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New Set: Number Properties!!! [#permalink]
25 Mar 2013, 03:50

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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.

Re: New Set: Number Properties!!! [#permalink]
25 Mar 2013, 06:10

1

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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 factorsn 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!!! [#permalink]
25 Mar 2013, 06:24

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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]
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!!! [#permalink]
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 = |1-y|

This is a GOOD one. IMO C

(1) |x - 2| < 2 - y

x-2>0, x>2 case 1)x>2 x-2<2-y x+y<4

case 2)0<x<=2 ( x is positive ) -x+2<2-y x>y

NOT SUFFICIENT

(2) x + y - 3 = |1-y|

case 1)y>1 x+y-3=1-y x+2y=4

case 2)0<y<=1 ( y is positive) x+y-3=-1+y x=2

NOT SUFFICIENT

Combining 1 and 2 we obtain that

------0------------1----------2---------------- ------|~~~~~~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!!! [#permalink]
25 Mar 2013, 07:23

1

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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 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!!! [#permalink]
25 Mar 2013, 07:34

1

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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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Re: New Set: Number Properties!!! [#permalink]
25 Mar 2013, 08:35

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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 = |1-y|

We know that x >0 and y>0 and they are integers.

From F.S 1, we have 2-y>=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+y-3 = y-1 or x=2(prime). Thus, Sufficient.

Re: New Set: Number Properties!!! [#permalink]
25 Mar 2013, 09:06

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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.

Re: New Set: Number Properties!!! [#permalink]
25 Mar 2013, 10:02

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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.

Re: New Set: Number Properties!!! [#permalink]
25 Mar 2013, 16:58

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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 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!!! [#permalink]
25 Mar 2013, 17:13

1

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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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Re: New Set: Number Properties!!! [#permalink]
26 Mar 2013, 02:57

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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 numbers 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 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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Last edited by Rock750 on 29 Mar 2013, 03:20, edited 1 time in total.