GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Oct 2019, 07:30 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  New Set: Number Properties!!!

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58434
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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 --> $$2\sqrt{x^2}=2\sqrt{1^2}=2\sqrt{1}=2$$.
_________________
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9704
Location: Pune, India
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1
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.

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/
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern  B
Joined: 11 Sep 2016
Posts: 1
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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?
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3074
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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. Thanks,
Saquib
Quant Expert
e-GMAT

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts  _________________
Senior Manager  G
Status: love the club...
Joined: 24 Mar 2015
Posts: 269
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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 ... Math Expert V
Joined: 02 Sep 2009
Posts: 58434
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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 ... 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.
_________________
Manager  B
Status: Preparing
Joined: 05 May 2016
Posts: 53
Location: India
Concentration: International Business, Finance
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58434
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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.
_________________
Intern  B
Joined: 30 Aug 2017
Posts: 13
Concentration: Real Estate, Operations
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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, [-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.

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?
Math Expert V
Joined: 02 Aug 2009
Posts: 7988
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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, [-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.

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?

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
_________________
Manager  B
Joined: 02 Jan 2017
Posts: 67
Location: Pakistan
Concentration: Finance, Technology
GMAT 1: 650 Q47 V34 GPA: 3.41
WE: Business Development (Accounting)
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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
Math Expert V
Joined: 02 Sep 2009
Posts: 58434
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.
_________________
Retired Moderator V
Joined: 28 Mar 2017
Posts: 1195
Location: India
GMAT 1: 730 Q49 V41 GPA: 4
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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
_________________
Math Expert V
Joined: 02 Aug 2009
Posts: 7988
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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...
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58434
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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.
_________________
Manager  B
Joined: 16 Jan 2018
Posts: 59
Concentration: Finance, Technology
GMAT 1: 600 Q40 V33 Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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)?
Math Expert V
Joined: 02 Sep 2009
Posts: 58434
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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:

$$\sqrt{x^2}=\sqrt{1^2}=\sqrt{1}=1$$;
$$\sqrt{x^2}=\sqrt{(-1)^2}=\sqrt{1}=1$$.
_________________
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9704
Location: Pune, India
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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/
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern  B
Joined: 03 May 2014
Posts: 47
Concentration: General Management, Other
GMAT 1: 460 Q42 V13 GPA: 3.83
WE: Consulting (Investment Banking)
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58434
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

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.

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, ...
_________________ Re: New Set: Number Properties!!!   [#permalink] 14 Apr 2018, 05:17

Go to page   Previous    1   2   3   4   5   6   7   8   9    Next  [ 165 posts ]

Display posts from previous: Sort by

New Set: Number Properties!!!

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  