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New Set: Number Properties!!!
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25 Mar 2013, 03:50
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: http://gmatclub.com/forum/newsetnumbe ... l#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: http://gmatclub.com/forum/newsetnumbe ... l#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: https://gmatclub.com/forum/newsetnumb ... l#p12090154. 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: http://gmatclub.com/forum/newsetnumbe ... l#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: http://gmatclub.com/forum/newsetnumbe ... l#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: http://gmatclub.com/forum/newsetnumbe ... l#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: http://gmatclub.com/forum/newsetnumbe ... l#p12053788. 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 Solution: http://gmatclub.com/forum/newsetnumbe ... l#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: http://gmatclub.com/forum/newsetnumbe ... l#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: http://gmatclub.com/forum/newsetnumbe ... l#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: http://gmatclub.com/forum/newsetnumbe ... l#p1205398Kudos points for each correct solution!!!
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Re: New Set: Number Properties!!!
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29 Mar 2013, 04:06
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.
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Re: New Set: Number Properties!!!
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28 Mar 2013, 18:27
2] n has only two positive factorss, therefore n is prime.
(1) n/2 is one of the factors of n. If n is prime, and n/2 is a factor of n, then n = 2. Sufficient
(2) As Bunuel always says, the greatest rule about LCM is that for two numbers a,b, we have: a.b =LCM(a,b).GCF(a,b). If n is prime, let's suppose n>2 (in this case, n will have to be odd since it is prime). If n is odd, then n+10 is odd as well. From the rule we just mentioned:
n.(n+10) = odd*odd = odd = LCM.GCF > Therefore, LCM(n,n+10) cannot be even (otherwise, the product would be even).
So, n has to be 2. Sufficient
Answer: D




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Re: New Set: Number Properties!!!
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25 Mar 2013, 03:50
Please suggest on what category would you like the next set to be. Thank you!
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Re: New Set: Number Properties!!!
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25 Mar 2013, 06:01
word problems Thanks for the set
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Re: New Set: Number Properties!!!
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25 Mar 2013, 06:03
If x is an integer, what is the value of x?
(1) 23x is a prime number Since 23 si prime,\(x\)can be \(+1\) or \(1\) not sufficient
(2) 2\sqrt{x^2} is a prime number. once again \(x\) can be \(+1\) or \(1\) not sufficient
And since 1)+2) provides no new info IMO E



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Re: New Set: Number Properties!!!
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25 Mar 2013, 06:10
2. If a positive integer n has exactly two positive factors what is the value of n?
Number of factors of a number is \(a+1\) where the number is \(n^a\) And since n has 2 factors\(n\) must be prime and\(>1\)
(1) n/2 is one of the factors of n Only \(2\) fits these conditions, so \(n=2\) Sufficient
(2) The lowest common multiple of n and n + 10 is an even number. Only \(2\) fits these conditions, so \(n=2\) once again. \(n\) IMO is prime so the only prime that respect statement (2) is 2 because all other prime are odd, and odd+even = odd, so the LCM of an odd and an odd is odd in all cases except n=2
IMO D



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Re: New Set: Number Properties!!!
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25 Mar 2013, 06:24
6. Set S consists of more than two integers. Are all the numbers in set S negative?
(1) The product of any three integers in the list is negative Not sufficient Example: \(S = {1,3,5}\) the product is always <0 but 2 numbers are positive Example: \(S = {1,3,5}\) the product is always <0 and all numbers are negative
(2) The product of the smallest and largest integers in the list is a prime number. Not sufficient Example: \(S = {1,3,5}\) \(1*5=5\) prime but all positive Example: S = \({1,3,5}\) \(1*5=5\) prime but all negative
(1)+(2) Sufficient IMO C Using statement 2 we know that all are positive or all are negative, using statement 1 we know that "at least" 1 is negative=> so all are negative



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Re: New Set: Number Properties!!!
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25 Mar 2013, 06:30
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
A,B and D cannot be the answer clearly. What about C?
\(x=12q+6\) \(x=14z+2\)
\(12q+6=14z+2\) \(62=14z12q\) \(4=2(7z6q)\) \(2=7z6q\)
\(z=2,q=2\)
\(x=12*2+6=30\) \(x=14*2+2=30\)
so x is not the square of an integer.IMO C



