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New Set: Number Properties!!!
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Bunuel
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New Set: Number Properties!!! [#permalink] New post  25 Mar 2013, 03:50
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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.


2. If a positive integer n has exactly two positive factors what is the value of n?

(1) n/2 is one of the factors of n
(2) The lowest common multiple of n and n + 10 is an even number.


3. If 0 < x < y and x and y are consecutive perfect squares, what is the remainder when y is divided by x?

(1) Both x and y is have 3 positive factors.
(2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers



4. Each digit of the three-digit integer K is a positive multiple of 4, what is the value of K?

(1) The units digit of K is the least common multiple of the tens and hundreds digit of K
(2) K is NOT a multiple of 3.


5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

(1) The median of {a!, b!, c!} is an odd number.
(2) c! is a prime number


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


7. Is x the square of an integer?

(1) When x is divided by 12 the remainder is 6
(2) When x is divided by 14 the remainder is 2


8. List A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the list less than 1/5?

(1) Reciprocal of the median is a prime number
(2) The product of any two terms of the list is a terminating decimal


9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

(1) ab = 2
(2) 0 < a < b < 2


10. If N = 3^x*5^y, where x and y are positive integers, and N has 12 positive factors, what is the value of N?

(1) 9 is NOT a factor of N
(2) 125 is a factor of N


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

(1) |x - 2| < 2 - y
(2) x + y - 3 = |1-y|


CHECK SOLUTIONS BELOW, ON PAGE 1!
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Bunuel
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Re: New Set: Number Properties!!! [#permalink] New post  25 Sep 2014, 07:23
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anupamadw wrote:
Bunuel wrote:
SOLUTIONS:

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) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Answer: E.


Hi Bunuel
sorry to ask a silly doubt,but prime numbers are always positive so can't we discard -1 ?
Thanks
Anupama


(2) \(2\sqrt{x^2}\) is a prime number, not x. For \(2\sqrt{x^2}\) to a prime number, x must be 1 or -1. So, we have two possible values of x, which makes the statement insufficient.
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Re: New Set: Number Properties!!! [#permalink] New post  30 Nov 2014, 00:41
Bunuel wrote:
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
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Re: New Set: Number Properties!!! [#permalink] New post  01 Dec 2014, 04:49
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bhatiavai wrote:
Bunuel wrote:
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.

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hope it helps.
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Re: New Set: Number Properties!!! [#permalink] New post  28 May 2015, 01:00
Bunuel wrote:
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.
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Re: New Set: Number Properties!!! [#permalink] New post  28 May 2015, 02:37
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gmatriser wrote:
Bunuel wrote:
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.


Hi gmatriser,

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.

Hope it's clear :)

Regards
Harsh
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Re: New Set: Number Properties!!! [#permalink] New post  02 Nov 2015, 07:20
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GMATPASSION wrote:
lets take 1/512 then??


1/512 = 0.001953125 = terminating decimal. Any fraction with 2-s and/or 5-s in the denominator will be a terminating decimal. Please re-read the theory given in the solution.
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Re: New Set: Number Properties!!! [#permalink] New post  20 Jan 2016, 19:06
Bunuel wrote:
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:

would (1)+(2) still not give different solutions i.e. (negative, negative, negative) and (positive, negative, positive)?
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Re: New Set: Number Properties!!! [#permalink] New post  20 Jan 2016, 20:42
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WilDThiNg wrote:
Bunuel wrote:
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:

would (1)+(2) still not give different solutions i.e. (negative, negative, negative) and (positive, negative, positive)?


In case of (positive, negative, positive), the smallest number there would be negative and the largest will be positive --> their product will be negative, so it cannot be a prime, which violates the second statement. By the way this is explained several times in this thread.
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Re: New Set: Number Properties!!! [#permalink] New post  22 Aug 2016, 07:35
Bunuel wrote:
9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

Answer: A.


Bunnel, are we not supposed to consider fractions for this? It's nowhere mentioned a and b are integers.

