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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. Solution: newsetnumberproperties14977540.html#p12053412. 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. Solution: newsetnumberproperties14977540.html#p12053553. 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 Solution: newsetnumberproperties14977560.html#p12053584. Each digit of the threedigit 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. Solution: newsetnumberproperties14977560.html#p12053615. 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 Solution: newsetnumberproperties14977560.html#p12053646. 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. Solution: newsetnumberproperties14977560.html#p12053737. 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 Solution: newsetnumberproperties14977560.html#p12053788. 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 (2) The product of any two terms of the set is a terminating decimal Solution: newsetnumberproperties14977560.html#p12053829. 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 Solution: newsetnumberproperties14977560.html#p120538910. 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 Solution: newsetnumberproperties14977560.html#p1205392BONUS QUESTION: 11. If x and y are positive integers, is x a prime number?(1) x  2 < 2  y (2) x + y  3 = 1y Solution: newsetnumberproperties14977560.html#p1205398Kudos points for each correct solution!!!
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Re: New Set: Number Properties!!! [#permalink]
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07 Apr 2013, 22:17
imhimanshu 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 \(y2\leq{0}\) and it cannot be greater than x  2). Next, since given that y is a positive integer, then y=1.
Answer: D. Hi Bunuel, Thanks for the awesome questions that you ve posted. I have a doubt in the above question. Could you please elaborate why can't we solve the Statement A as below  x  2 < 2  y > y2< X2 < 2y I tried solving above, by removing the Modulus, however I am not getting any specific value. Kindly respond. Thanks Yes, to get the value of y its better to use the approach shown in my solution.
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Re: New Set: Number Properties!!! [#permalink]
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08 Apr 2013, 20:13
imhimanshu 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 \(y2\leq{0}\) and it cannot be greater than x  2). Next, since given that y is a positive integer, then y=1.
Answer: D. Hi Bunuel, Thanks for the awesome questions that you ve posted. I have a doubt in the above question. Could you please elaborate why can't we solve the Statement A as below  x  2 < 2  y > y2< X2 < 2y I tried solving above, by removing the Modulus, however I am not getting any specific value. Kindly respond. Thanks Responding to a pm: If you must solve it using algebra, then do this: x  2 < 2  y Take the two cases: Case 1: x  2 >= 0 i.e. x >= 2 x  2 < 2  y x + y < 4 Since x must be 2 or greater than 2 and x and y both are positive integers, x must be 2 and y must be 1 (x cannot be 3 or greater since y must be at least 1 but the sum of x and y in that case will not be less than 4). Case 2: x  2 < 0 i.e. x < 2 The only value that x can take in this case is 1. 1  2 < 2  y 1 < 2  y y < 1 But we know that y must be a positive integer so x cannot take value 1. The only value that x can take is 2 which is prime. Hence this statement alone is sufficient. Mind you, using algebra too you will have to use logic so you might as well look at Bunuel's solution again.
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Re: New Set: Number Properties!!! [#permalink]
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09 Apr 2013, 04:18
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. A small question, when the outcome of {a!, b!, c!} is an odd number ...is ...{1, 1, c!}, should we need to check what C is ? as whatever the C is it cant be consecutive integers. Regards Sanjiv



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Re: New Set: Number Properties!!! [#permalink]
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09 Apr 2013, 04:31
casanjiv 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. A small question, when the outcome of {a!, b!, c!} is an odd number ...is ...{1, 1, c!}, should we need to check what C is ? as whatever the C is it cant be consecutive integers. Regards Sanjiv If we knew for sure that the set is {1, 1, c!}, then yes we would have a definite NO answer to the question. But {1, 1, c!} is not possible since a < b < c implies that the integers in the set are distinct. (1) implies that the set is {0, 1, c!} ({0!=1, 1!=1, c!}) and if c=2, then the answer would be YES, but if c=3, then the answer would be NO. Hope it's clear.
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10 Apr 2013, 20:02
amazing set bunuel!



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10 Jul 2013, 06:41
Thanks Bunuel,
ONE CORRECTION:
set queston no. 9 the option should be D because you are using the condition given in II 0<a<b<2 to sufficient



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09 Sep 2013, 04:02
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 viseversa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or viseversa). 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. Hi Bunuel, Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = 2 for any negative a and b? i took the case of a=2 and b=1 and hence ab=2. In this case i am getting [a]+[b] = (2) + (1) = 3 Let me know if i am missing something.



