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# New Set: Number Properties!!!

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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 11:57
1
1. E
2. D
3. E
4. E
5. C
6. C
7. A
8. E
9. C
10. D
11. D
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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 12:09
1
1
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

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]

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Updated on: 28 Mar 2013, 16:28
1
Bunuel PS comment on method used- If you have the time of course.
Attachments

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Gmatclub_01.jpg [ 60.39 KiB | Viewed 25713 times ]

Originally posted by samsonfred76 on 28 Mar 2013, 16:16.
Last edited by samsonfred76 on 28 Mar 2013, 16:28, edited 1 time in total.
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Re: New Set: Number Properties!!!  [#permalink]

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Updated on: 29 Mar 2013, 08:16
Bunuel PS comment on method used- If you have the time of course.
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Gmatclub_03.jpg [ 54.37 KiB | Viewed 25665 times ]

Originally posted by samsonfred76 on 28 Mar 2013, 16:18.
Last edited by samsonfred76 on 29 Mar 2013, 08:16, edited 2 times in total.
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Re: New Set: Number Properties!!!  [#permalink]

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Updated on: 28 Mar 2013, 16:29
1
Bunuel PS comment on method used- If you have the time of course.
Attachments

Gmatclub_06.jpg [ 58.12 KiB | Viewed 25668 times ]

Gmatclub_05.jpg [ 69.29 KiB | Viewed 25678 times ]

Originally posted by samsonfred76 on 28 Mar 2013, 16:25.
Last edited by samsonfred76 on 28 Mar 2013, 16:29, edited 1 time in total.
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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 16:27
1
Bunuel PS comment on method used- If you have the time of course.
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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 18:49
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

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.

I Completely agree with you

Thank you
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Re: New Set: Number Properties!!!  [#permalink]

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Updated on: 28 Mar 2013, 19:27
2
1
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

Originally posted by caioguima on 28 Mar 2013, 19:24.
Last edited by caioguima on 28 Mar 2013, 19:27, edited 1 time in total.
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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 19:30
3
3]

(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

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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 19:37
1
4]
(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

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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 19:43
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

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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 19:54
2
6]
(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

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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 20:05
2
7]

(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

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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 20:31
1
1
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

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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 20:47
1
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

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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 20:54
2
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

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Re: New Set: Number Properties!!!  [#permalink]

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28 Mar 2013, 21:02
2
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

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Re: New Set: Number Properties!!!  [#permalink]

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29 Mar 2013, 00:50
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]

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29 Mar 2013, 02:19
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.

A.
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Re: New Set: Number Properties!!!  [#permalink]

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29 Mar 2013, 03:30
6
15
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.

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Re: New Set: Number Properties!!!   [#permalink] 29 Mar 2013, 03:30

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# New Set: Number Properties!!!

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