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Re: New Set: Number Properties!!!
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29 Mar 2013, 03:55
2. If a positive integer n has exactly two positive factors what is the value of n?Notice that, n has exactly two positive factors simply means that n is a prime number, so its factors are 1 and n itself. (1) n/2 is one of the factors of n. Since n/2 cannot equal to n, then n/2=1, thus n=2. Sufficient. (2) The lowest common multiple of n and n + 10 is an even number. If n is an odd prime, then n+10 is also odd. The LCM of two odd numbers cannot be even, therefore n is an even prime, so 2. Sufficient. Answer: D.
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29 Mar 2013, 04:12
3. If 0 < x < y and x and y are consecutive perfect squares, what is the remainder when y is divided by x?Notice that since x and y are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers. (1) Both x and y have 3 positive factors. This statement implies that \(x=(prime_1)^2\) and \(y=(prime_2)^2\). From above we have that \(\sqrt{x}=prime_1\) and \(\sqrt{y}=prime_2\) are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, \(x=(prime_1)^2=4\) and \(y=(prime_2)^2=9\). The remainder when 9 is divided by 4 is 1. Sufficient. (2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. The same here: \(\sqrt{x}=2\) and \(\sqrt{y}=3\). Sufficient. Answer: D.
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Re: New Set: Number Properties!!!
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29 Mar 2013, 04:29
4. Each digit of the threedigit 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. K could be 444, 488, 848, or 888. Not sufficient. (2) K is NOT a multiple of 3. K could be 448, 484, 488, 844, 848, or 884. Not sufficient. (1)+(2) From above K could be 488 or 848. Not sufficient. Answer: E.
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Re: New Set: Number Properties!!!
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29 Mar 2013, 04:43
5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?Note that: A. The factorial of a negative number is undefined. B. 0!=1. C. Only two factorials are odd: 0!=1 and 1!=1. D. Factorial of a number which is prime is 2!=2. (1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient. (2) c! is a prime number. This implies that c=2. Not sufficient. (1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient. Answer: C.
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Re: New Set: Number Properties!!!
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29 Mar 2013, 05:06
6. Set S consists of more than two integers. Are all the integers in set S negative?(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient. (2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient. (1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient. Answer: C.
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Re: New Set: Number Properties!!!
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29 Mar 2013, 05:25
7. Is x the square of an integer?The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square). Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even. (1) When x is divided by 12 the remainder is 6. Given that \(x=12q+6=6(2q+1)=2*3*(2q+1)\). Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient. (2) When x is divided by 14 the remainder is 2. Given that \(x=14p+2\). So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient. Answer: A.
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29 Mar 2013, 05:52
9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [1.5]=2, ... (1) ab = 2. First of all this means that a and b are of the same sign. If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b. So, this case is out. If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or viseversa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or viseversa). But in this case ab cannot be equal to 2. So, this case is also out. We have that the answer to the question is NO. Sufficient. (2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient. Answer: A.
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Re: New Set: Number Properties!!!
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29 Mar 2013, 06:03
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?\(N = 3^x*5^y\) has 12 positive factors means that (x+1)(y+1)=12=2*6=6*2=3*4=4*3. We can have the following cases: \(N = 3^1*5^5\); \(N = 3^5*5^1\); \(N = 3^2*5^3\); \(N = 3^3*5^2\). (1) 9 is NOT a factor of N. This implies that the power of 3 is less than 2, thus N could only be \(3^1*5^5\). Sufficient. (2) 125 is a factor of N. This implies that the power of 5 is more than or equal to 3, thus N could be \(3^1*5^5\) or \(3^2*5^3\). Not sufficient. Answer: A. THEORY. Finding the Number of Factors of an IntegerFirst make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers. The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\) Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
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Re: New Set: Number Properties!!!
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29 Mar 2013, 06:11
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 \(2y\leq{0}\) 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 \(1 < x2 < 1\), or \(1 < x < 3\), thus \(x=2=prime\). Sufficient. (2) x + y  3 = 1y. Since y is a positive integer, then \(1y\leq{0}\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient. Answer: D.
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Re: New Set: Number Properties!!!
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07 Apr 2013, 02:07
Bunuel wrote: 4. Each digit of the threedigit integer N 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. K could be 444, 488, 848, or 888. Not sufficient.
(2) K is NOT a multiple of 3. K could be 448, 484, 488, 844, 848, or 884. Not sufficient.
(1)+(2) From above K could be 488 or 848. Not sufficient.
Answer: E. Hi Bunuel, Could you please clarify if 0 is a multiple of every number in GMAT land. Thanks



