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New Set: Number Properties!!!

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New post 25 Mar 2013, 04:50
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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.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205341


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.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205355

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

Solution: https://gmatclub.com/forum/new-set-numb ... l#p1209015


4. Each digit of the three-digit 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/new-set-numbe ... l#p1205361


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

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205364


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
(2) The product of the smallest and largest integers in the list is a prime number.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205373


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

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205378


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

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205382


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

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205389


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

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205392


BONUS QUESTION:
11. If x and y are positive integers, is x a prime number?

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

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205398


Kudos points for each correct solution!!!
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New post 29 Mar 2013, 06:11
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11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 2 (if y is more than or equal to 2, then \(2-y\leq{0}\) and it cannot be greater than |x - 2|). Next, since given that y is a positive integer, then y=1.

So, we have that: \(|x - 2| < 1\), which implies that \(-1 < x-2 < 1\), or \(1 < x < 3\), thus \(x=2=prime\). Sufficient.

(2) x + y - 3 = |1-y|. Since y is a positive integer, then \(1-y\leq{0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

Answer: D.
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Collection of Questions:
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Re: New Set: Number Properties!!!  [#permalink]

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New post 28 Mar 2013, 19:27
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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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New post 25 Mar 2013, 04:50
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New post 25 Mar 2013, 07:03
1
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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New post 25 Mar 2013, 07:10
1
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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New post 25 Mar 2013, 07:24
1
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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New post 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?

\(x=12q+6\)
\(x=14z+2\)

\(12q+6=14z+2\)
\(6-2=14z-12q\)
\(4=2(7z-6q)\)
\(2=7z-6q\)

\(z=2,q=2\)

\(x=12*2+6=30\)
\(x=14*2+2=30\)

so x is not the square of an integer.IMO C
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New post 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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New post Updated on: 25 Mar 2013, 08:57
1
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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Originally posted by Zarrolou on 25 Mar 2013, 07:49.
Last edited by Zarrolou on 25 Mar 2013, 08:57, edited 1 time in total.
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New post 25 Mar 2013, 08:10
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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New post 25 Mar 2013, 08:23
1
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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New post 25 Mar 2013, 08:34
1
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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New post 25 Mar 2013, 09:35
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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.

D.
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New post 25 Mar 2013, 09:49
My Answers:
1-B
2-A
3-E
4-E
5-B
6-C
7-A
8-D
9-B
10-A
11-D
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New post 25 Mar 2013, 10:06
2
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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New post 25 Mar 2013, 11:02
2
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.

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New post 25 Mar 2013, 17:58
1
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]

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New post 25 Mar 2013, 18:13
1
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!!! &nbs [#permalink] 25 Mar 2013, 18:13

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