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New Set: Number Properties!!!

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New Set: Number Properties!!! [#permalink] New post 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.

Solution: new-set-number-properties-149775-40.html#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: new-set-number-properties-149775-40.html#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: new-set-number-properties-149775-60.html#p1205358


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: new-set-number-properties-149775-60.html#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: new-set-number-properties-149775-60.html#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: new-set-number-properties-149775-60.html#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: new-set-number-properties-149775-60.html#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: new-set-number-properties-149775-60.html#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: new-set-number-properties-149775-60.html#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: new-set-number-properties-149775-60.html#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: new-set-number-properties-149775-60.html#p1205398


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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 15:16
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 15:18
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 15:25
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 15:27
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 17:49
jonnyhh wrote:
Rock750 wrote:
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|

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

If x<2 --> x=1 because x is a positive integer then x is prime
If x=2 --> x is prime

If x>2 --> x - 2 < 2 - y
In this case :
if y<2 --> y = 1 -> x < 3 so x can take two values : 1 and 2 --> both are primes ; thus x is prime
if y = 2 -> x - 2 < 0 -> x<2 so x = 1 and thus x is prime
if y>2 hence 2 - y < 0 -> x - 2 < 2 - y < 0 -> x - 2 < 0 -> x < 2 -> x = 1 -> thus x is prime

(1) ALONE IS SUFFICIENT

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

y is a positive integer
--> if y = 1 --> x = 2 thus x is prime
--> if y>1 --> |1-y| = y - 1 --> x + y - 3 = y - 1 --> x = 2 and thus x is prime

(2) ALONE IS SUFFICIENT

Answer : D



Though your answer is correct, the reasoning you provided is incorrect.

1. FYI - 1 is NOT a prime number.
2. For statement 1: y<2 => 1<x<3 so x cannot be =1. Also, y>2 is not possible as |x-2| cannot be less than 0.

HTH.


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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 18:24
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1] (1) |23.x| is a prime number. Since 23 is prime, and x is an integer, any other value of x other than +1 or -1
will result in a multiple of 23, therefore, nonprime number. Therefore, x = -1 or x = 1
Insufficient


(2) 2.|x| is a prime number. Any other value of x other than +1 or -1 will result in an even number other than 2. Therefore, x=-1 or x =1
Insufficient

(1)+(2) still gives x = 1 or x = -1 Insufficient


Answer: E

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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 18:27
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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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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 18:30
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(1) Both x and y have 3 factors. Which means that x and y are not only perfect squares, but also perfect squares of PRIME numbers. Let's remember that the only 2 consecutive prime numbers are 2 and 3, therefore: Sufficient


(2) This already gives us directly the same information that sqrt(x)=2, and sqrt(y) = 3, therefore: Sufficient

Answer: D
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 18:37
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(1) 444, 488 and 848 all fit the description with this information. Theferore, insufficient


(2) K is a multiple 3 if, and only if, the sum of its digits is a multiple of 3. Note that, applying this rule, only
444 and 888 are multiple of 3, from the possibilities that fit the stem description. All other possibilites are not multiples of 3.
Therefore, insufficient

(1) + (2): 488 and 848 are still a possibility. Insufficient


Answer: E

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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 18:43
5]

(1) If a<b<c, then a! < b! < c! , which means that the median of the set {a!, b!,c!} is b!. If b! = odd, then b! = 1, which means b = 0 or b = 1. However, c can be any number greater than 1. Therefore, insufficient

(2) c! is prime, than c = 2. Which means, b=1, and a=0 (assuming that GMAT is not defining factorial functions for negative integers). Sufficient


[b]Answer: B

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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 18:54
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(1) Consider the sets: {-1,-2,-3} and {1,2,-3}, which both fit the stem description and statement 1. For one the answer is YES, for the other, the answer is NO. Therefore, Insufficient.


(2) Consider the sets: {1,2,3} and {-3,-2,-1}, which both fit the stem description and statement 2. For one the answer is YES, for the other, the answer is NO. Therefore, Insufficient

(1) + (2):
From the information in (2), we have that the largest and smallest number HAVE to be of the same sign. Therefore, for a three integer set, the middle number will have to be of the same sign. From statement (1), they all have to be negative.
Note that for sets with more than 3 integers, the fact that the product of any 3 integers have to be negative, forces the other numbers to be negative as well. Therefore, Sufficient

Answer: C
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 19:05
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(1) Look for patterns. For perfect squares, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169,... the remainders when divided by 12 are: 1, 4, 9,4,1,4,9,0, 1,4,9.... Therefore, the answer will always be no. Sufficient
(keep in mind, this is not a trivial nor formal passage, but this pattern check will do for gmat)

(2) Consider the numbers 16, and 100 (both perfect squares), and both have remainder 2 when divided by 14. Therefore, insufficient

Answer: A
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 19:31
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8] Since we have an even number of elements in the set, the median will be the average of the two central numbers. Therefore, the median will be of type: (1/p + 1/q)/2 = (q+p)/(2.p.q), where p and q are primes.