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Re: New Set: Number Properties!!!
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25 Mar 2013, 06:40
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
The question can be seen as (given statement 2): \([a] + [b] = 1\) case 1:\(0<(=)a<1\) => \([a]=0\) and \(1(=)<b<2\) => \([b]=1\) so \([a] + [b] = 1\) or the opposite case 2: \(0<(=)b<1\) => \([b]=0\) and \(1(=)<a<2\) => \([a]=1\) so \([a] + [b] = 1\)
But as ab=2 we know that one term is \(\frac{1}{2}\) and the other is \(2\) So we are in one of the two senarios above, IMO C



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Re: New Set: Number Properties!!!
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Updated on: 25 Mar 2013, 07:57
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?
Number of factors of \(N = (x+1)(y+1) = 12\) the combinations are \(3*4\) with \(x= 2\) and \(y=3\) \(6*2\) with \(x=5\) and\(y = 1\) and the "other way round" of each one
(1) 9 is NOT a factor of N So x must be 1, \(x=1\) because \((1+1)(y+1)=12\) \(y=5\) Sufficient
(2) 125 is a factor of N So \(y>=3\), y can be 3 or 5, NOT sufficient \(y=3, (3+1)(x+1)=12, x=2\) \(y=5, (5+1)(x+1)=12, x=1\)
IMO A
Originally posted by Zarrolou on 25 Mar 2013, 06:49.
Last edited by Zarrolou on 25 Mar 2013, 07:57, edited 1 time in total.



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Re: New Set: Number Properties!!!
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25 Mar 2013, 07:10
11. If x and y are positive integers, is x a prime number?
(1) x  2 < 2  y (2) x + y  3 = 1y
This is a GOOD one. IMO C
(1) x  2 < 2  y
\(x2>0, x>2\) case 1)\(x>2\) \(x2<2y\) \(x+y<4\)
case 2)\(0<x<=2\) ( x is positive ) \(x+2<2y\) \(x>y\)
NOT SUFFICIENT
(2) x + y  3 = 1y
case 1)\(y>1\) \(x+y3=1y\) \(x+2y=4\)
case 2)\(0<y<=1\) ( y is positive) \(x+y3=1+y\) \(x=2\)
NOT SUFFICIENT
Combining 1 and 2 we obtain that
012 ~~~~~~x>y~~~~~~~~~~x+y<4 for the first one ~~x=2~~~~~~~~~x+2y=4~~~~for the second one
And combining all the cases together we obtain 1)\(0<x<=2\) with \(0<y<=1\) \(x=2\) and \(x>y\) so \(x=2\) and \(y=1\) 2)\(0<x<=2\) with \(y>1\) \(x>y\) and \(x+2y=4\), given that x and y are positive \(x=2, y=1\) 3)\(x>2\) with \(0<y<=1\) \(x+y<4\) and \(x=2\) so \(x=2,y=1\) 4)x>2 with y>1 \(x+y<4\) and \(x+2y=4\), \(x=2,y=1\) In each case \(x=2\) so x is prime



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Re: New Set: Number Properties!!!
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25 Mar 2013, 07:23
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. the median of three elements is the one in the middle, so b! is odd there are only 2 cases in which n! is odd and are if n=1 or if n=0 so b is 0 , 1 Not Sufficient
(2) c! is a prime number c can once again be 0,1 or in this case 2. Not sufficient
This is a weak passage, I don't know if I'm right n! is possible only for positive number so given that a < b < c c must be 2, b must be 1, and (because of my weak hypothesis a>=0) a must be 0
IMO C



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Re: New Set: Number Properties!!!
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25 Mar 2013, 07:34
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 Not sufficient
(2) The product of any two terms of the set is a terminating decimal Because \(\frac{1}{prime}\) is not a terminating decimal, with the only exception of 1/4, 1/1, 1/5 and 1/2 any set made by these three CANNOT have a median < 1/5, it can be = 1/5 but NEVER < Some examples: A={1,1,1,1,1/5,1/5,1/5,1/5,1/5,1/5} the median is 1/5 = 1/5 A={1,1,1,1,1/2,1/2,1/2,1/2,1/2,1/2} the median is 1/2 > 1/5 SUFFICIENT
IMO B



