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New Set: Number Properties!!!

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New post 27 Feb 2019, 01:57
Hi Bunuel,

In 2) it says 2√x^2 is a prime number. Does that not mean that x is a 1 since -2 cannot be a prime number. I think answer is B
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New post 27 Feb 2019, 02:01
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New post 20 Mar 2019, 18:14
caioguima wrote:
8] Since we have an even number of elements in the set, the median will be the average of the two central numbers. Therefore, the median will be of type: (1/p + 1/q)/2 = (q+p)/(2.p.q), where p and q are primes.

(1) The reciprocal of the median, X = 2pq/(p+q) , is prime. So, pq/(p+q) = 1 (let me know if this passage was not trivial for you). Therefore: pq = p + q , which means p = q = 2. Since 1/2 is also the largest possibility of an element for S, we must have all numbers of the set equal to 1/2. Therefore, Sufficient

(2) The only terminating decimals of type 1/n, where n is a prime number are: 1/2 and 1/5. Set S has to have only elements equal to 1/2 and 1/5, which means the median can be equal to 1/2, 1/5 or something between 1/2 and 1/5. Therefore, the median is always greater than 1/5. Sufficient

Answer: D

if all the numbers were 1/5 , the median would have been (1/5 + 1/5)/2= 1/5.
the reciprocal of 1/5 is 5 a prime number. so we can have all 1/2 or all 1/5 . in both the cases the numbers are not less than 1/5.Please correct me where I am going wrong. option A is sufficient.
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New post 02 Apr 2019, 22:20
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 Bunuel,
Agree with the analysis about (2) , max and min numbers shall have the same sign. However, for case of positive signs, the max and min can only be 2 and 1, with information of set consists of more than 2 integers, we can eliminate the case of both are positive signs, thus the sets have all negative integers. Ans : B

What do you think about this, is there any flaw in my thinking?
Thanks!
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New post 02 Apr 2019, 22:58
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greenbear 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 Bunuel,
Agree with the analysis about (2) , max and min numbers shall have the same sign. However, for case of positive signs, the max and min can only be 2 and 1, with information of set consists of more than 2 integers, we can eliminate the case of both are positive signs, thus the sets have all negative integers. Ans : B

What do you think about this, is there any flaw in my thinking?
Thanks!


Cannot a set be say {1, 2, 3, 4, 5, 7} --> the product of the smallest and largest = 1*7 = 7 = prime.
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New post 12 Apr 2019, 06:48
Bunuel wrote:
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.



With respect to this question, are we making an assumption that n doesn't have any negative factors? The question stem is silent on the negative factors of n.
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New post 12 Apr 2019, 07:06
tejasvis7 wrote:
Bunuel wrote:
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.



With respect to this question, are we making an assumption that n doesn't have any negative factors? The question stem is silent on the negative factors of n.


Factors (at least on GMAT) are positive divisors.
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New post 04 Jul 2019, 21:54
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.


Hi Bunuel ,
From S2 we get c=2 and we know a<b<c and that a,b,c are non negetive int. So, it's quite evident theat a,b have to be 0,1 ? Hence, sufficient? Am i missing something here ?
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New post 04 Jul 2019, 22:05
Karmesh 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.


Hi Bunuel ,
From S2 we get c=2 and we know a<b<c and that a,b,c are non negetive int. So, it's quite evident theat a,b have to be 0,1 ? Hence, sufficient? Am i missing something here ?


How do you know when considering (2) that a, b, and c are non-negative?
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New post 04 Jul 2019, 22:11
Bunuel wrote:
Karmesh 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.


Hi Bunuel ,
From S2 we get c=2 and we know a<b<c and that a,b,c are non negetive int. So, it's quite evident theat a,b have to be 0,1 ? Hence, sufficient? Am i missing something here ?


How do you know when considering (2) that a, b, and c are non-negative?


As you rightly pointed out - The factorial of a negative number is undefined.
We have a set [a!,b!,c!]. Now if c! is 2. Then that leaves a,b with 0,1.
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New post 04 Jul 2019, 22:14
Karmesh wrote:
Hi Bunuel ,
From S2 we get c=2 and we know a<b<c and that a,b,c are non negetive int. So, it's quite evident theat a,b have to be 0,1 ? Hence, sufficient? Am i missing something here ?


How do you know when considering (2) that a, b, and c are non-negative?[/quote]

As you rightly pointed out - The factorial of a negative number is undefined.
We have a set [a!,b!,c!]. Now if c! is 2. Then that leaves a,b with 0,1.[/quote]

My question still remains. How do you know that when considering (2)??? You cannot use info from (1), when considering (2).
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New post 04 Jul 2019, 23:14
Bunuel wrote:
Karmesh wrote:
Hi Bunuel ,
From S2 we get c=2 and we know a<b<c and that a,b,c are non negetive int. So, it's quite evident theat a,b have to be 0,1 ? Hence, sufficient? Am i missing something here ?


How do you know when considering (2) that a, b, and c are non-negative?


