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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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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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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 = |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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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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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 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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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

(1) When x is divided by 12 the remainder is 6

when x is divided by 12 the remainder is 6 --> x = 12q + 6 = 3*2*(2q+1) ; So the answer is NO, x cannot be the square of any integer

For example :
q = 1 : x = 18
q=2 : x = 30 ...

(1) SUFFICIENT

(2) When x is divided by 14 the remainder is 2

When x is divided by 14 the remainder is 2 --> x = 14p + 2 = 2*(7p+1)

For example :
p = 1 --> x = 16 = 4^2
p = 2 --> x = 30

So, we dont know --> (2) NOT SUFFICIENT

Answer : A
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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?

(1) 9 is NOT a factor of N
(2) 125 is a factor of N

(x,y) two positive integers and N has 12 postive factors --> (x+1)(y+1)= 12

x + 1 = 2 and y + 1 = 6 --> (x,y) = (1,5)
x + 1 = 6 and y + 1 = 2 --> (x,y) = (5,1)
x + 1 = 3 and y + 1 = 4 --> (x,y) = (2,3)
x + 1 = 4 and y + 1 = 3 --> (x,y) = (3,2)

(1) 9 is NOT a factor of N

Since x is a positive integer and 9 is not a factor of N, then dicard x=2 and x=3 and x=5 --> x=1

x=1 --> y=5 --> N = 3^x*5^y = 3^1 * 5^5 --> Statement 1 ALONE SUFFICENT

(2) 125 is a factor of N

we know that : 5^3 = 125 , hence discard all possible values less than 3 for y --> two possibilties remain (x,y) = (1,5) and (x,y) = (2,3) --> Statement 2 ALONE INSUFFICIENT

Answer : A
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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 not prime
If x=2 --> x is prime

If x>2 --> x - 2 < 2 - y
In this case :
if y<2 --> y = 1 -> x < 3 and because x>2 so this can't be true --> y must be greater than or equalt to 2
if y = 2 -> x - 2 < 0 -> x<2 so x = 1 and thus x is not prime
if y>2 hence 2 - y < 0 -> x - 2 < 2 - y < 0 -> x - 2 < 0 -> x < 2 -> x = 1 -> thus x is not 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
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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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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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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

Answer: D
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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


Answer: E
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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

Answer: C
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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

Answer: A
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