December 20, 2018 December 20, 2018 10:00 PM PST 11:00 PM PST This is the most inexpensive and attractive price in the market. Get the course now! December 22, 2018 December 22, 2018 07:00 AM PST 09:00 AM PST Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
Author 
Message 
TAGS:

Hide Tags

Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 237
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
Updated on: 29 Mar 2013, 03:20
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 (1) Both x and y is have 3 positive factors.Consecutive perfect squares could be : 4 , 9 , 16 , 25 , 36 , 49 , 64 .... Among these numbers, only 4 and 9 are consecutive perfect squares that have 3 positive factors ( for instance 16 = 4*4 = 2*2*2*2 > 5 factors and SO ON ) Hence, y=9 and x=4 > 9 = 4.2 +1 > R = 1 Thus, (1) SUFFICIENT (2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbersConsecutive perfect squares could be : 4 , 9 , 16 , 25 , 36 , 49 , 64 .... Among these numbers, only 4 and 9 are consecutive perfect squares that have their square roots as prime numbers ( for example : \(\sqrt{16}\) = 4, which is not a prime number and SO ON ) Hence, y=9 and x=4 > 9 = 4.2 > R = 1 Thus, (2) SUFFICIENT Answer : D
_________________
KUDOS is the good manner to help the entire community.
"If you don't change your life, your life will change you"
Originally posted by Rock750 on 26 Mar 2013, 02:57.
Last edited by Rock750 on 29 Mar 2013, 03:20, edited 1 time in total.



Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 237
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 03:27
4. Each digit of the threedigit integer N is a 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. Given that each digit of the threedigit integer N is a multiple of 4 , K could be : 444 or 448 or 484 or 488 ... (1) The units digit of K is the least common multiple of the tens and hundreds digit of KSo , K should b equal to 444 (LCM(4,4) = 4) OR equalt to 888 (LCM(8,8) = 8) OR equalt to 488 (LCM(4,8) = 8) .... Hence, (1) NOT SUFFICIENT (2) K is NOT a multiple of 3.So, K could be equal to 448 or 484 ... Hence, (2) NOT SUFFICIENT (1) + (2) Both 488 and 848 have their units digit as the LCM of the tens and hundreds and are not a mulitple of 3 Hence, (1) + (2) NOT SUFFICIENT Answer : E
_________________
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: 237
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 03:57
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 (1) The median of {a!, b!, c!} is an odd number.The median of {a!, b!, c!} is an odd number > b! is the median > b! is odd > b can take two values : 0 or 1 If b = 0 and b>a then a is a negative integer > impossible Hence b must be equal to 1 , and because b>a and The median of {a!, b!, c!} is an odd number., then a = 0 and c = 2 because the median must be equal to b! wich is 1. (1) SUFFICIENT (2) c! is a prime numberIn this case, c can be equal to 0, 1 or 2 (2) INSUFFICIENT Answer : A
_________________
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: 237
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 05:43
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. (1) The product of any three integers in the list is negativeConsidering S with 3 integers : the set can be constitued with + +  OR    , So this statement is INSUFFICIENT (2) The product of the smallest and largest integers in the list is a prime number.By the usual definition of prime for integers, negative integers can not be prime. So, for the product of the smallest and largest integers in the list to be a prime number, the largest and the smallest both must be either positive or negative. For example : Set = {13, 8 , 6 , 1} : the product of the largest and smallest = 13*1 = 13prime And Set {1, 5 9 , 13} : the product of the largest and smallest = 13 * 1 = 13 prime So, we cannot conclude if all the numbers in set S are negative or not , this statement alone is INSUFFICIENT (1) + (2) > Given by the second statement that ALL THE NUMBERS IN SET S MUST HAVE THE SAME SIGN, and given by the first statement that ANY THREE INTEGERS in the list is negative > THE ANSWER IS YES : ALL THE NUMBERS IN SET S ARE NEGATIVE Answer : C
_________________
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: 237
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 06:07
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 (1) When x is divided by 12 the remainder is 6when x is divided by 12 the remainder is 6 > x = 12q + 6 = 3*2*(2q+1) ; So the answer is NO, x cannot be the square of any integer For example : q = 1 : x = 18 q=2 : x = 30 ... (1) SUFFICIENT (2) When x is divided by 14 the remainder is 2When x is divided by 14 the remainder is 2 > x = 14p + 2 = 2*(7p+1) For example : p = 1 > x = 16 = 4^2 p = 2 > x = 30 So, we dont know > (2) NOT SUFFICIENT Answer : A
_________________
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: 237
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 06:38
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 Set A consist of 10 terms A = { x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 , x9 , x10 } where xi >_ xi1 ( i = 1.2....10 ) and xi = 1/Pj Pj: prime number Median A = (x5 + x6) / 2 < 1/5 ?? (1) Reciprocal of the median is a prime numberwe don't know anything about x5 and x6 so this statement is INSUFFICIENT (2) The product of any two terms of the set is a terminating decimal Given that statement > the set A is composed only by 1/2 or 1/5 or both many times > the mean will always be greater than or equalt to 1/5 but never less than 1/5 > the answer to the question is NO > this statement ALONE is SUFFICIENT Answer : B
_________________
KUDOS is the good manner to help the entire community.
"If you don't change your life, your life will change you"



