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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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11 May 2014, 05:08



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23 May 2014, 09:57
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. a doubt t on explanation for statement 1 if a set S={A(+ve) , B(ve), C(+ve) , D (+ve)} and we take product ACD>it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing



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23 May 2014, 10:02
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. a doubt on the statement 2 \sqrt{x*x}=1 0r 1 but since its given that 2*(sqrtX^2)=prime; x can only take positive value as 2 would not be a prime no.. thus according to me statement two should be sufficient



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New Set: Number Properties!!! [#permalink]
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23 May 2014, 10:17
somyakalra 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. a doubt on the statement 2 \sqrt{x*x}=1 0r 1 but since its given that 2*(sqrtX^2)=prime; x can only take positive value as 2 would not be a prime no.. thus according to me statement two should be 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 \(2\sqrt{x^2}=2\sqrt{1}=2=prime\). Writing Mathematical Formulas on the Forum: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628
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23 May 2014, 10:26
somyakalra 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. a doubt t on explanation for statement 1 if a set S={A(+ve) , B(ve), C(+ve) , D (+ve)} and we take product ACD>it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing You misinterpreted the statement. It says: the product of ANY three integers in the set is negative. So, the set cannot be {positive, positive, positive, negative} because the product of the first 3 terms there is NOT negative. Basically if the set has 4 or more terms, and not all of them are negative, then this statement will be violated, because we'll be able to pick two negative terms and one nonnegative, which will not give negative product. Hope it's clear.
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23 May 2014, 10:30
somyakalra 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. a doubt t on explanation for statement 1 if a set S={A(+ve) , B(ve), C(+ve) , D (+ve)} and we take product ACD>it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing Hello somyakalra, For st1: it is given that product of any 3 integers is negative. The word "any" is of prime importance here The case that you have taken for ACD will not be part of this set since it does not meet the condition in st 1...for a set of 4 integers which should satisfy st1 all 4 nos have to be negative only.You cannot question the given statements in DS questions. For st 2: Note prime nos are always positive Consider a set of negative nos 1 to 5 Largest no: 1 , smallest: 5 Product of these no: 5 which is prime. All the members in this set will have to be in this range only of 1 to 5 only Consider another case meeting condition given in st2 Another group with smallest integer :1 and largest integer:5 Their product is 5. Again we get 2 cases ( all members negative and all members positive) for st2 and hence it not sufficient Combining you get Ans C as you are left with only 1 criteria which satisfies st1 and st2 and that all nos in the set are negative Hope it helps Posted from my mobile device Posted from my mobile device Posted from my mobile device
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23 May 2014, 10:34
somyakalra 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. a doubt t on explanation for statement 1 if a set S={A(+ve) , B(ve), C(+ve) , D (+ve)} and we take product ACD>it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing As for the second statement. (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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13 Jul 2014, 07:20
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, In (2), by definition negative Z's are not prime. So 1 ruled out. B then? Can you please clarify!



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24 Jul 2014, 03:39
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. Bunuel how we do get the above highlighted part ? How about when a and b are equal to 4 and 0.5



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24 Jul 2014, 03:45
himanshujovi 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. Bunuel how we do get the above highlighted part ? How about when a and b are equal to 4 and 0.5 Please read the whole thread: newsetnumberproperties14977580.html#p1265170newsetnumberproperties14977580.html#p1265177As for your example, if a = 4 and b= 0.5, then [a] + [b] = 4  1 = 5, which is LESS than 2.
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03 Sep 2014, 08:29
Hi,
Can you please suggest in question number 9 why can't we have values a = 0.1 and b = 20 so a*b will be 2 and equation won't be satisfied



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14 Sep 2014, 17:01
Hi Bunuel,
Good day. Need your help,
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.
I took your posibilities here for this problem as follows. S(smallest), L(largest) S,,L +S,,+L
Now here, (1) & (2) satisfies as prod of 3 will be ve and prod of S&L will be +ve.. So we have 2 solutions even after combining the 2 equations. So E should be the answer. Please clarify..
Thanks & Regards, Lokesh.



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14 Sep 2014, 17:04
sheolokesh wrote: Hi Bunuel,
Good day. Need your help,
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.
I took your posibilities here for this problem as follows. S(smallest), L(largest) S,,L +S,,+L
Now here, (1) & (2) satisfies as prod of 3 will be ve and prod of S&L will be +ve.. So we have 2 solutions even after combining the 2 equations. So E should be the answer. Please clarify..
Thanks & Regards, Lokesh. How can positive number in your second set be the smallest one in the set if you have a negative number there?
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14 Sep 2014, 17:12
Bunuel wrote: sheolokesh wrote: Hi Bunuel,
Good day. Need your help,
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.
I took your posibilities here for this problem as follows. S(smallest), L(largest) S,,L +S,,+L
Now here, (1) & (2) satisfies as prod of 3 will be ve and prod of S&L will be +ve.. So we have 2 solutions even after combining the 2 equations. So E should be the answer. Please clarify..
Thanks & Regards, Lokesh. How can positive number in your second set be the smallest one in the set if you have a negative number there? Ahhh... Yes.. Missed it.. Thanks Bro..



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25 Sep 2014, 07:05
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



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25 Sep 2014, 07:23



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07 Oct 2014, 16:28
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 nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). 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 2s and/or 5s 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.
I believe you meant to write "1/5 + 1/2 = 7/10", not 7/20 and that "250 = 2 * 5^3" not 2 * 5^2? just want to be clear. thank you.



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Re: New Set: Number Properties!!! [#permalink]
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08 Oct 2014, 01:01
bsmith37 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 nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). 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 2s and/or 5s 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.
I believe you meant to write "1/5 + 1/2 = 7/10", not 7/20 and that "250 = 2 * 5^3" not 2 * 5^2? just want to be clear. thank you. It's (1/5 + 1/2) /2 = 7/20, so no typo there. You are right about 250 = 2 * 5^3, though. Edited. Thank you.
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Re: New Set: Number Properties!!!
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08 Oct 2014, 01:01



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