(1) The reciprocal of the median, X = 2pq/(p+q) , is prime. So, pq/(p+q) = 1 (let me know if this passage was not trivial for you). Therefore: pq = p + q , which means p = q = 2. Since 1/2 is also the largest possibility of an element for S, we must have all numbers of the set equal to 1/2. Therefore, Sufficient

(2) The only terminating decimals of type 1/n, where n is a prime number are: 1/2 and 1/5. Set S has to have only elements equal to 1/2 and 1/5, which means the median can be equal to 1/2, 1/5 or something between 1/2 and 1/5. Therefore, the median is always greater than 1/5. Sufficient

Answer: D
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 19:47
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9]
(1) a.b = 2 , therefore b = 2/a.
If a < 0 , [a]+[2/a] will be negative. Therefore [a]+[ b ] is not equal to 1.
If 0<a<1, [a] = 0, [2/a]>2. Therefore, [a]+[ b ] is not equal to 1.
If 1< a<2, [a] =1, [2/a]=1. Therefore, [a]+[ b ] is not equal to 1
If a>2, [2/a] = 0, [a] >=2. Therefore, [a]+[ b ] is not equal to 1.

Therefore, [a]+[ b ] is never iqual to 1. Sufficient


(2) 0 < a < b < 2
Consider a = 1/2, b = 3/2. In this case, the answer is yes, since [a] +[ b ] = 0 + 1 = 1
Consider a = 1/2, b= 3/4. In this case, the answer is no, since [a] + [ b ] = 0 + 0 = 0

Therefore, insufficient


Answer: A
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 19:54
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10] (x+1).(y+1) = 12, results in the possible (x,y) pairs: (1,5), (2,3), (3,2), (5,1)


(1) If 9 is not a factor of N, then x < 2. Therefore N = 3.5^5. Sufficient

(2) If 125 is a factor of N, then y >=3. Therefore (x,y) can be either (1,5) or (2,3). Insufficient

Answer: A
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 20:02
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11] (BONUS QUESTION)

(1) |x-2| < 2-y
=> y-2 < x - 2 < 2 - y
=> y < x < 4 -y
=> y = 1 (try, plugging y >1, to see what happens to the inequality)
=> 1 < x < 3
=> x = 2
=> x is prime

Sufficient


(2) x + y - 3 = |1-y|
As y is a positive integer, y>=1 always, which means |1-y| = y-1. Therefore, x + y - 3 = y -1, which implies x = 2.
Sufficient

Answer: D
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Re: New Set: Number Properties!!! [#permalink] New post 28 Mar 2013, 23:50
Zarrolou wrote:
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


Though I too think that the correct answer is "C", but the explanation is a little confusing..
as the stmt 1 says that the product of ab=2, so its not necessary that one will be 2 and the other be 1/2. It could be either(-1*-2) or (1/2*4). Hope you got my point here.
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Re: New Set: Number Properties!!! [#permalink] New post 29 Mar 2013, 01:19
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lahca007 wrote:
Zarrolou wrote:
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


Though I too think that the correct answer is "C", but the explanation is a little confusing..
as the stmt 1 says that the product of ab=2, so its not necessary that one will be 2 and the other be 1/2. It could be either(-1*-2) or (1/2*4). Hope you got my point here.


From F.S 1 , we know that a,b both are not equal to zero. Also, it is being asked whether, [a]+[b] =1. b = 2/a.

Thus, the question is asking whether, [a] + [2/a] = 1? --> 1-[a] = [2/a]

Notice, for 0<a<1, we have [a] = 0, and [2/a] is atleast 2. Thus, 1-[a] is not equal to [2/a].

Similarly, for 1<a<2, we have [a] = 1, and [2/a] = 1. Thus, 1-[a] is not equal to [2/a].

For a>2, we have 1-[a] as negative and [2/a] as 0. Hence, not equal again.

Similarly, for a<0, we would always have 1-[a] as positive but [2/a] will always be negative and hence not equal.

One can check for a=1,2,-1,-2 and would still find they are not equal. Sufficient.

From F.S 2, we have 0<a<b<2. For a=0.1 and b=1, we have [a]+[b] =1, so a YES. However, for a=1 and b = 1.5, we have [a]+[b] = 2, hence a NO. Insufficient.

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Re: New Set: Number Properties!!! [#permalink] New post 29 Mar 2013, 02:30
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SOLUTIONS:

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

(2) 2\sqrt{x^2} is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Answer: E.
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Re: New Set: Number Properties!!! [#permalink] New post 29 Mar 2013, 02:55
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2. If a positive integer n has exactly two positive factors what is the value of n?

Notice that, n has exactly two positive factors simply means that n is a prime number, so its factors are 1 and n itself.

(1) n/2 is one of the factors of n. Since n/2 cannot equal to n, then n/2=1, thus n=2. Sufficient.

(2) The lowest common multiple of n and n + 10 is an even number. If n is an odd prime, then n+10 is also odd. The LCM of two odd numbers cannot be even, therefore n is an even prime, so 2. Sufficient.

Answer: D.
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Re: New Set: Number Properties!!!   [#permalink] 29 Mar 2013, 02:55
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