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25 Mar 2013, 11:02
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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.
(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.
25 Mar 2013, 18:13
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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
(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
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
(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
