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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
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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
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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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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
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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. 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
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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.

(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
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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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Last edited by Rock750 on 29 Mar 2013, 03:20, edited 1 time in total.

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