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09 Sep 2013, 04:06
emailmkarthik 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 viseversa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or viseversa). 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. Hi Bunuel, Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = 2 for any negative a and b? i took the case of a=2 and b=1 and hence ab=2. In this case i am getting [a]+[b] = (2) + (1) = 3 Let me know if i am missing something. "If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b." Since, 3<2, then this statement holds for this set of numbers too. Does this make sense?
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09 Sep 2013, 04:08
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Bunuel wrote: emailmkarthik 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 viseversa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or viseversa). 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. Hi Bunuel, Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = 2 for any negative a and b? i took the case of a=2 and b=1 and hence ab=2. In this case i am getting [a]+[b] = (2) + (1) = 3 Let me know if i am missing something. "If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b." Since, 3<2, then this statement holds for this set of numbers too. Does this make sense? Oh yes! It does make sense. I understand that the statement does not fall apart. But I was just interested in knowing if there is a case possible where we can get 2 as the maximum value.



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09 Sep 2013, 04:20
emailmkarthik wrote: Bunuel wrote: emailmkarthik 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 viseversa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or viseversa). 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.
Hi Bunuel,
Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = 2 for any negative a and b?
i took the case of a=2 and b=1 and hence ab=2. In this case i am getting [a]+[b] = (2) + (1) = 3
Let me know if i am missing something. "If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b." Since, 3<2, then this statement holds for this set of numbers too. Does this make sense? Oh yes! It does make sense. I understand that the statement does not fall apart. But I was just interested in knowing if there is a case possible where we can get 2 as the maximum value. What I meant was that generally if a and b are negative, then the maximum value of [a] + [b] is 1+(1)=2. But you cannot get 2, if ab=2.
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09 Sep 2013, 04:24
Thanks for the clarification and also for the really good questions!



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13 Sep 2013, 06:43
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 \(y2\leq{0}\) 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 \(1 < x2 < 1\), or \(1 < x < 3\), thus \(x=2=prime\). Sufficient.
(2) x + y  3 = 1y. Since y is a positive integer, then \(1y\leq{0}\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D. Hi Can any one explain me the Why Statement 1 is not sufficient. . I am not able to fellow the Bunuel concept and i doubt even in exam will i be able to think like it. Please explain in algebric method. Thanks in advance, Rrsnathan.



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13 Sep 2013, 09:01
rrsnathan 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 \(y2\leq{0}\) 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 \(1 < x2 < 1\), or \(1 < x < 3\), thus \(x=2=prime\). Sufficient.
(2) x + y  3 = 1y. Since y is a positive integer, then \(1y\leq{0}\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D. Hi Can any one explain me the Why Statement 1 is not sufficient. . I am not able to fellow the Bunuel concept and i doubt even in exam will i be able to think like it. Please explain in algebric method. Thanks in advance, Rrsnathan. The first statement is sufficient. Alternative explanation is here: newsetnumberproperties14977580.html#p1209577Hope it helps.
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15 Sep 2013, 07:54
Quote: 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



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15 Sep 2013, 08:34
fido wrote: Quote: 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.
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Re: New Set: Number Properties!!! [#permalink]
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16 Sep 2013, 11:43
Yes,Bunuel.Thank you



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Re: New Set: Number Properties!!! [#permalink]
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26 Sep 2013, 07:09
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. Bunuel A question, in Statement 1 if we take the value of x = 1, than 23x = 23 I think 23 is not a prime. So x can only and only be +1 Therefore, Statement (1) is sufficient. What's your take on it. Regards



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Re: New Set: Number Properties!!! [#permalink]
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26 Sep 2013, 08:24
SUK123 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. Bunuel A question, in Statement 1 if we take the value of x = 1, than 23x = 23 I think 23 is not a prime. So x can only and only be +1 Therefore, Statement (1) is sufficient. What's your take on it. Regards Notice that in the first statement we have the absolute value of 23x > 23x. If x=1, then 23x=23*(1)=23=prime. Hope it's clear.
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Re: New Set: Number Properties!!!
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