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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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New Set: Number Properties!!!
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29 Mar 2013, 04:34
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. 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 list is a terminating decimal. This statement implies that the list 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 \(2*5^3\). 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. Questions testing this concept: http://gmatclub.com/forum/doesthedeci ... 89566.htmlhttp://gmatclub.com/forum/anydecimalt ... 01964.htmlhttp://gmatclub.com/forum/ifabcdan ... 25789.htmlhttp://gmatclub.com/forum/700question94641.htmlhttp://gmatclub.com/forum/isrs2isa ... 91360.htmlhttp://gmatclub.com/forum/plexplain89566.htmlhttp://gmatclub.com/forum/whichofthe ... 88937.html
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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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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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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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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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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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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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