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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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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, 07:10
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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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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!!! [#permalink]
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25 Mar 2013, 07:30
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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25 Mar 2013, 07:40
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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25 Mar 2013, 07:49
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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
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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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25 Mar 2013, 08:23
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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 rightn! 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, 08:34
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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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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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My Answers: 1B 2A 3E 4E 5B 6C 7A 8D 9B 10A 11D

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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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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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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 numberFor \(x=1\) \(23x = 23*1\) > 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, 18:13
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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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3. 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 (1) Both x and y is have 3 positive factors.Consecutive perfect squares could be : 4 , 9 , 16 , 25 , 36 , 49 , 64 .... Among these numbers, only 4 and 9 are consecutive perfect squares that have 3 positive factors ( for instance 16 = 4*4 = 2*2*2*2 > 5 factors and SO ON ) Hence, y=9 and x=4 > 9 = 4.2 +1 > R = 1 Thus, (1) SUFFICIENT (2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbersConsecutive perfect squares could be : 4 , 9 , 16 , 25 , 36 , 49 , 64 .... Among these numbers, only 4 and 9 are consecutive perfect squares that have their square roots as prime numbers ( for example : \(\sqrt{16}\) = 4, which is not a prime number and SO ON ) Hence, y=9 and x=4 > 9 = 4.2 > R = 1 Thus, (2) SUFFICIENT Answer : D
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26 Mar 2013, 04:27
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4. Each digit of the threedigit integer N is a 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. Given that each digit of the threedigit integer N is a multiple of 4 , K could be : 444 or 448 or 484 or 488 ... (1) The units digit of K is the least common multiple of the tens and hundreds digit of KSo , K should b equal to 444 (LCM(4,4) = 4) OR equalt to 888 (LCM(8,8) = 8) OR equalt to 488 (LCM(4,8) = 8) .... Hence, (1) NOT SUFFICIENT (2) K is NOT a multiple of 3.So, K could be equal to 448 or 484 ... Hence, (2) NOT SUFFICIENT (1) + (2) Both 488 and 848 have their units digit as the LCM of the tens and hundreds and are not a mulitple of 3 Hence, (1) + (2) NOT SUFFICIENT Answer : E
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Re: New Set: Number Properties!!!
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26 Mar 2013, 04:27



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