You are wrong. This is addressed several times on previous pages. I'll try again: if x = -1, then \(2\sqrt{x^2}=2=prime\). Also, \(\sqrt{x^2}\) cannot be negative for any value of x.
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Oh I can see the mistake now. Thanks.

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Hi Bunuel,

Why Can't a be 1 in this case? This will imply that they are not consecutive integers.

There is no condition defined for a in these 2 statements.

The median of 1, 1 and 2 is also 1.

If \(0 \lt x \lt y\) and \(x\) and \(y\) are consecutive perfect squares, what is the remainder when \(y\) is divided by \(x\)?

Notice that since \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers.

(1) Both \(x\) and \(y\) have 3 positive factors. This statement implies that \(x=(prime_1)^2\) and \(y=(prime_2)^2\). From above we have that \(\sqrt{x}=prime_1\) and \(\sqrt{y}=prime_2\) are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, \(x=(prime_1)^2=4\) and \(y=(prime_2)^2=9\). The remainder when 9 is divided by 4 is 1. Sufficient.

(2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. The same here: \(\sqrt{x}=2\) and \(\sqrt{y}=3\). Sufficient.


Answer: D[/quote]

Bunuel I could not understand this. (1) states, both are having 3 positive factors. The consecutive prime numbers which are 2 & 3, both are having 2factors only, i.e 1 and the number itself. Moreover, Prime Numbers does not have 3 factors. Pls tell, where am I getting wrong
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Bunuel I could not understand this. (1) states, both are having 3 positive factors. The consecutive prime numbers which are 2 & 3, both are having 2factors only, i.e 1 and the number itself. Moreover, Prime Numbers does not have 3 factors. Pls tell, where am I getting wrong[/quote]

Primes have 2 factors, yes, but the square of primes have 3 factors. So, from (1) \(x=(prime_1)^2\) and \(y=(prime_2)^2\).
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Bunuel

I'm still having difficulty understanding when to consider only the positive root when GMAT gives the even square root. It seems like when the value is defined that this rule will apply (e.g. sqrt(25) is 5, not -5); when a variable is provided instead, the answer is the abs(x), and 'x' itself can be +/-.

Can you help me better understand? I only considered the pos. root in this problem and got the wrong answer as a result.

Thanks!

Perfect. Thanks Bunuel.

Can't find the solution to Question 2


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.



As per me the ans is D

As each statement along with the given info says that the number is 2

Can an expert confirm is the answer is correct? The link to the solution is not working

Here is the solution: https://gmatclub.com/forum/new-set-numb ... l#p1205355
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can't find the solution to question 8

After question 7 I can directly see solution to question 9

Bunuel pls help

Bunuel, a question about notation. In the first question, the statement says |23x|. What does this mean? Are we to assume that x is just a placeholder here rather than | 23 x (x) |? If so, can't x be prime integers rather than 1 or -1? The statement is still insufficient b/c, say, 233 and 239 are both prime.

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number
(2) 2x2‾‾‾√2x2 is a prime number.
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23x means 23*x. If 23x were a three-digit number, that would be mentioned explicitly.
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Hi Bunuel, I am sorry but I don't quit get the explanation of (1), even if it is odd, why can't x be a square of an odd integer? Also, why do we need to add perfect square in this case?

Thanks in advance!

Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: 36=2^2*3^2, powers of prime factors 2 and 3 are even.

From (1) we get that x=2*3*(2q+1). Since 2q + 1 is odd then 2 in x will be in power of 1 (odd). So, x cannot be a perfect square.
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1.
St1 -- x could be 1 or -1
St2 -- again x could be 1 or -1

Answer - E

2. As n has only 2 positive factors, n must be a prime number as it has 1 and itself as its factors.
St1 -- n/2 is one of the factors of n, n can be only be 2. SUFFICIENT
St2 -- As LCM of n and n+10 is an even number. Apart from 2, all other prime numbers are even. So, if n is not 2, we will have n (Odd) and n+10 (another odd number), their LCM cannot be even. Hence n = 2.

Answer - D

3. x and y are consecutive perfect squares
St1 - Both x & y have 3 positive factors -> x & y are squares of prime numbers. If x = 2^2, and y =3^2, then remainder is 1, but if x=3^2 and y = 5^2, then remainder is 7. NOT SUFFICIENT
St2 - Again x & y are square of prime numbers, which is same as St1.

