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New Set: Number Properties!!!

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

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New post 11 Apr 2017, 07:39
sindhugclub 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.


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.


Yes, the square root cannot give negative result but this has little to do with the problem.

Both x=1 and x=-1 when substituted into \(2\sqrt{x^2}\) give 2.

x = -1 --> \(2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1}=2\).

x = 1 --> \(2\sqrt{x^2}=2\sqrt{1^2}=2\sqrt{1}=2\).
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Re: New Set: Number Properties!!!  [#permalink]

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New post 27 Apr 2017, 20:54
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, 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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Re: New Set: Number Properties!!!  [#permalink]

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New post 27 Apr 2017, 23:00
rachitasetiya wrote:

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.

I hope the above explanation is clear. :)


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

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New post 23 Aug 2017, 09:03
Bunuel wrote:
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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New post 23 Aug 2017, 11:26
gmatcracker2017 wrote:
Bunuel wrote:
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


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

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New post 27 Aug 2017, 05:39
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

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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New post 27 Aug 2017, 06:17
nishantt7 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

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.

Hope it's clear.
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Re: New Set: Number Properties!!!  [#permalink]

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New post 25 Sep 2017, 10:59
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.


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.
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New post 25 Sep 2017, 17:47
1
Ace800 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. 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.

Hope it helps
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New Set: Number Properties!!!  [#permalink]

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New post 27 Dec 2017, 16:39
VeritasPrepKarishma wrote:
sindhugclub 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.


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.


In addition to what Bunuel said, check out this post: https://www.veritasprep.com/blog/2016/0 ... oots-gmat/



Hi
Could you explain why this would be wrong here?

statement 2 tells us \(2 \sqrt{x^2}\) is prime number

the expression \(2 * [x^2 ] ^ \frac{1}{2}\) becomes \(2 x\)[ square canceling out with the square root ]

if the expression summarises to \(2x\)then statement 2 becomes\(2x\) is a prime number and \(x =1\)is the only value which satisfies.

If this is wrong, could you explain the why? It will be really helpful.
Thankyou
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New post 27 Dec 2017, 21:13
mtk10 wrote:
VeritasPrepKarishma wrote:
sindhugclub wrote:

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.


In addition to what Bunuel said, check out this post: https://www.veritasprep.com/blog/2016/0 ... oots-gmat/



Hi
Could you explain why this would be wrong here?

statement 2 tells us \(2 \sqrt{x^2}\) is prime number

the expression \(2 * [x^2 ] ^ \frac{1}{2}\) becomes \(2 x\)[ square canceling out with the square root ]

if the expression summarises to \(2x\)then statement 2 becomes\(2x\) is a prime number and \(x =1\)is the only value which satisfies.

If this is wrong, could you explain the why? It will be really helpful.
Thankyou


You could see that this is wrong by considering the values given in the solution: both x=1 or x=-1 make \(2\sqrt{x^2}\) prime number.

That's because \(\sqrt{x^2}=|x|\) NOT x.

The point here is that since square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

Hope it's clear.
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Re: New Set: Number Properties!!!  [#permalink]

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New post 12 Jan 2018, 02:05
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.

Questions testing this concept:
http://gmatclub.com/forum/does-the-deci ... 89566.html
http://gmatclub.com/forum/any-decimal-t ... 01964.html
http://gmatclub.com/forum/if-a-b-c-d-an ... 25789.html
http://gmatclub.com/forum/700-question-94641.html
http://gmatclub.com/forum/is-r-s2-is-a- ... 91360.html
http://gmatclub.com/forum/pl-explain-89566.html
http://gmatclub.com/forum/which-of-the- ... 88937.html


Hi @Bunuel/ chetan2u,

The question is essentially asking whether "median of the set is less than 1/5?". So if we have median EQUALto 1/5 the answer CAN'T be B.

Please elucidate: what am I missing here? IMHO the answer should be "E".

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

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New post 12 Jan 2018, 03:42
gmatexam439 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.

