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

 It is currently 21 Oct 2019, 01:09 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: 58413
New Set: Number Properties!!!  [#permalink]

Show Tags

55
227
The next set of medium/hard DS number properties questions. I'll post OA's with detailed explanations on Friday. Please, post your solutions along with the answers.

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number
(2) $$2\sqrt{x^2}$$ is a prime number.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205341

2. If a positive integer n has exactly two positive factors what is the value of n?

(1) n/2 is one of the factors of n
(2) The lowest common multiple of n and n + 10 is an even number.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205355

3. If 0 < x < y and x and y are consecutive perfect squares, what is the remainder when y is divided by x?

(1) Both x and y is have 3 positive factors.
(2) Both $$\sqrt{x}$$ and $$\sqrt{y}$$ are prime numbers

Solution: https://gmatclub.com/forum/new-set-numb ... l#p1209015

4. Each digit of the three-digit integer K is a positive multiple of 4, what is the value of K?

(1) The units digit of K is the least common multiple of the tens and hundreds digit of K
(2) K is NOT a multiple of 3.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205361

5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

(1) The median of {a!, b!, c!} is an odd number.
(2) c! is a prime number

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205364

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

(1) The product of any three integers in the list is negative
(2) The product of the smallest and largest integers in the list is a prime number.

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205373

7. Is x the square of an integer?

(1) When x is divided by 12 the remainder is 6
(2) When x is divided by 14 the remainder is 2

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205378

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
(2) The product of any two terms of the set is a terminating decimal

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205382

9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

(1) ab = 2
(2) 0 < a < b < 2

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205389

10. If N = 3^x*5^y, where x and y are positive integers, and N has 12 positive factors, what is the value of N?

(1) 9 is NOT a factor of N
(2) 125 is a factor of N

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205392

BONUS QUESTION:
11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y
(2) x + y - 3 = |1-y|

Solution: http://gmatclub.com/forum/new-set-numbe ... l#p1205398

Kudos points for each correct solution!!!
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58413
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

13
46
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:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html
_________________
Manager  Joined: 02 Jan 2013
Posts: 54
GMAT 1: 750 Q51 V40 GPA: 3.2
WE: Consulting (Consulting)
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

6
2
2] n has only two positive factorss, therefore n is prime.

(1) n/2 is one of the factors of n. If n is prime, and n/2 is a factor of n, then n = 2. Sufficient

(2) As Bunuel always says, the greatest rule about LCM is that for two numbers a,b, we have: a.b =LCM(a,b).GCF(a,b). If n is prime, let's suppose n>2 (in this case, n will have to be odd since it is prime). If n is odd, then n+10 is odd as well. From the rule we just mentioned:

n.(n+10) = odd*odd = odd = LCM.GCF -> Therefore, LCM(n,n+10) cannot be even (otherwise, the product would be even).

So, n has to be 2. Sufficient

General Discussion
Math Expert V
Joined: 02 Sep 2009
Posts: 58413
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1
1
Please suggest on what category would you like the next set to be. Thank you!
_________________
Board of Directors D
Joined: 01 Sep 2010
Posts: 3405
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1
word problems Thanks for the set
_________________
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1019
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1
If x is an integer, what is the value of x?

(1) |23x| is a prime number
Since 23 si prime,$$x$$can be $$+1$$ or $$-1$$
not sufficient

(2) 2\sqrt{x^2} is a prime number.
once again $$x$$ can be $$+1$$ or $$-1$$
not sufficient

And since 1)+2) provides no new info IMO E
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1019
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1
2. If a positive integer n has exactly two positive factors what is the value of n?

Number of factors of a number is $$a+1$$ where the number is $$n^a$$
And since n has 2 factors$$n$$ must be prime and$$>1$$

(1) n/2 is one of the factors of n
Only $$2$$ fits these conditions, so $$n=2$$
Sufficient

(2) The lowest common multiple of n and n + 10 is an even number.
Only $$2$$ fits these conditions, so $$n=2$$ once again. $$n$$ IMO is prime so the only prime that respect statement (2) is 2 because all other prime are odd, and odd+even = odd, so the LCM of an odd and an odd is odd in all cases except n=2

IMO D
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1019
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1
6. Set S consists of more than two integers. Are all the numbers in set S negative?

