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The next set of medium/hard DS number properties questions. I'll post OA's with detailed explanations on Friday. Please, post your solutions along with the answers.

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number (2) \(2\sqrt{x^2}\) is a prime number.

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

Hi

{-1,-2,-3} --> Here (-1)*(-3) = 3 = prime. also, {1,-2,3} --> here 1*3 = 3 = prime

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

Hi

{-1,-2,-3} --> Here (-1)*(-3) = 3 = prime. also, {1,-2,3} --> here 1*3 = 3 = prime

so shouldn't the answer be E ?

Are 1 and 3 the smallest and largest integers of {1, -2, 3}?
_________________

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

Hi

{-1,-2,-3} --> Here (-1)*(-3) = 3 = prime. also, {1,-2,3} --> here 1*3 = 3 = prime

so shouldn't the answer be E ?

Are 1 and 3 the smallest and largest integers of {1, -2, 3}?

11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 2 (if y is more than or equal to 2, then \(y-2\leq{0}\) and it cannot be greater than |x - 2|). Next, since given that y is a positive integer, then y=1.

So, we have that: \(|x - 2| < 1\), which implies that \(-1 < x-2 < 1\), or \(1 < x < 3\), thus \(x=2=prime\). Sufficient.

(2) x + y - 3 = |1-y|. Since y is a positive integer, then \(1-y\leq{0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

Answer: D.

Hi Bunuel,

What the best way to approach Absolute questions.

My current logic is that |x| = |-x|

and using the above I try to get my range/value of x

11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 2 (if y is more than or equal to 2, then \(y-2\leq{0}\) and it cannot be greater than |x - 2|). Next, since given that y is a positive integer, then y=1.

So, we have that: \(|x - 2| < 1\), which implies that \(-1 < x-2 < 1\), or \(1 < x < 3\), thus \(x=2=prime\). Sufficient.

(2) x + y - 3 = |1-y|. Since y is a positive integer, then \(1-y\leq{0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

Answer: D.

Hi Bunuel,

What the best way to approach Absolute questions.

My current logic is that |x| = |-x|

and using the above I try to get my range/value of x

|x-2| < 2-y

x-2 < 2-y or -x+2 < 2-y

x+y < 4 or x-y > 0

Is this the correct way to approach it ?

Cheers

Yes, when x >=2, then |x - 2| = x - 2, so in this case we get x - 2 < 2 - y and when x < 2, then |x - 2| = -(x - 2), so in this case we get -(x - 2) < 2 - y. But this does not help us to answer the question. To do so, you should approach the question shown in my solution.

As for the best way solve modulus questions - it depends on a question, multiple approaches are possible. Check the links below for practice.

(1) 9 is NOT a factor of N. This implies that the power of 3 is less than 2, thus N could only be \(3^1*5^5\). Sufficient.

(2) 125 is a factor of N. This implies that the power of 5 is more than or equal to 3, thus N could be \(3^1*5^5\) or \(3^2*5^3\). Not sufficient.

Answer: A.

THEORY. Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

bunuel,

would the gmat ever test the situation where they list the prime factorization (as in this problem) but try to trick you with a number being to the 0 power? in other words, wouldnt 3 also be to the 0 power here, leaving 1 x 5^y ?

6. Set S consists of more than two integers. Are all the numbers in set S negative?

(1) The product of any three integers in the list is negative Not sufficient Example: \(S = {-1,3,5}\) the product is always <0 but 2 numbers are positive Example: \(S = {-1,-3,-5}\) the product is always <0 and all numbers are negative

(2) The product of the smallest and largest integers in the list is a prime number. Not sufficient Example: \(S = {1,3,5}\) \(1*5=5\) prime but all positive Example: S = \({-1,-3,-5}\) \(-1*-5=5\) prime but all negative

(1)+(2) Sufficient IMO C Using statement 2 we know that all are positive or all are negative, using statement 1 we know that "at least" 1 is negative=> so all are negative

Sry... Am confused.....wat if the set S={1,3,5,7} ....its not given that the set must contain only 3 numbers....can some one explain plzzzzzzz

6. Set S consists of more than two integers. Are all the numbers in set S negative?

(1) The product of any three integers in the list is negative Not sufficient Example: \(S = {-1,3,5}\) the product is always <0 but 2 numbers are positive Example: \(S = {-1,-3,-5}\) the product is always <0 and all numbers are negative

(2) The product of the smallest and largest integers in the list is a prime number. Not sufficient Example: \(S = {1,3,5}\) \(1*5=5\) prime but all positive Example: S = \({-1,-3,-5}\) \(-1*-5=5\) prime but all negative

(1)+(2) Sufficient IMO C Using statement 2 we know that all are positive or all are negative, using statement 1 we know that "at least" 1 is negative=> so all are negative

Sry... Am confused.....wat if the set S={1,3,5,7} ....its not given that the set must contain only 3 numbers....can some one explain plzzzzzzz

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

Dear Bunuel, Thanks for the excellent set of problems. However, I have one query with regard the above solution. I have highlighted the text in yellow - with regard the statement (2), how can we assume that all integers are either positive or negative. It is possible that only the smallest and the largest integers have similar signs and that the other integers have different signs; Or am i misunderstanding something here? Please confirm.

