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Bunuel
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.
Show more


Hi Bunuel,

Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = -2 for any negative a and b?

i took the case of a=-2 and b=-1 and hence ab=2. In this case i am getting [a]+[b] = (-2) + (-1) = -3

Let me know if i am missing something.
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Bunuel
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.


Hi Bunuel,

Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = -2 for any negative a and b?

i took the case of a=-2 and b=-1 and hence ab=2. In this case i am getting [a]+[b] = (-2) + (-1) = -3

Let me know if i am missing something.
Show more

"If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b." Since, -3<-2, then this statement holds for this set of numbers too.

Does this make sense?
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6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.
Show more


Dear Bunuel,
First of all,i would like to thank you for all your sincere efforts that help innumerable gMAT takers like me.

In the explanation provided for statement 2,i did not quite understand how all the numbers of the set would have the same sign,if the smallest and largest terms have the same sign.The set contains>2 terms and from statement,we can comment only about 2 terms.Can you help me understand?

Thank you,
Fido
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Quote:
6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.


Dear Bunuel,
First of all,i would like to thank you for all your sincere efforts that help innumerable gMAT takers like me.

In the explanation provided for statement 2,i did not quite understand how all the numbers of the set would have the same sign,if the smallest and largest terms have the same sign.The set contains>2 terms and from statement,we can comment only about 2 terms.Can you help me understand?

Thank you,
Fido
Show more

(2) says: The product of the smallest and largest integers in the set is a prime number.

Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign.

So either: smallest * greatest = negative * negative and in this case as both the smallest and the greatest are negative then ALL integers in the list are negative OR smallest * greatest = positive * positive and in this case as both the smallest and the greatest are positive then ALL integers in the list are positive.

Hope it's clear.
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Hi,
In question 3 above, it says that X and Y have 3 postive factors. But it does not say only positive factors. Could the number not have other factors such as -3,-1,-9 apart from 3,1,9? What am I missing out?
It will be great if someone can point what I am doing incorrectly.
Thanks!
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Hi,
In question 3 above, it says that X and Y have 3 postive factors. But it does not say only positive factors. Could the number not have other factors such as -3,-1,-9 apart from 3,1,9? What am I missing out?
It will be great if someone can point what I am doing incorrectly.
Thanks!
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In GMAT, factors are always positive.

On the other hand, multiples can be positive as well as negative.
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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.

Hi Bunuel
sorry to ask a silly doubt,but prime numbers are always positive so can't we discard -1 ?
Thanks
Anupama
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(2) \(2\sqrt{x^2}\) is a prime number, not x. For \(2\sqrt{x^2}\) to a prime number, x must be 1 or -1. So, we have two possible values of x, which makes the statement insufficient.
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Bunuel
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.
Show more

Hi Bunuel,

What the best way to approach Absolute questions.

My current logic is that |x| = |-x|

and using the above I try to get my range/value of x

|x-2| < 2-y

x-2 < 2-y or -x+2 < 2-y

x+y < 4 or x-y > 0

Is this the correct way to approach it ?

Cheers
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Bunuel
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,

What the best way to approach Absolute questions.

My current logic is that |x| = |-x|

and using the above I try to get my range/value of x

|x-2| < 2-y

x-2 < 2-y or -x+2 < 2-y

x+y < 4 or x-y > 0

Is this the correct way to approach it ?

Cheers
Show more

Yes, when x >=2, then |x - 2| = x - 2, so in this case we get x - 2 < 2 - y and when x < 2, then |x - 2| = -(x - 2), so in this case we get -(x - 2) < 2 - y. But this does not help us to answer the question. To do so, you should approach the question shown in my solution.

As for the best way solve modulus questions - it depends on a question, multiple approaches are possible. Check the links below for practice.

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hope it helps.
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Bunuel
6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.
Show more


Dear Bunuel,
Thanks for the excellent set of problems. However, I have one query with regard the above solution. I have highlighted the text in yellow - with regard the statement (2), how can we assume that all integers are either positive or negative. It is possible that only the smallest and the largest integers have similar signs and that the other integers have different signs; Or am i misunderstanding something here? Please confirm.
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Bunuel
6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.


Dear Bunuel,
Thanks for the excellent set of problems. However, I have one query with regard the above solution. I have highlighted the text in yellow - with regard the statement (2), how can we assume that all integers are either positive or negative. It is possible that only the smallest and the largest integers have similar signs and that the other integers have different signs; Or am i misunderstanding something here? Please confirm.
Show more

Hi gmatriser,

Statement-II specifically talks about the smallest and the largest integer. Since their product is positive, there may be two possibilities:

1. Smallest integer and largest integer both are negative - Since the largest integer is negative there can't be a number greater than the largest integer. As any positive integer is greater than a negative integer, there can't be a positive integer in the set.

2. Smallest integer and largest integer both are positive - Similarly as the smallest integer is positive, there can't be a negative integer in the set because the negative integer then will become the smallest integer.

Hence the set consists of only negative or positive integers.

Hope it's clear :)

Regards
Harsh
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Bunuel
6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.
Show more
Hi:

would (1)+(2) still not give different solutions i.e. (negative, negative, negative) and (positive, negative, positive)?
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Bunuel
6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.
Hi:

would (1)+(2) still not give different solutions i.e. (negative, negative, negative) and (positive, negative, positive)?
Show more

In case of (positive, negative, positive), the smallest number there would be negative and the largest will be positive --> their product will be negative, so it cannot be a prime, which violates the second statement. By the way this is explained several times in this thread.
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Bunuel
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.
Show more

Bunnel, are we not supposed to consider fractions for this? It's nowhere mentioned a and b are integers.

Similarly for question 6, It says "Are all the numbers in set S negative"?? However it's modified in the explanation post as "Are all the integers in set S negative".
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Bunuel
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. Ifa=1/2and 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.

Bunnel, are we not supposed to consider fractions for this? It's nowhere mentioned a and b are integers.

Similarly for question 6, It says "Are all the numbers in set S negative"?? However it's modified in the explanation post as "Are all the integers in set S negative".
Show more

a and b can be negative (for example check highlighted part) but [a] and [b] cannot since [] is a function which rounds DOWN a number to the nearest integer.
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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.
Show more

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?
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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?
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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.

I hope the above explanation is clear. :)


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7. Is x the square of an integer?

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.
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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 ... :cry:

thanks in advance
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Bunuel
7. Is x the square of an integer?

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 ... :cry:

thanks in advance
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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.
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Bunuel
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.
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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?
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