Questions testing this concept:
http://gmatclub.com/forum/does-the-deci ... 89566.html
http://gmatclub.com/forum/any-decimal-t ... 01964.html
http://gmatclub.com/forum/if-a-b-c-d-an ... 25789.html
http://gmatclub.com/forum/700-question-94641.html
http://gmatclub.com/forum/is-r-s2-is-a- ... 91360.html
http://gmatclub.com/forum/pl-explain-89566.html
http://gmatclub.com/forum/which-of-the- ... 88937.html


Hi @Bunuel/ chetan2u,

The question is essentially asking whether "median of the set is less than 1/5?". So if we have median EQUALto 1/5 the answer CAN'T be B.

Please elucidate: what am I missing here? IMHO the answer should be "E".

Regards


Hi..
If you ask Q is 1/5 less than 1/5?... Ans will be NO
If you ask .. is 1/2 less than 1/5 .... Ans is again NO
So evwrytime and is NO..
Sufficient

But say the Q was - is x >1/5?
If 1/2 .Ans will be Yes..
If 1/5.. Ans will be NO
So insufficient...
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Re: New Set: Number Properties!!!  [#permalink]

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New post 12 Jan 2018, 03:51
gmatexam439 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.

Questions testing this concept:
http://gmatclub.com/forum/does-the-deci ... 89566.html
http://gmatclub.com/forum/any-decimal-t ... 01964.html
http://gmatclub.com/forum/if-a-b-c-d-an ... 25789.html
http://gmatclub.com/forum/700-question-94641.html
http://gmatclub.com/forum/is-r-s2-is-a- ... 91360.html
http://gmatclub.com/forum/pl-explain-89566.html
http://gmatclub.com/forum/which-of-the- ... 88937.html


Hi @Bunuel/ chetan2u,

The question is essentially asking whether "median of the set is less than 1/5?". So if we have median EQUALto 1/5 the answer CAN'T be B.

Please elucidate: what am I missing here? IMHO the answer should be "E".

Regards


For (2) all possible cases (including the median of 1/5) give a NO answer to the question. A definite No answer to the question means that the statement is sufficient.
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Re: New Set: Number Properties!!!  [#permalink]

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New post 27 Feb 2018, 19:55
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.


For II, square root sign is given, I thought we only take the +ve value when the sign is present in the stem. so, is the value not just x (and not + - x)?
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Re: New Set: Number Properties!!!  [#permalink]

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New post 27 Feb 2018, 21:07
krish76 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.


For II, square root sign is given, I thought we only take the +ve value when the sign is present in the stem. so, is the value not just x (and not + - x)?


The value of \(\sqrt{x^2}\) is still positive for both 1 and -1:

\(\sqrt{x^2}=\sqrt{1^2}=\sqrt{1}=1\);
\(\sqrt{x^2}=\sqrt{(-1)^2}=\sqrt{1}=1\).
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Re: New Set: Number Properties!!!  [#permalink]

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New post 28 Feb 2018, 03:35
krish76 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.


For II, square root sign is given, I thought we only take the +ve value when the sign is present in the stem. so, is the value not just x (and not + - x)?


Also, check this post: https://www.veritasprep.com/blog/2016/0 ... oots-gmat/
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Re: New Set: Number Properties!!!  [#permalink]

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New post 14 Apr 2018, 04:50
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 Bunuel
Thanks for explanation.
I have doubt. Some say consecutive prime numbers are 2 and 3 and some say consecutive prime numbers are 2, 3, 5, 7, 11, 13.
Could you clarify which one to consider if question asks consecutive prime number?
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Re: New Set: Number Properties!!!  [#permalink]

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New post 14 Apr 2018, 05:17
Andysingh 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 Bunuel
Thanks for explanation.
I have doubt. Some say consecutive prime numbers are 2 and 3 and some say consecutive prime numbers are 2, 3, 5, 7, 11, 13.
Could you clarify which one to consider if question asks consecutive prime number?


Consecutive primes are 2, 3, 5, 7, ...
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