New Set: Number Properties!!!

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

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

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.

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.

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

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

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

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

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

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

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

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

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

\(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;
Similarly \(\sqrt{\frac{1}{16}} = \frac{1}{4}\), NOT +1/4 or -1/4.


Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

If we knew for sure that the set is {1, 1, c!}, then yes we would have a definite NO answer to the question. But {1, 1, c!} is not possible since a < b < c implies that the integers in the set are distinct.

(1) implies that the set is {0, 1, c!} ({0!=1, 1!=1, c!}) and if c=2, then the answer would be YES, but if c=3, then the answer would be NO.

Hope it's clear.

Dear Bunuel,
First of all,i would like to thank you for all your sincere efforts that help innumerable gMAT takers like me.

In the explanation provided for statement 2,i did not quite understand how all the numbers of the set would have the same sign,if the smallest and largest terms have the same sign.The set contains>2 terms and from statement,we can comment only about 2 terms.Can you help me understand?

Thank you,
Fido

In GMAT, factors are always positive.

On the other hand, multiples can be positive as well as negative.