(1) The product of any three integers in the list is negative
Not sufficient
Example: $$S = {-1,3,5}$$
the product is always <0 but 2 numbers are positive
Example: $$S = {-1,-3,-5}$$
the product is always <0 and all numbers are negative

(2) The product of the smallest and largest integers in the list is a prime number.
Not sufficient
Example: $$S = {1,3,5}$$
$$1*5=5$$ prime but all positive
Example: S = $${-1,-3,-5}$$
$$-1*-5=5$$ prime but all negative

(1)+(2) Sufficient IMO C
Using statement 2 we know that all are positive or all are negative, using statement 1 we know that "at least" 1 is negative=> so all are negative
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1019
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

7. Is x the square of an integer?

(1) When x is divided by 12 the remainder is 6
(2) When x is divided by 14 the remainder is 2

A,B and D cannot be the answer clearly. What about C?

$$x=12q+6$$
$$x=14z+2$$

$$12q+6=14z+2$$
$$6-2=14z-12q$$
$$4=2(7z-6q)$$
$$2=7z-6q$$

$$z=2,q=2$$

$$x=12*2+6=30$$
$$x=14*2+2=30$$

so x is not the square of an integer.IMO C
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1019
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

2
9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

(1) ab = 2
(2) 0 < a < b < 2

The question can be seen as (given statement 2):
$$[a] + [b] = 1$$
case 1:$$0<(=)a<1$$ => $$[a]=0$$ and $$1(=)<b<2$$ => $$[b]=1$$ so $$[a] + [b] = 1$$
or the opposite
case 2: $$0<(=)b<1$$ => $$[b]=0$$ and $$1(=)<a<2$$ => $$[a]=1$$ so $$[a] + [b] = 1$$

But as ab=2 we know that one term is $$\frac{1}{2}$$ and the other is $$2$$
So we are in one of the two senarios above, IMO C
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1019
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1
10. If N = 3^x*5^y, where x and y are positive integers, and N has 12 positive factors, what is the value of N?

Number of factors of $$N = (x+1)(y+1) = 12$$
the combinations are
$$3*4$$ with $$x= 2$$ and $$y=3$$
$$6*2$$ with $$x=5$$ and$$y = 1$$
and the "other way round" of each one

(1) 9 is NOT a factor of N
So x must be 1, $$x=1$$
because $$(1+1)(y+1)=12$$
$$y=5$$
Sufficient

(2) 125 is a factor of N
So $$y>=3$$, y can be 3 or 5, NOT sufficient
$$y=3, (3+1)(x+1)=12, x=2$$
$$y=5, (5+1)(x+1)=12, x=1$$

IMO A
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Originally posted by Zarrolou on 25 Mar 2013, 07:49.
Last edited by Zarrolou on 25 Mar 2013, 08:57, edited 1 time in total.
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1019
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y
(2) x + y - 3 = |1-y|

This is a GOOD one. IMO C

(1) |x - 2| < 2 - y

$$x-2>0, x>2$$
case 1)$$x>2$$
$$x-2<2-y$$
$$x+y<4$$

case 2)$$0<x<=2$$ ( x is positive )
$$-x+2<2-y$$
$$x>y$$

NOT SUFFICIENT

(2) x + y - 3 = |1-y|

case 1)$$y>1$$
$$x+y-3=1-y$$
$$x+2y=4$$

case 2)$$0<y<=1$$ ( y is positive)
$$x+y-3=-1+y$$
$$x=2$$

NOT SUFFICIENT

Combining 1 and 2 we obtain that

------0------------1----------2----------------
------|~~~~~~x>y~~~~~~~|~~~x+y<4 for the first one
------|~~x=2~~~|~~~~~~x+2y=4~~~~for the second one

And combining all the cases together we obtain
1)$$0<x<=2$$ with $$0<y<=1$$
$$x=2$$ and $$x>y$$ so $$x=2$$ and $$y=1$$
2)$$0<x<=2$$ with $$y>1$$
$$x>y$$ and $$x+2y=4$$, given that x and y are positive $$x=2, y=1$$
3)$$x>2$$ with $$0<y<=1$$
$$x+y<4$$ and $$x=2$$ so $$x=2,y=1$$
4)x>2 with y>1
$$x+y<4$$ and $$x+2y=4$$, $$x=2,y=1$$
In each case $$x=2$$ so x is prime
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1019
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1
5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

(1) The median of {a!, b!, c!} is an odd number.
the median of three elements is the one in the middle, so b! is odd
there are only 2 cases in which n! is odd and are if n=1 or if n=0
so b is 0 , 1
Not Sufficient

(2) c! is a prime number
c can once again be 0,1 or in this case 2.
Not sufficient

-This is a weak passage, I don't know if I'm right-

n! is possible only for positive number so given that
a < b < c
c must be 2, b must be 1, and (because of my weak hypothesis a>=0) a must be 0

IMO C
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1019
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1
1
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
Not sufficient

(2) The product of any two terms of the set is a terminating decimal
Because $$\frac{1}{prime}$$ is not a terminating decimal, with the only exception of 1/4, 1/1, 1/5 and 1/2
any set made by these three CANNOT have a median < 1/5, it can be = 1/5 but NEVER <
Some examples:
A={1,1,1,1,1/5,1/5,1/5,1/5,1/5,1/5} the median is 1/5 = 1/5
A={1,1,1,1,1/2,1/2,1/2,1/2,1/2,1/2} the median is 1/2 > 1/5
SUFFICIENT

IMO B
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
Verbal Forum Moderator B
Joined: 10 Oct 2012
Posts: 590
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

2
11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y
(2) x + y - 3 = |1-y|

We know that x >0 and y>0 and they are integers.