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

Dear Bunuel, Thanks for the excellent set of problems. However, I have one query with regard the above solution. I have highlighted the text in yellow - with regard the statement (2), how can we assume that all integers are either positive or negative. It is possible that only the smallest and the largest integers have similar signs and that the other integers have different signs; Or am i misunderstanding something here? Please confirm.

Statement-II specifically talks about the smallest and the largest integer. Since their product is positive, there may be two possibilities:

1. Smallest integer and largest integer both are negative - Since the largest integer is negative there can't be a number greater than the largest integer. As any positive integer is greater than a negative integer, there can't be a positive integer in the set.

2. Smallest integer and largest integer both are positive - Similarly as the smallest integer is positive, there can't be a negative integer in the set because the negative integer then will become the smallest integer.

Hence the set consists of only negative or positive integers.

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

Dear Bunuel, Thanks for the excellent set of problems. However, I have one query with regard the above solution. I have highlighted the text in yellow - with regard the statement (2), how can we assume that all integers are either positive or negative. It is possible that only the smallest and the largest integers have similar signs and that the other integers have different signs; Or am i misunderstanding something here? Please confirm.

for question 1 to have 2 * root (x^2) to be a prime number don't we think we have to x as positive only. if we consider x as negative -2 is not a prime. please clarify

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

for question 1 to have 2 * root (x^2) to be a prime number don't we think we have to x as positive only. if we consider x as negative -2 is not a prime. please clarify

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

for question 1 to have 2 * root (x^2) to be a prime number don't we think we have to x as positive only. if we consider x as negative -2 is not a prime. please clarify

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

\(2\sqrt{x^2}\) won't give negative value for any x, because \(\sqrt{nonnegative \ number}\geq{0}\): the square root function cannot give negative result.

If x=1 or x=-1, then \(2\sqrt{x^2}=2\sqrt{1}=2=prime\).
_________________

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

Dear Bunuel, First of all,i would like to thank you for all your sincere efforts that help innumerable gMAT takers like me.

In the explanation provided for statement 2,i did not quite understand how all the numbers of the set would have the same sign,if the smallest and largest terms have the same sign.The set contains>2 terms and from statement,we can comment only about 2 terms.Can you help me understand?

Thank you, Fido

(2) says: The product of the smallest and largest integers in the set is a prime number.

Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign.

So either: smallest * greatest = negative * negative and in this case as both the smallest and the greatest are negative then ALL integers in the list are negative OR smallest * greatest = positive * positive and in this case as both the smallest and the greatest are positive then ALL integers in the list are positive.

Hope it's clear.

Doesn't this give us 2 answers?

One is that all numbers could be positive; and the other being, that all numbers could be negative.

There is no we can say all are negative; or all are positive. at best, we are ruling out a mixed set.

So we get no unique answer.

we would only have a unique answer if the question was: "Are all the integers in set S either >0 or <0?"

Dear Bunuel, First of all,i would like to thank you for all your sincere efforts that help innumerable gMAT takers like me.

In the explanation provided for statement 2,i did not quite understand how all the numbers of the set would have the same sign,if the smallest and largest terms have the same sign.The set contains>2 terms and from statement,we can comment only about 2 terms.Can you help me understand?

Thank you, Fido

(2) says: The product of the smallest and largest integers in the set is a prime number.

Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign.

So either: smallest * greatest = negative * negative and in this case as both the smallest and the greatest are negative then ALL integers in the list are negative OR smallest * greatest = positive * positive and in this case as both the smallest and the greatest are positive then ALL integers in the list are positive.

Hope it's clear.

Doesn't this give us 2 answers?

One is that all numbers could be positive; and the other being, that all numbers could be negative.

There is no we can say all are negative; or all are positive. at best, we are ruling out a mixed set.

So we get no unique answer.

we would only have a unique answer if the question was: "Are all the integers in set S either >0 or <0?"

HOW is it possible the set to have all positive numbers and the product of any three integers in the set to be negative (first statement)??? Please re-read the question and solution.
_________________

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient. Can another option here be {Positive, Negative, Positive} (product is still negative)

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient. Based on the above option, even after joining two statements, the answer is inconclusive. What am i missing here?

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.

"The product of any two terms of the set is a terminating decimal" : IF THIS IS TRUE THE ACCORDING TO THIS 1/2 AND 1/5 can be in the set. But any two means that we can take 1/2 and 1/2 and that will not be necessarily a terminating decimal........ I hope my doubt is clear....

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.

"The product of any two terms of the set is a terminating decimal" : IF THIS IS TRUE THE ACCORDING TO THIS 1/2 AND 1/5 can be in the set. But any two means that we can take 1/2 and 1/2 and that will not be necessarily a terminating decimal........ I hope my doubt is clear....

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.

"The product of any two terms of the set is a terminating decimal" : IF THIS IS TRUE THE ACCORDING TO THIS 1/2 AND 1/5 can be in the set. But any two means that we can take 1/2 and 1/2 and that will not be necessarily a terminating decimal........ I hope my doubt is clear....

Why not? 1/2*1/2 = 1/4 = terminating decimal.

Bunuel wrote:

GMATPASSION wrote:

Bunuel wrote:

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.

"The product of any two terms of the set is a terminating decimal" : IF THIS IS TRUE THE ACCORDING TO THIS 1/2 AND 1/5 can be in the set. But any two means that we can take 1/2 and 1/2 and that will not be necessarily a terminating decimal........ I hope my doubt is clear....

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