Director
Joined: 25 Apr 2012
Posts: 683
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 07:01
1. If x is an integer, what is the value of x? (1) 23x is a prime number (2)2\sqrt{x2}is a prime number Hellooz, St 1, we have 23x is a prime no. Possible values of x are 1 and 1 and hence not sufficient St 2 Now 2\sqrt{x2} is equal 2 x is equal 2 x. Since in GMAT even root is considered postive only therefore we have 2\sqrt{x2} = 2x and 2x is a prime no and hence value of x is 1 Ans should be B
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Director
Joined: 25 Apr 2012
Posts: 683
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 07:06
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. We need to find the value of n and n has exactly 2 postive factors This implies that n is of the form n= a^1 where a is prime factor of n. From St 1, we have n/2 is one of the factors of n. Now any prime no has only 1 and the no itself as its factor therefore n can be any of the prime nos ie. 2,3,5,7.... and so on. However when n = 2, n/2 = 1 and 1 is factor of n and hence the value n will be 2 Note that any other prime no when divided by 2 will not give an integer as 2 is the only even prime no So B,C and E ruled out St 2. We have multiple options here The lowest common multiple of n and n + 10 is an even number. n =2 and n+ 10 =12, LCM is 12( Even no) value of n =2> n is prime n=4 and n+10 = 14 , LCM is 28 (even no)value of n =4 > n is not prime Hence D ruled out, Ans should be A
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Director
Joined: 25 Apr 2012
Posts: 683
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 07:12
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 From the Q stem we get x and y are consecutive perfect squares ie. x=4, y=9 or x=9 then y=16 and so on We need to find y/x St 1. Both x and y have 3 postive factors Consider X , Since X has 3 psotive factors and x is a perfect square x is of the form x = a^2 where is "a" is a prime no. Then possible factors of x are 1,a, and a2 Similarly for y= b^2 where b is prime no, factors will be 1,b,and b^2 Clearly we can multiple remainders for ex x=4 , y =9 , R (y/x)= 1 x= 25, y=36, R (y/x)= 11 Hence not sufficient and therefore A and D ruled out St 2. \sqrt{x} and \sqrt{y} are prime no which means x=a^2 and y= b^2 where a and b are prime no and x and y are sqaure of prime nos. This is same information as St1 and hence B and C ruled out Ans should be E
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Director
Joined: 25 Apr 2012
Posts: 683
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 07:18
4. Each digit of the threedigit integer N is a 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. I think K should be N is statement choices Lets look at the possible 3 digit no options based on the given Question stem. We have 444, 888,484, 448,844 and so on....the digit integer N will have only 4' s and 8's St 1 implies that no is of the form 48 8> where 8 in unit place is LCM of 4 and 8 (tens place) or it could be 88 8> Again 8 in unit place is LCM of 8 and 8 in tens and hundreds place We have 2 options hence A alone not sufficient therefore A and ruled out 2. St N is not a multiple of 3> means N can be 448,844,884,848 and so on.. B alone is not sufficent combining we get following choices 484,848...Again we have 2 possible values of N and therefore Ans should be E
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Director
Joined: 25 Apr 2012
Posts: 683
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 07:29
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 We need to find if a,b and c are consecutive integers or not st 1 says The median of {a!, b!, c!} is an odd number. We know that 2! =2 and 3!=6 and 4!=24 > for any factorial greater then equal to 2! will always be even no as it will have 2 This implies possible options of a,b and c are 0,1 and 2 0!=1 and 1!=1 and 2!= 2 so our set is (0,1 and 2) and median is 1 which is odd. This implies that a,b and c are consecutive integers Now also we can have any other no in place of 2!, > we can have 3! or 4! our median will be odd...a,b and c are not CI So A and D ruled out St 2 says c! is a prime number c can only be 2 as factorial of other nos greater then 2 will have some factor and a and b can be any nos a=3 , b=2 and c= 2> a,b and c are not CI but again if c=2, b=1 and a=0 then they are CI So B alone not sufficient Combining we get only case of c=2,b=1 and a=0 and therefore Ans should be C
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Director
Joined: 25 Apr 2012
Posts: 683
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 09:58
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. Let Set S has n integers so n>2 From St1,we have 2 possibilities for All 3 nos are negative ie. **= Negative no Or 2 nos are positive and one negative > +**+ So A and ruled out St2 Since n> 2, Lets take n=4 and the nos be n1<n2<n3<n4 Let all nos be positive. Since the product of smallest and largest is a prime no then smallest no will have to be 1 and largest not can be any prime no greater than n2 and n3. n2 and n3 can be any nos Another case will be n4<n3<n2<n1....in this case let us assume all nos are negative and n1 is the largest nos and n4 is the smallest in this case as well n4*n1=n1n4> which can be a prime no Ex 7*1=7 is a prime no Since we have 2 cases so Option is ruled out. Now Combining we get a case where all nos are negative the 2 statement conditions are met and hence should be C Thanks Mridul
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Director
Joined: 25 Apr 2012
Posts: 683
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 10:08
