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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.

(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.

From what I heard from the experts, GMAT's general rule is that square root of a number is always positive on gmat. By this, i can get B as an answer. since root 1 will be 1 and not -1. Please explain.

5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

Note that: A. The factorial of a negative number is undefined. B. 0!=1. C. Only two factorials are odd: 0!=1 and 1!=1. D. Factorial of a number which is prime is 2!=2.

(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient.

(2) c! is a prime number. This implies that c=2. Not sufficient.

(1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient.

Answer: C.

Hi Bunuel, thanks for putting up such amazing questions. My Question: STatement 2: we know that c=2. Now the question states that a<b<c; (integers) and we know that the factorial of a negative number is not defined. thus, automatically a=0, b=1. so shouldnt the answer be B?

Hi Bunuel, thanks for putting up such amazing questions. My Question: STatement 2: we know that c=2. Now the question states that a<b<c; (integers) and we know that the factorial of a negative number is not defined. thus, automatically a=0, b=1. so shouldnt the answer be B?

Hey rachitasetiya,

The second statement only focuses on the factorial of c. So, we can definitely conclude that c = 2. But there is no information in the second statement, which can help us conclude that a,b and c are consecutive integers.

I guess when you said that " the factorial of a negative number is not defined", you also considered the information given in the first statement, to draw your conclusion.

Please remember, in DS, when we are solving a question using a particular statement we should focus only on the information given in that statement independently.

Hence, from the second statement, we get to know that c = 2 and there is nothing mentioned about a and b. Hence the second statement is not sufficient to answer the question.

It is only by combining, you can conclude that a = 0, b= 1 and c =2.

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The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square).

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.

(1) When x is divided by 12 the remainder is 6. Given that \(x=12q+6=6(2q+1)=2*3*(2q+1)\). Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient.

(2) When x is divided by 14 the remainder is 2. Given that \(x=14p+2\). So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient.

Answer: A.

hi

2 * 3 * (2q + 1) obviously, (2q + 1) is an odd number, however, since (2q + 1) is an odd number, the power of 2 in x will be odd... maybe it is very obvious to you, but I am totally stumped here... please help me understand it ...

The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square).

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.

(1) When x is divided by 12 the remainder is 6. Given that \(x=12q+6=6(2q+1)=2*3*(2q+1)\). Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient.

(2) When x is divided by 14 the remainder is 2. Given that \(x=14p+2\). So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient.

Answer: A.

hi

2 * 3 * (2q + 1) obviously, (2q + 1) is an odd number, however, since (2q + 1) is an odd number, the power of 2 in x will be odd... maybe it is very obvious to you, but I am totally stumped here... please help me understand it ...

thanks in advance

x=2*3*odd. Since an odd number does not have 2 in it, then 2 in x (highlighted) will be the only 2 in x. Since a perfect square always has even powers of its prime factors and 2 is in odd power (1), then x cannot be a perfect square.
_________________

The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square).

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.

(1) When x is divided by 12 the remainder is 6. Given that \(x=12q+6=6(2q+1)=2*3*(2q+1)\). Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient.

(2) When x is divided by 14 the remainder is 2. Given that \(x=14p+2\). So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient.

Answer: A.

hi

2 * 3 * (2q + 1) obviously, (2q + 1) is an odd number, however, since (2q + 1) is an odd number, the power of 2 in x will be odd... maybe it is very obvious to you, but I am totally stumped here... please help me understand it ...

thanks in advance

x=2*3*odd. Since an odd number does not have 2 in it, then 2 in x (highlighted) will be the only 2 in x. Since a perfect square always has even powers of its prime factors and 2 is in odd power (1), then x cannot be a perfect square.

2. If a positive integer n has exactly two positive factors what is the value of n?

Notice that, n has exactly two positive factors simply means that n is a prime number, so its factors are 1 and n itself.

(1) n/2 is one of the factors of n. Since n/2 cannot equal to n, then n/2=1, thus n=2. Sufficient.

(2) The lowest common multiple of n and n + 10 is an even number. If n is an odd prime, then n+10 is also odd. The LCM of two odd numbers cannot be even, therefore n is an even prime, so 2. Sufficient.

Answer: D.

hi man

okay with your manipulation ....but

can this problem be seen as under ....

n / n / 2 = integer

or n * 2/n = integer or 2n/n = integer

now, whichever prime "n" can take, the equation will hold true....

under circumstance as such, statement 1 is not sufficient .....