Similarly for question 6, It says "Are all the numbers in set S negative"?? However it's modified in the explanation post as "Are all the integers in set S negative".
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Re: New Set: Number Properties!!! [#permalink] New post  22 Aug 2016, 07:43
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Argo wrote:
Bunuel wrote:
9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. Ifa=1/2and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

Answer: A.


Bunnel, are we not supposed to consider fractions for this? It's nowhere mentioned a and b are integers.

Similarly for question 6, It says "Are all the numbers in set S negative"?? However it's modified in the explanation post as "Are all the integers in set S negative".


a and b can be negative (for example check highlighted part) but [a] and [b] cannot since [] is a function which rounds DOWN a number to the nearest integer.
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Re: New Set: Number Properties!!! [#permalink] New post  22 Aug 2016, 08:31
Bunuel wrote:
Argo wrote:
Bunuel wrote:
9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. Ifa=1/2and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

Answer: A.


Bunnel, are we not supposed to consider fractions for this? It's nowhere mentioned a and b are integers.

Similarly for question 6, It says "Are all the numbers in set S negative"?? However it's modified in the explanation post as "Are all the integers in set S negative".


a and b can be negative (for example check highlighted part) but [a] and [b] cannot since [] is a function which rounds DOWN a number to the nearest integer.


So, say a=6/5 and b=5/3, in that case the answer would be E. Correct me if I am wrong.
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Re: New Set: Number Properties!!! [#permalink] New post  22 Aug 2016, 08:39
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Argo wrote:
Bunuel wrote:
Argo wrote:
9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. Ifa=1/2and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

Answer: A.

Bunnel, are we not supposed to consider fractions for this? It's nowhere mentioned a and b are integers.

Similarly for question 6, It says "Are all the numbers in set S negative"?? However it's modified in the explanation post as "Are all the integers in set S negative".


a and b can be negative (for example check highlighted part) but [a] and [b] cannot since [] is a function which rounds DOWN a number to the nearest integer.


So, say a=6/5 and b=5/3, in that case the answer would be E. Correct me if I am wrong.


If a = 6/5 and b =5/3, then [a] + [b] = 1 + 1 = 2, which gives the same NO answer as in the solution.
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Re: New Set: Number Properties!!! [#permalink] New post  17 Oct 2016, 08:44
Bunuel wrote:
5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

Note that:
A. The factorial of a negative number is undefined.
B. 0!=1.
C. Only two factorials are odd: 0!=1 and 1!=1.
D. Factorial of a number which is prime is 2!=2.

(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient.

(2) c! is a prime number. This implies that c=2. Not sufficient.

(1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient.

Answer: C.


Statement 1: Implies b! is odd. The only positive odd factorial is for number 1. And it is mentioned in the question that a,b and c are consecutive. Therefore can only be 0,1, 2.
Statement 2: Implies that c! is prime which means that c can be 2! only. And it is mentioned in the question that a,b and c are consecutive. Therefore can only be 0,1, 2.

So shouldn't the answer be D? Please help out.
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Re: New Set: Number Properties!!! [#permalink] New post  17 Oct 2016, 08:49
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prvy wrote:
Bunuel wrote:
5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

Note that:
A. The factorial of a negative number is undefined.
B. 0!=1.
C. Only two factorials are odd: 0!=1 and 1!=1.
D. Factorial of a number which is prime is 2!=2.

(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient.

(2) c! is a prime number. This implies that c=2. Not sufficient.

(1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient.

Answer: C.


Statement 1: Implies b! is odd. The only positive odd factorial is for number 1. And it is mentioned in the question that a,b and c are consecutive. Therefore can only be 0,1, 2.
Statement 2: Implies that c! is prime which means that c can be 2! only. And it is mentioned in the question that a,b and c are consecutive. Therefore can only be 0,1, 2.

So shouldn't the answer be D? Please help out.


For (1) note that 0! = 1.

