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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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New Set: Number Properties!!! [#permalink]
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17 Nov 2015, 20:18
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. Shouldn't \(\sqrt{x^2}\) always give a positive value. In that case, B is sufficient!? I read somewhere that, on the GMAT, sqrt(25) is only +5. Although x^2=25 has different solutions.
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17 Nov 2015, 21:01
hdwnkr 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. Shouldn't \(\sqrt{x^2}\) always give a positive value. In that case, B is sufficient!? I read somewhere that, on the GMAT, sqrt(25) is only +5. Although x^2=25 has different solutions. PLEASE, read the whoel thread before posting a question.Chances are that your doubt is already addressed there: newsetnumberproperties149775120.html#p1367573newsetnumberproperties14977580.html#p1271393\(2\sqrt{x^2}\) 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 \(2\sqrt{x^2}=2\sqrt{1}=2=prime\).
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Re: New Set: Number Properties!!! [#permalink]
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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]
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20 Jan 2016, 20:42
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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20 Jan 2016, 20:54
Bunuel wrote: 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. Thanks!... sorry, inadvertently my reasoning got muddled up somewhere



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20 Jan 2016, 21:07
Hi, the Q is Quote: 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 (2) The product of any two terms of the set is a terminating decimal.. the main point in the line "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?" is that the answer can be YES or NO if the numbers can be same, but the answer will always be YES if a statement would have told us that all numbers are differentWhile doing Q, we should keep watching out for info that can be gathered from the main statements.
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21 Aug 2016, 22:02
Bunuel wrote: 4. 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. K could be 444, 488, 848, or 888. Not sufficient.
(2) K is NOT a multiple of 3. K could be 448, 484, 488, 844, 848, or 884. Not sufficient.
(1)+(2) From above K could be 488 or 848. Not sufficient.
Answer: E. Bunnel, can we also consider 400 for statement 1?



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21 Aug 2016, 23:56
Argo wrote: Bunuel wrote: 4. 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. K could be 444, 488, 848, or 888. Not sufficient.
(2) K is NOT a multiple of 3. K could be 448, 484, 488, 844, 848, or 884. Not sufficient.
(1)+(2) From above K could be 488 or 848. Not sufficient.
Answer: E. Bunnel, can we also consider 400 for statement 1? No, because we are told that the digits must be positive multiples of 4, 0 is neither positive nor negative.
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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 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. 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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22 Aug 2016, 07:43
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 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. 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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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 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. 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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22 Aug 2016, 08:39
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 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. 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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29 Aug 2016, 02:31
Quote: 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 Can we consider 3^0 * (5^11) in (1)? I chose E cause it gives 12 factors as 3^1*5^5 does. Thanks!
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29 Aug 2016, 02:44



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29 Aug 2016, 02:47
Bunuel wrote: Akela wrote: Quote: 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 Can we consider 3^0 * (5^11) in (1)? Gives the 12 factors. Thanks! No, because we are told that x and y are positive integers, while 0 is neither positive nor negative. Ah, ok. Thanks!
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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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17 Oct 2016, 08:49
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]
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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]
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11 Apr 2017, 06:39
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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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: New Set: Number Properties!!!
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