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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.

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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.

(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.

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.

(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 _________________

KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you"

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 _________________

KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you"

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 _________________

KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you"

Last edited by Rock750 on 29 Mar 2013, 04:20, edited 1 time in total.

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