As you rightly pointed out - The factorial of a negative number is undefined.
We have a set [a!,b!,c!]. Now if c! is 2. Then that leaves a,b with 0,1.[/quote]

My question still remains. How do you know that when considering (2)??? You cannot use info from (1), when considering (2).[/quote]

@#$%! Dammm!!!! Sorry. In my head the factorials were a part of the original statement. Need to be more careful
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New post 18 Jul 2019, 02:16
Bunuel wrote:
3. If 0 < x < 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.



Hi There Bunel,

Was looking at this solution, the question says that the X and Y are consecutive perfect squares. One could interpret it to mean that they are perfect squares that follow each other ie 4, 9, 16, 25 etc, that is the perfect squares or consecutive primes, which should be different in meaning from consecutive integers. How come the solution assumes that they are consecutive integers? while solving it, I was able to tell that the numbers are prime, but thought the answer cannot be ascertained because it could be any primes, any of the sets that follow each other, so I chose E instead. Please clarify.
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New post 18 Jul 2019, 03:06
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 Bunel,

in the solution for statement 1, I need some clarity. The question says that the product of "any" 3 integers is negative. Does that not mean that any three integers selected and multiplied will "certainly" be negative? Bearing in mind that we were not told the number in the set, it just says the number is more than 2. So, any three integers selected could turn out to be positive, positive, positive. Therefore, in my opinion, the only way this statement will always be true is if all the items are negative, such that "any 3" will surely mean negative. Based on this, I'ld say 1 is sufficient. Please let me know where I got this wrong.

I don't have any issues with statement 2 not being sufficient.
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New post 18 Jul 2019, 03:39
[quote="Bunuel"]8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set 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.

Answer: B.

Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.



Hi Bunel,

For your solution in 1, if all the terms are 1/2, wouldn't the median be 1/2+1/2 divided by 2? since it says median, not mean. Same applies to 1/7, Such that the reciprocal in both cases cannot be prime.
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New post 18 Jul 2019, 03:58
Gigie wrote:
Bunuel wrote:
3. If 0 < x < 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.



Hi There Bunel,

Was looking at this solution, the question says that the X and Y are consecutive perfect squares. One could interpret it to mean that they are perfect squares that follow each other ie 4, 9, 16, 25 etc, that is the perfect squares or consecutive primes, which should be different in meaning from consecutive integers. How come the solution assumes that they are consecutive integers? while solving it, I was able to tell that the numbers are prime, but thought the answer cannot be ascertained because it could be any primes, any of the sets that follow each other, so I chose E instead. Please clarify.


Please read the highlighted part.

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

x and y are consecutive perfect squares, so x and y could be:
\(x = 1\) and \(y = 4\) --> \(\sqrt{x}=1\) and \(\sqrt{y}=2\), consecutive integers;
\(x = 4\) and \(y = 9\) --> \(\sqrt{x}=2\) and \(\sqrt{y}=3\), consecutive integers;
\(x = 9\) and \(y = 16\) --> \(\sqrt{x}=3\) and \(\sqrt{y}=4\), consecutive integers;
...
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New post 18 Jul 2019, 04:02
Gigie 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 Bunel,

in the solution for statement 1, I need some clarity. The question says that the product of "any" 3 integers is negative. Does that not mean that any three integers selected and multiplied will "certainly" be negative? Bearing in mind that we were not told the number in the set, it just says the number is more than 2. So, any three integers selected could turn out to be positive, positive, positive. Therefore, in my opinion, the only way this statement will always be true is if all the items are negative, such that "any 3" will surely mean negative. Based on this, I'ld say 1 is sufficient. Please let me know where I got this wrong.

I don't have any issues with statement 2 not being sufficient.


Just to clarify: all solutions here are correct 100%.

As it's explained in the solution, the product of any three integers in the set to be negative, we can have TWO cases:
a. the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}.
b. the set consists of more than 3 terms, then the set can only have negative numbers.
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New post 18 Jul 2019, 04:04
Gigie wrote:
Bunuel wrote:
8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set 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.

Answer: B.

Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.



Hi Bunel,

For your solution in 1, if all the terms are 1/2, wouldn't the median be 1/2+1/2 divided by 2? since it says median, not mean. Same applies to 1/7, Such that the reciprocal in both cases cannot be prime.


1/2+1/2 divided by 2 is 1/2. It's reciprocal is 2, which is a prime.
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New post Updated on: 18 Jul 2019, 07:47
In reference to Q1: In second statement although √x² can have value x=1 and x=-1, yet the 2√x² will be prime only when it's value is positive; There are no negative primes. Therefore x can take just value=1.
In my view Option B is correct answer.

Posted from my mobile device

Originally posted by anuMishra on 18 Jul 2019, 07:40.
Last edited by anuMishra on 18 Jul 2019, 07:47, edited 1 time in total.
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New post 18 Jul 2019, 07:46
anuMishra wrote:
In second statement although √x² can have value x=1 and x=-1, yet the 2√x² will be prime only when it's value is positive; There are no negative primes. Therefore x can take just value=1.
In my view Option B is correct answer.

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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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