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Re: New Set: Number Properties!!!
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Updated on: 29 Mar 2013, 04:20
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
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Originally posted by Rock750 on 26 Mar 2013, 03:57.
Last edited by Rock750 on 29 Mar 2013, 04:20, edited 1 time in total.



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Re: New Set: Number Properties!!!
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26 Mar 2013, 04:27
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 04:57
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 06:43
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 07:07
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 07:38
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 08:01
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 08:06
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 08:12
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 08:18
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 08:29
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 10:58
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 11:08
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 11:42
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 11:56
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
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Re: New Set: Number Properties!!!
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26 Mar 2013, 12:12
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
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Re: New Set: Number Properties!!!
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27 Mar 2013, 04:46
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
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Re: New Set: Number Properties!!!
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Updated on: 28 Mar 2013, 17:24
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
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Re: New Set: Number Properties!!!
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Updated on: 28 Mar 2013, 19:07
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
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Originally posted by Rock750 on 28 Mar 2013, 02:43.
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Re: New Set: Number Properties!!!
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28 Mar 2013, 03:08
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 !
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Re: New Set: Number Properties!!!
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