From F.S 1, we have 2-y>=0 or y<=2. Thus y can only be 2 or 1. Now if y=2, we would have 0>some thing positive or 0>0(when x also equal to 2). Either case is not possible. Thus, y can only be 1. For y=1, we can only have x = 2. Which is prime. Sufficient.

From F.S 2, we have either y>1 or y<1. Now as y is a positive integer, y can't be less than 1.For y=1, we anyways have x=2(prime).In the first case, we have y>1--> x+y-3 = y-1 or x=2(prime). Thus, Sufficient.

D.
_________________
Manager  Joined: 26 Feb 2013
Posts: 51
Concentration: Strategy, General Management
GMAT 1: 660 Q50 V30 WE: Consulting (Telecommunications)
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1-B
2-A
3-E
4-E
5-B
6-C
7-A
8-D
9-B
10-A
11-D
Verbal Forum Moderator B
Joined: 10 Oct 2012
Posts: 590
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

2
7.Is x the square of an integer?

(1) When x is divided by 12 the remainder is 6
(2) When x is divided by 14 the remainder is 2

From F.S 1, we have x = 12q+6 --> 6(2q+1). For x to be a square of an integer, we should have 2q+1 of the form 6^pk^2, where both q,p and k are integers and p is odd. Now we know that 2q+1 is an odd number and 6^pk^2 is even. Thus they can never be equal and hence x can never be the square of an integer. Sufficient.

From F.S 2, we have x = 14q+2 --> 2(7q+1).Just as above, we should have 7q+1 = 2^pk^2. Now for q=1, k=2 and p=1, we have 8=8, thus x is the square of an integer. But for q=0, x is not. Insufficient.

Basically for the second fact statement, we can plug in easily. No need for the elaborate theory.

A.
_________________
Verbal Forum Moderator B
Joined: 10 Oct 2012
Posts: 590
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

2
5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

(1) The median of {a!, b!, c!} is an odd number
(2) c! is a prime number

From F.S 1, we have b! = odd, thus b can be 0 or 1.But, as factorial notation is only for positive integers, thus, if b=0, then a would become negative and thus b is only equal to 1.Now, a can only be 0 as we are given that a! exists. But nothing has been mentioned about c. All we know is that c>1 and an integer. Insufficient.

From F.S 2, we have c! is a prime number. Again, c has to be positive and c can only be 2.However, a and b can take any values, even negative. Insufficient.

Taking both together, we have a=0, b=1 and c=2. Sufficient.

C.
_________________
Manager  Status: Final Lap
Joined: 25 Oct 2012
Posts: 227
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

1
1. If x is an integer, what is the value of x?

(1) |23x| is a prime number
(2)$$2 \sqrt{x^2}$$ is a prime number.

(1) |23x| is a prime number

For $$x=1$$ $$|23x| = |23*1|$$ --> 23 is prime

For $$x=-1$$ $$|23x| = |23*(-1)|]$$ --> $$|-23| = 23$$ 23 is prime also

Thus, this holds true for two values of x and because of that, the value of x cannot be determined.

(1) INSUFFICIENT

(2)$$2 \sqrt{x^2}$$ is a prime number.
x=-1 --> $$2 \sqrt{x^2}$$ =2 prime
x=1 --> $$2 \sqrt{x^2}$$ =2 prime

Thus, x can take the value of either 1 or -1
(2) INSUFFICIENT

|23x| is a prime number AND $$2 \sqrt{x^2}$$ is a prime number.
For $$x=1$$ 23 is prime and 2 is prime
For $$x=-1$$ 23 is prime and 2 is prime
(1) +(2) INSUFFICIENT

_________________
KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you"
Manager  Status: Final Lap
Joined: 25 Oct 2012
Posts: 227
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)
Re: New Set: Number Properties!!!  [#permalink]

Show Tags

2
2. If a positive integer n has exactly two positive factors what is the value of n?

(1) n/2 is one of the factors of n
(2) The lowest common multiple of n and n + 10 is an even number.

n is a positive integer that has exactly two positive factors --> 1 must be one of its factor --> n is a prime number and not equal to 1 (because 1 has only one positive factors, itself)

So these two factors could be (1,2) or (1,3) or (1,5) ... and n could be 2,3,5 .......

(1) n/2 is one of the factors of n

n/2 is a factor of n --> n/2 is an integer and from the pairs (1,2), (1,3) ... only 2 is divisible by 2
Hence , n = 2 --> (1) SUFFICIENT

(2) The lowest common multiple of n and n + 10 is an even number.

LCM (n,n+10) = EVEN

If n = 2 then LCM(2,12) = 12, which is EVEN
If n = 3 then LCM (3,13) = 39, which is ODD
Except for n=2, Like n=3, n = 5,7,11 .... LCM (n,n+10) will be ALWAYS ODD.
Hence, n = 2 --> (2) SUFFICIENT

_________________
KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you" Re: New Set: Number Properties!!!   [#permalink] 25 Mar 2013, 18:13

Go to page    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  