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 Question stem, we have whether x=I^2 where I is some Integer From St1, Possible values of x will be 6,18,30,42... In General we have x= 12a+ R where R=6 and a is some integer I^2= 12a+6> I^2= 2^2*3a+6 Now the RHS expression cannot be a square of any Integer I and therefore X is not a square of any integer. So option B,C and E ruled out St 2 When x is divided by 14 the remainder is 2 x= 14b+2 >Possible values of x are 2,16, 30,44,58... We have 2 cases ie. X=16 a perfect sqaure and X = 2 or 30 or 44 not a perfect square Hence B alone is not sufficient So Ans should be A
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Director
Joined: 25 Apr 2012
Posts: 683
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 10:42
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 From St 1, we have ab=2 (Note that a and b are of same sign) Possible values are a= 2, b= 1, Is exp [a] + [b] = 1 Clearly no as [2] + [1] = 3 =1 a=1, b=2, Is exp [a] + [b] = 1 No as > Expression equals 3 ([1]+[2] = 3) a=Sqt 2 , b= Sqt 2, Exp [Sqt2] + [sqt 2] = 1+1 =2 hence no For any value of a and b the expression does not seem to be true and hence [a]+[b]=1 is not true for all cases (I hope this doesn't come back to bite me) Hence ans should be A so B,C and E ruled out St2 0 < a < b < 2 Let a = 0.3, b =1.3 so we have [0.3]+[1.3]= 0+1 and hence expression is true but if a= 1.1 and b = 1.6 then we have [1.1]+[1.6] = 1+1 =2 not equal to 1 Since 2 cases are possible with st2 therefore ans should be A
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Director
Joined: 25 Apr 2012
Posts: 683
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 10:56
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 Since x and y are postive and N has 12 postive factors which means we can have more than 1 value for N for different combination of prime factors N = 3^x*5^y  >Now No. factors is given by (x+1)(y+1)= 12 Possible combinations (x+1)(y+1) 1 12 2 6 3 4 and vice versa St1 : 9 is not a factor of N, Now if X =2 or 3 then 9 becomes a factor of N therefore possible values of x are 0 and 1 For x =0 the number N will be 5^11 For x =1, then the number will be 3^1 *5^5 (Factors will be 3, 5, 3*5, 5^5, 5^4, 5^3, 5^2, 3*5^5 , 3*5^4, 3*5^3,3*5^2 and 1 ) So 2 values of N possible hence not sufficient. Option A & D ruled out St 2 :125 is a factor of N Now if y= 3 then 125 becomes factor of N therefore possible values of y are 3 or 6 When y=3, x= 2 then no N will be = 3^2*5^3 or when y=5, x=1 then no N will be = 3^1*5^5 2 values of N possible so option B ruled out Combining 2 statements we get N = 3^1*5^5Hence Ans should be C
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Director
Joined: 25 Apr 2012
Posts: 683
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
26 Mar 2013, 11:12
BONUS QUESTION: 11. If x and y are positive integers, is x a prime number? (1) x  2 < 2  y (2) x + y  3 = 1y X and Y are +Integers, Is X a prime no St1 x  2 < 2  y Can be rewritten as x  2 +y < 2 Solving for Modulus we get Case 1 x2 +y <2 > x+y< 4 Lets say x=2 , y= 1 then is X prime> Yes But if X=1 and Y=2 then is X prime > no (However if we put these values in the given expression we get 3< 2 which is not the case ) and hence this case is not considered as it goes against the given condition Case 2 Taking value in Mod as negative we get (x2) +y < 4> x+2 +y < 2, we get x> y. and therefore for the expression we will have x=2, y = 1. Thus X is prime SO option B,C and E ruled out St 2 : x + y  3 = 1y x+y = 1y +3 Solving for mod cases we get Case 1 : Term in modulus is positive, we get x+y= (1y) +3 x+y =1 y + 3 > x+2y= 4 Again x=2, y= 2, Is X prime yes Case 2 Term in Modulus is negative, we get x+y= (1y) + 3 x+y = 1 +y +3 We get x =2 Therefore St2 alone is also sufficient Hence Ans should be D
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 237
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
27 Mar 2013, 03:46
9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + = 1 ?
(1) ab = 2 (2) 0 < a < b < 2
[b](1) ab = 2ab = 2 > b = 2/a If a<0 > b<0 > [ a ] + [ b ] <0 > Answer to the question is NO If 0 < a _< 1 > b >_ 2 > [ a ] + [ b ] > 1 > Answer to the question is NO If a>1 > [ a ] + [ b ] > 1 > Answer to the question is NO (1) SUFFICIENT (2) 0 < a < b < 2If a=0.3 and b=0.5 , therefore [a] = 0 and [b] = 0, and it follows that [a] + [b] = 0 ; the answer is NO If a = 0.2 and b = 1.1 therefore [a] = 0 and [b] = 1, and if follows that [a] + [b] = 1 , the answer is YES So , the second statement is INSUFFICIENT Answer : A
_________________
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: 237
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
Updated on: 28 Mar 2013, 16:24
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 (x,y) two positive integers and N has 12 postive factors > (x+1)(y+1)= 12 x + 1 = 2 and y + 1 = 6 > (x,y) = (1,5) x + 1 = 6 and y + 1 = 2 > (x,y) = (5,1) x + 1 = 3 and y + 1 = 4 > (x,y) = (2,3) x + 1 = 4 and y + 1 = 3 > (x,y) = (3,2) (1) 9 is NOT a factor of NSince x is a positive integer and 9 is not a factor of N, then dicard x=2 and x=3 and x=5 > x=1 x=1 > y=5 > N = 3^x*5^y = 3^1 * 5^5 > Statement 1 ALONE SUFFICENT (2) 125 is a factor of Nwe know that : 5^3 = 125 , hence discard all possible values less than 3 for y > two possibilties remain (x,y) = (1,5) and (x,y) = (2,3) > Statement 2 ALONE INSUFFICIENT Answer : A
_________________
KUDOS is the good manner to help the entire community.
"If you don't change your life, your life will change you"
Originally posted by Rock750 on 27 Mar 2013, 06:58.
Last edited by Rock750 on 28 Mar 2013, 16:24, edited 1 time in total.



Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 237
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
Updated on: 28 Mar 2013, 18:07
BONUS QUESTION: 11. If x and y are positive integers, is x a prime number? (1) x  2 < 2  y (2) x + y  3 = 1y (1) x  2 < 2  yIf x<2 > x=1 because x is a positive integer then x is not prime If x=2 > x is prime If x>2 > x  2 < 2  y In this case : if y<2 > y = 1 > x < 3 and because x>2 so this can't be true > y must be greater than or equalt to 2 if y = 2 > x  2 < 0 > x<2 so x = 1 and thus x is not prime if y>2 hence 2  y < 0 > x  2 < 2  y < 0 > x  2 < 0 > x < 2 > x = 1 > thus x is not prime (1) ALONE IS SUFFICIENT 2) x + y  3 = 1yy is a positive integer > if y = 1 > x = 2 thus x is prime > if y>1 > 1y = y  1 > x + y  3 = y  1 > x = 2 and thus x is prime (2) ALONE IS SUFFICIENT Answer : D
_________________
KUDOS is the good manner to help the entire community.
"If you don't change your life, your life will change you"
Originally posted by Rock750 on 28 Mar 2013, 01:43.
Last edited by Rock750 on 28 Mar 2013, 18:07, edited 1 time in total.



Manager
Status: Final Lap
Joined: 25 Oct 2012
Posts: 237
Concentration: General Management, Entrepreneurship
GPA: 3.54
WE: Project Management (Retail Banking)

Re: New Set: Number Properties!!!
[#permalink]
Show Tags
28 Mar 2013, 02:08
As carcass suggested, next set word problems Bunnel i wonder if you could add geometry besides so that we can have two sets for the next week. Thanks !
_________________
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!!! &nbs
[#permalink]
28 Mar 2013, 02:08



Go to page
Previous
1 2 3 4 5 6 7
Next
[ 140 posts ]