2. If a positive integer n has exactly two positive factors what is the value of n?

Notice that, n has exactly two positive factors simply means that n is a prime number, so its factors are 1 and n itself.

(1) n/2 is one of the factors of n. Since n/2 cannot equal to n, then n/2=1, thus n=2. Sufficient.

(2) The lowest common multiple of n and n + 10 is an even number. If n is an odd prime, then n+10 is also odd. The LCM of two odd numbers cannot be even, therefore n is an even prime, so 2. Sufficient.

Answer: D.

hi man

okay with your manipulation ....but

can this problem be seen as under ....

n / n / 2 = integer

or n * 2/n = integer or 2n/n = integer

now, whichever prime "n" can take, the equation will hold true....

under circumstance as such, statement 1 is not sufficient .....

please say to me what is wrong here...

thanks in advance ...

n divided by n/2 gives 2 irrespective of n, so the way you are doing cannot help to get sufficiency or insufficiency of the statement.
_________________

2. If a positive integer n has exactly two positive factors what is the value of n?

Notice that, n has exactly two positive factors simply means that n is a prime number, so its factors are 1 and n itself.

(1) n/2 is one of the factors of n. Since n/2 cannot equal to n, then n/2=1, thus n=2. Sufficient.

(2) The lowest common multiple of n and n + 10 is an even number. If n is an odd prime, then n+10 is also odd. The LCM of two odd numbers cannot be even, therefore n is an even prime, so 2. Sufficient.

Answer: D.

hi man

okay with your manipulation ....but

can this problem be seen as under ....

n / n / 2 = integer

or n * 2/n = integer or 2n/n = integer

now, whichever prime "n" can take, the equation will hold true....

under circumstance as such, statement 1 is not sufficient .....

please say to me what is wrong here...

thanks in advance ...

n divided by n/2 gives 2 irrespective of n, so the way you are doing cannot help to get sufficiency or insufficiency of the statement.

11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 2 (if y is more than or equal to 2, then \(y-2\leq{0}\) and it cannot be greater than |x - 2|). Next, since given that y is a positive integer, then y=1.

So, we have that: \(|x - 2| < 1\), which implies that \(-1 < x-2 < 1\), or \(1 < x < 3\), thus \(x=2=prime\). Sufficient.

(2) x + y - 3 = |1-y|. Since y is a positive integer, then \(1-y\leq{0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

Answer: D.

Hi Bunuel

Can you please elaborate the solution of this one? I did it with critical point method and ended with 2 equations in case of x>2(x+y=4) and x<2(x+y=0) and 2 equations with x<1(x+2y=4) and x>1(x=2). How to proceed further?

11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 2 (if y is more than or equal to 2, then \(y-2\leq{0}\) and it cannot be greater than |x - 2|). Next, since given that y is a positive integer, then y=1.

So, we have that: \(|x - 2| < 1\), which implies that \(-1 < x-2 < 1\), or \(1 < x < 3\), thus \(x=2=prime\). Sufficient.

(2) x + y - 3 = |1-y|. Since y is a positive integer, then \(1-y\leq{0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

Answer: D.

Hi Bunuel

Can you please elaborate the solution of this one? I did it with critical point method and ended with 2 equations in case of x>2(x+y=4) and x<2(x+y=0) and 2 equations with x<1(x+2y=4) and x>1(x=2). How to proceed further?

For (1): In \(|x - 2| \lt 2 - y\), the left hand side of the inequality (|x - 2|) is an absolute value. An absolute value of a number cannot be negative, so the left hand side of the inequality (|x - 2|) is not negative, which means that the right hand side of the inequality, which is greater than the left hand side, also cannot be negative. This gives \(0 \lt 2 - y\).

For (2): We know from the stem that y is a positive integer. Since y is a positive integer (1, 2, 3, ...), then 1 - y = 1 - positive = non-positive, which measn that |1-y|=-(1-y). Therefore, for (2) we'd have \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

Answer: A.

Can someone explain the reasoning behind "If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b"? how did you arrive at this maximum value? Thanks in advance.

9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

Answer: A.

Can someone explain the reasoning behind "If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b"? how did you arrive at this maximum value? Thanks in advance.

Hi To get the max value, take a and b as the smallest possible negative values.. So a =b= -0.00000000000....1 Least INTEGER for both a and b will be -1.. Thus [a]+[b]=-1+(-1)=-2 So -2 is the max value of a and b are NEGATIVE numbers.

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