(prime_1)^2=4\) and \(y=(prime_2)^2=9\). The remainder when 9 is divided by 4 is 1. Sufficient.

(2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. The same here: \(\sqrt{x}=2\) and \(\sqrt{y}=3\). Sufficient.


Answer: D[/quote]

Bunuel I could not understand this. (1) states, both are having 3 positive factors. The consecutive prime numbers which are 2 & 3, both are having 2factors only, i.e 1 and the number itself. Moreover, Prime Numbers does not have 3 factors. Pls tell, where am I getting wrong
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Bunuel I could not understand this. (1) states, both are having 3 positive factors. The consecutive prime numbers which are 2 & 3, both are having 2factors only, i.e 1 and the number itself. Moreover, Prime Numbers does not have 3 factors. Pls tell, where am I getting wrong[/quote]

Primes have 2 factors, yes, but the square of primes have 3 factors. So, from (1) \(x=(prime_1)^2\) and \(y=(prime_2)^2\).
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