Answer - E

4. As each digit is a positive multiple of 4, it can be 4 or 8
St1 - We can have unit digit as 4, then tens and hundreds digit will be also 4 & 4 (number = 444), but, we can also have unit digit as 8, and tens and hundreds digit can be 4 & 8, 8 & 4 and 8 & 8. NOT SUFFICIENT
St2 - Sum of digits of multiple of 3 is also divisible by 3. Hence sum of digits of the number should not be divisible by 3. So, 444 and 888 are out. We can have any combination of 8,8,4 or 4,4,8 . NOT SUFFICIENT

Combining two statements, we are still left with 848 and 488. Hence E

5.
St1 -- As a<b<c, median of a!, b! & c! will be b!. As b! is odd, we only have 0! and 1! as odd number (i.e. 1). So b can be 0 or 1, but if b will be 0, then a will be -ve integer, and we do not have factorial of -ve number defines, thus a = 0, and b= 1, but c can be any integer. NOT SUFFICIENT
St2 -- c! is a prime number, thus c=2, as 2! = 2, and that is the only prime possible for factorials. but, a & b can be any integer.

Combining, we have a=0, b=1 and c=2. Hence C.

6.
St1 -- we can have {2,3,-2} or {-2,-3,-2}. NOT SUFFICIENT
St2 -- We can have {-3,-2,-1} or {1,2,3}. NOT SUFFICIENT

Combining, as we cannot have all elements +ve due to St1, and St2 restricts all elements to be of same sign. Thus, all elements will be -ve. SUFFICIENT

7.
St1 -- x = 12k + 6 = 6(2k+1), it will be a square number if 2k+1 is an odd power of 6, as 2k+1 will always be a odd number. x is not a square of an integer. SUFFICIENT
St2 -- x = 14k + 2 = 2(7k+1). For k=1, x is a square, but for k=2, it is not. NOT SUFFICIENT

Hence, A

8.
St1 -- That is already given in the question, no additional info provided. NOT SUFFICIENT
St2 -- A terminating decimal is one, which only has 2 and 5 or their multiple in denominator. So, all of them can be 1/3, thus median will be 1/3 > 1/5. Also, all elements can be 1/7, thus median will be 1/7<1/5 NOT SUFFICIENT

Combination, has same result as St2, cause St1 provided no additional info. Hence E

9.
St1 - As ab=2, a & b can be both +ve or both -ve. If both are -ve, we will have [a] or [b] to be <0, thus [a] + [b] cannot be 1. If both are +ve, then for [a] + [b] = 1, we need to have one of them as 0, and other 1, but if one of the number is less than 1 to have [] as 0, we need to have other number >2 to have ab=2, thus [a]+[b] != 1. SUFFICIENT
St2 -- 0<a<b<2, we can have multiple values for a &b, which can provide [a]+[b] as 0 or 1 or 2. NOT SUFFICIENT.
Hence A

10.
St1 -- if 9 is not a factor of N, x can be 1 only. If x=1, then y=5 to have 12 positive factors. SUFFICIENT
St2 -- if 125 is a factor of N, then y>=3, we can have y=3, x=2, or y=5, x=1. NOT SUFFICIENT
Hence A

11.
St1 -- |x - 2| < 2 - y. As |x-2| >=0, thus 2-y>0 -> y<2 -> y=1 (+ve integer). If y=1, then |x-2| < 1 -> 1< x 3 -> x=2 9 (prime number). SUFFICIENT
St2 -- x+y-3 = |1-y|. For y<=1 -> x+y-3 = 1-y -> x+2y=4. As x & y are +ve integers, only possible solution is x=2 & y=1. For y>1 -> x+y-3 = y-1 -> x=2. SUFFICIENT.
Hence C

I read on gmat club that the value of square root of x is considered positive value only (We consider + or - for absolute values only). So, in statement 2 why are we considering square root of x square as +1 or -1?

Hi Bunnel,

could you please tell how the list of 10 terms will look like with product of any two term resulting to terminating decimal ? can we have all the numbers in list as 1/2 ?

1) can my list look like : 1/2,1/2, 1/2.... as per me this also satisfy the statement 2 and median is equal to 0.5 or can list have all the term as 1/5 , in that median will be 0.2?




8. List A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the list less than 1/5?

(2) The product of any two terms of the list is a terminating decimal. This statement implies that the list must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

i didn't understood this statement If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out. How can it be -2 if i consider a = -1 and b = -2 then the sum comes to -3 and since both negative are integer they will remain unchanged

What I meant was that generally if a and b are negative, then the maximum value of [a] + [b] is -1+(-1)=-2. But you cannot get -2, if ab=2.

For example, [-0.5] + [-0.5] = -1 + (-1) = -2.
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[quote="Bunuel"]SOLUTIONS:

1. If x is an integer, what is the value of x?

(1) \(|23x|\) is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

(2) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Answer: E

I have a doubt. In the second statement, no mod sign is given and it says that the result is a prime number. Since prime numbers are not negative, isn't the answer supposed to be 1 and thus B???

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