For (2) c = 2 but we are NOT told that a, b, and c are consecutive integers. The question asks whether a, b, and are c consecutive integers.
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Re: New Set: Number Properties!!! [#permalink] New post  11 Apr 2017, 06:29
Bunuel wrote:
SOLUTIONS:

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) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Answer: E.


From what I heard from the experts, GMAT's general rule is that square root of a number is always positive on gmat. By this, i can get B as an answer. since root 1 will be 1 and not -1. Please explain.
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Re: New Set: Number Properties!!! [#permalink] New post  11 Apr 2017, 06:39
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sindhugclub wrote:
Bunuel wrote:
SOLUTIONS:

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) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Answer: E.


From what I heard from the experts, GMAT's general rule is that square root of a number is always positive on gmat. By this, i can get B as an answer. since root 1 will be 1 and not -1. Please explain.


Yes, the square root cannot give negative result but this has little to do with the problem.

Both x=1 and x=-1 when substituted into \(2\sqrt{x^2}\) give 2.

x = -1 --> \(2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1}=2\).

x = 1 --> \(2\sqrt{x^2}=2\sqrt{1^2}=2\sqrt{1}=2\).
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Re: New Set: Number Properties!!! [#permalink] New post  11 Apr 2017, 22:26
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sindhugclub wrote:
Bunuel wrote:
SOLUTIONS:

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) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Answer: E.


From what I heard from the experts, GMAT's general rule is that square root of a number is always positive on gmat. By this, i can get B as an answer. since root 1 will be 1 and not -1. Please explain.


In addition to what Bunuel said, check out this post: https://www.veritasprep.com/blog/2016/0 ... oots-gmat/
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Re: New Set: Number Properties!!! [#permalink] New post  27 Apr 2017, 19:54
Bunuel wrote:
5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

Note that:
A. The factorial of a negative number is undefined.
B. 0!=1.
C. Only two factorials are odd: 0!=1 and 1!=1.
D. Factorial of a number which is prime is 2!=2.

(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient.

(2) c! is a prime number. This implies that c=2. Not sufficient.

(1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient.

Answer: C.


Hi Bunuel, thanks for putting up such amazing questions.
My Question:
STatement 2: we know that c=2.
Now the question states that a<b<c; (integers) and we know that the factorial of a negative number is not defined.
thus, automatically a=0, b=1.
so shouldnt the answer be B?
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Re: New Set: Number Properties!!! [#permalink] New post  27 Apr 2017, 22:00
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rachitasetiya wrote:

Hi Bunuel, thanks for putting up such amazing questions.
My Question:
STatement 2: we know that c=2.
Now the question states that a<b<c; (integers) and we know that the factorial of a negative number is not defined.
thus, automatically a=0, b=1.
so shouldnt the answer be B?



Hey rachitasetiya,

The second statement only focuses on the factorial of c. So, we can definitely conclude that c = 2. But there is no information in the second statement, which can help us conclude that a,b and c are consecutive integers.

I guess when you said that " the factorial of a negative number is not defined", you also considered the information given in the first statement, to draw your conclusion.

Please remember, in DS, when we are solving a question using a particular statement we should focus only on the information given in that statement independently.

Hence, from the second statement, we get to know that c = 2 and there is nothing mentioned about a and b. Hence the second statement is not sufficient to answer the question.

It is only by combining, you can conclude that a = 0, b= 1 and c =2.

I hope the above explanation is clear. :)


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Re: New Set: Number Properties!!! [#permalink] New post  23 Aug 2017, 08:03
Bunuel wrote:
7. Is x the square of an integer?

The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square).

Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

(1) When x is divided by 12 the remainder is 6. Given that \(x=12q+6=6(2q+1)=2*3*(2q+1)\). Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient.

(2) When x is divided by 14 the remainder is 2. Given that \(x=14p+2\). So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient.

Answer: A.


hi

2 * 3 * (2q + 1)
obviously, (2q + 1) is an odd number, however, since (2q + 1) is an odd number, the power of 2 in x will be odd...
maybe it is very obvious to you, but I am totally stumped here... please help me understand it ... :cry:

thanks in advance
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