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New Set: Number Properties!!!

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Re: New Set: Number Properties!!!  [#permalink]

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New post 13 Jul 2014, 08:20
Bunuel wrote:
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,

In (2), by definition negative Z's are not prime. So -1 ruled out. B then? Can you please clarify!
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New post 03 Sep 2014, 09:29
Hi,

Can you please suggest in question number 9 why can't we have values a = 0.1 and b = 20 so a*b will be 2 and equation won't be satisfied
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New post 03 Sep 2014, 09:44
ashish14 wrote:
Hi,

Can you please suggest in question number 9 why can't we have values a = 0.1 and b = 20 so a*b will be 2 and equation won't be satisfied


The point is that [a] + [b] = 1 won't be true for any a and b if ab = 2, so in all cases we have a NO answer to the question (not only for your case).
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Re: New Set: Number Properties!!!  [#permalink]

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New post 25 Sep 2014, 08:05
Bunuel wrote:
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
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New post 25 Sep 2014, 08:23
anupamadw wrote:
Bunuel wrote:
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


(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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Re: New Set: Number Properties!!!  [#permalink]

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New post 30 Nov 2014, 01:41
Bunuel wrote:
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
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Re: New Set: Number Properties!!!  [#permalink]

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New post 01 Dec 2014, 05:49
bhatiavai wrote:
Bunuel wrote:
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


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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Re: New Set: Number Properties!!!  [#permalink]

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New post 28 May 2015, 02:00
Bunuel wrote:
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.
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Re: New Set: Number Properties!!!  [#permalink]

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New post 28 May 2015, 03:37
1
gmatriser wrote:
Bunuel wrote:
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.


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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New post 20 Jan 2016, 20:06
Bunuel wrote:
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)?
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New post 20 Jan 2016, 21:42
WilDThiNg wrote:
Bunuel wrote:
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)?


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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Re: New Set: Number Properties!!!  [#permalink]

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New post 22 Aug 2016, 08:35
Bunuel wrote:
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.


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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New post 22 Aug 2016, 08:43
Argo wrote:
Bunuel wrote:
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".


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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Re: New Set: Number Properties!!!  [#permalink]

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New post 22 Aug 2016, 09:31
Bunuel wrote:
Argo wrote:
Bunuel wrote:
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".


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.


So, say a=6/5 and b=5/3, in that case the answer would be E. Correct me if I am wrong.
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New post 22 Aug 2016, 09:39
Argo wrote:
Bunuel wrote:
Argo wrote:
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".


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.


So, say a=6/5 and b=5/3, in that case the answer would be E. Correct me if I am wrong.


If a = 6/5 and b =5/3, then [a] + [b] = 1 + 1 = 2, which gives the same NO answer as in the solution.
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Re: New Set: Number Properties!!!  [#permalink]

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New post 17 Oct 2016, 09:44
Bunuel wrote:
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.


Statement 1: Implies b! is odd. The only positive odd factorial is for number 1. And it is mentioned in the question that a,b and c are consecutive. Therefore can only be 0,1, 2.
Statement 2: Implies that c! is prime which means that c can be 2! only. And it is mentioned in the question that a,b and c are consecutive. Therefore can only be 0,1, 2.

So shouldn't the answer be D? Please help out.
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Re: New Set: Number Properties!!!  [#permalink]

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New post 17 Oct 2016, 09:49
prvy wrote:
Bunuel wrote:
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.


Statement 1: Implies b! is odd. The only positive odd factorial is for number 1. And it is mentioned in the question that a,b and c are consecutive. Therefore can only be 0,1, 2.
Statement 2: Implies that c! is prime which means that c can be 2! only. And it is mentioned in the question that a,b and c are consecutive. Therefore can only be 0,1, 2.

So shouldn't the answer be D? Please help out.


For (1) note that 0! = 1.

For (2) c = 2 but we are NOT told that a, b, and c are consecutive integers. The question asks whether a, b, and are c consecutive integers.
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Re: New Set: Number Properties!!!  [#permalink]

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New post 11 Apr 2017, 07:29
Bunuel wrote:
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.


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.
Re: New Set: Number Properties!!! &nbs [#permalink] 11 Apr 2017, 07:29

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