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Re: New Set: Number Properties!!!
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25 Mar 2013, 08:35
11. If x and y are positive integers, is x a prime number? (1) x  2 < 2  y (2) x + y  3 = 1y We know that x >0 and y>0 and they are integers. From F.S 1, we have 2y>=0 or y<=2. Thus y can only be 2 or 1. Now if y=2, we would have 0>some thing positive or 0>0(when x also equal to 2). Either case is not possible. Thus, y can only be 1. For y=1, we can only have x = 2. Which is prime. Sufficient. From F.S 2, we have either y>1 or y<1. Now as y is a positive integer, y can't be less than 1.For y=1, we anyways have x=2(prime).In the first case, we have y>1> x+y3 = y1 or x=2(prime). Thus, Sufficient. D.
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Re: New Set: Number Properties!!!
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25 Mar 2013, 08:49
My Answers: 1B 2A 3E 4E 5B 6C 7A 8D 9B 10A 11D



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Re: New Set: Number Properties!!!
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25 Mar 2013, 09:06
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 From F.S 1, we have x = 12q+6 > 6(2q+1). For x to be a square of an integer, we should have 2q+1 of the form 6^pk^2, where both q,p and k are integers and p is odd. Now we know that 2q+1 is an odd number and 6^pk^2 is even. Thus they can never be equal and hence x can never be the square of an integer. Sufficient. From F.S 2, we have x = 14q+2 > 2(7q+1).Just as above, we should have 7q+1 = 2^pk^2. Now for q=1, k=2 and p=1, we have 8=8, thus x is the square of an integer. But for q=0, x is not. Insufficient. Basically for the second fact statement, we can plug in easily. No need for the elaborate theory. A.
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Re: New Set: Number Properties!!!
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25 Mar 2013, 10:02
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 From F.S 1, we have b! = odd, thus b can be 0 or 1.But, as factorial notation is only for positive integers, thus, if b=0, then a would become negative and thus b is only equal to 1.Now, a can only be 0 as we are given that a! exists. But nothing has been mentioned about c. All we know is that c>1 and an integer. Insufficient. From F.S 2, we have c! is a prime number. Again, c has to be positive and c can only be 2.However, a and b can take any values, even negative. Insufficient. Taking both together, we have a=0, b=1 and c=2. Sufficient. C.
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Re: New Set: Number Properties!!!
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25 Mar 2013, 16:58
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.
(1) 23x is a prime number
For \(x=1\) \(23x = 23*1\) > 23 is prime
For \(x=1\) \(23x = 23*(1)]\) > \(23 = 23\) 23 is prime also
Thus, this holds true for two values of x and because of that, the value of x cannot be determined.
(1) INSUFFICIENT
(2)\(2 \sqrt{x^2}\) is a prime number. x=1 > \(2 \sqrt{x^2}\) =2 prime x=1 > \(2 \sqrt{x^2}\) =2 prime Thus, x can take the value of either 1 or 1 (2) INSUFFICIENT
23x is a prime number AND \(2 \sqrt{x^2}\) is a prime number. For \(x=1\) 23 is prime and 2 is prime For \(x=1\) 23 is prime and 2 is prime (1) +(2) INSUFFICIENT
Answer : E



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Re: New Set: Number Properties!!!
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25 Mar 2013, 17:13
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.
n is a positive integer that has exactly two positive factors > 1 must be one of its factor > n is a prime number and not equal to 1 (because 1 has only one positive factors, itself)
So these two factors could be (1,2) or (1,3) or (1,5) ... and n could be 2,3,5 .......
(1) n/2 is one of the factors of n
n/2 is a factor of n > n/2 is an integer and from the pairs (1,2), (1,3) ... only 2 is divisible by 2 Hence , n = 2 > (1) SUFFICIENT
(2) The lowest common multiple of n and n + 10 is an even number.
LCM (n,n+10) = EVEN
If n = 2 then LCM(2,12) = 12, which is EVEN If n = 3 then LCM (3,13) = 39, which is ODD Except for n=2, Like n=3, n = 5,7,11 .... LCM (n,n+10) will be ALWAYS ODD. Hence, n = 2 > (2) SUFFICIENT
Answer : D




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