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Re: New Set: Number Properties!!!
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07 Apr 2013, 02:09
If \(0 \lt x \lt y\) and \(x\) and \(y\) are consecutive perfect squares, what is the remainder when \(y\) is divided by \(x\)? Notice that since \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers. (1) Both \(x\) and \(y\) have 3 positive factors. This statement implies that \(x=(prime_1)^2\) and \(y=(prime_2)^2\). From above we have that \(\sqrt{x}=prime_1\) and \(\sqrt{y}=prime_2\) are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, \(x=(prime_1)^2=4\) and \(y=(prime_2)^2=9\). The remainder when 9 is divided by 4 is 1. Sufficient. (2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. The same here: \(\sqrt{x}=2\) and \(\sqrt{y}=3\). Sufficient. Answer: D
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Re: New Set: Number Properties!!!
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07 Apr 2013, 03:59
Zarrolou wrote: imhimanshu wrote: Hi Bunuel, Could you please clarify if 0 is a multiple of every number in GMAT land. Thanks Yes, 0 is a multiple of every number! 0 is a multiple of every integer, except 0 itself.
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Re: New Set: Number Properties!!!
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08 Apr 2013, 20:13
imhimanshu 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 \(y2\leq{0}\) and it cannot be greater than x  2). Next, since given that y is a positive integer, then y=1.
Answer: D. Hi Bunuel, Thanks for the awesome questions that you ve posted. I have a doubt in the above question. Could you please elaborate why can't we solve the Statement A as below  x  2 < 2  y > y2< X2 < 2y I tried solving above, by removing the Modulus, however I am not getting any specific value. Kindly respond. Thanks Responding to a pm: If you must solve it using algebra, then do this: x  2 < 2  y Take the two cases: Case 1: x  2 >= 0 i.e. x >= 2 x  2 < 2  y x + y < 4 Since x must be 2 or greater than 2 and x and y both are positive integers, x must be 2 and y must be 1 (x cannot be 3 or greater since y must be at least 1 but the sum of x and y in that case will not be less than 4). Case 2: x  2 < 0 i.e. x < 2 The only value that x can take in this case is 1. 1  2 < 2  y 1 < 2  y y < 1 But we know that y must be a positive integer so x cannot take value 1. The only value that x can take is 2 which is prime. Hence this statement alone is sufficient. Mind you, using algebra too you will have to use logic so you might as well look at Bunuel's solution again.
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Re: New Set: Number Properties!!!
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09 Apr 2013, 04:18
Bunuel wrote: 5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?
Note that: A. The factorial of a negative number is undefined. B. 0!=1. C. Only two factorials are odd: 0!=1 and 1!=1. D. Factorial of a number which is prime is 2!=2.
(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient. (2) c! is a prime number. This implies that c=2. Not sufficient.
(1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient.
Answer: C. A small question, when the outcome of {a!, b!, c!} is an odd number ...is ...{1, 1, c!}, should we need to check what C is ? as whatever the C is it cant be consecutive integers. Regards Sanjiv



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Re: New Set: Number Properties!!!
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09 Apr 2013, 04:31
casanjiv wrote: Bunuel wrote: 5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?
Note that: A. The factorial of a negative number is undefined. B. 0!=1. C. Only two factorials are odd: 0!=1 and 1!=1. D. Factorial of a number which is prime is 2!=2.
(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient. (2) c! is a prime number. This implies that c=2. Not sufficient.
(1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient.
Answer: C. A small question, when the outcome of {a!, b!, c!} is an odd number ...is ...{1, 1, c!}, should we need to check what C is ? as whatever the C is it cant be consecutive integers. Regards Sanjiv If we knew for sure that the set is {1, 1, c!}, then yes we would have a definite NO answer to the question. But {1, 1, c!} is not possible since a < b < c implies that the integers in the set are distinct. (1) implies that the set is {0, 1, c!} ({0!=1, 1!=1, c!}) and if c=2, then the answer would be YES, but if c=3, then the answer would be NO. Hope it's clear.
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Re: New Set: Number Properties!!!
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09 Sep 2013, 04:02
Bunuel wrote: 9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?
Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [1.5]=2, ...
(1) ab = 2. First of all this means that a and b are of the same sign.
If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b. So, this case is out.
If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or viseversa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or viseversa). But in this case ab cannot be equal to 2. So, this case is also out.
We have that the answer to the question is NO. Sufficient.
(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.
Answer: A. Hi Bunuel, Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = 2 for any negative a and b? i took the case of a=2 and b=1 and hence ab=2. In this case i am getting [a]+[b] = (2) + (1) = 3 Let me know if i am missing something.



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Re: New Set: Number Properties!!!
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09 Sep 2013, 04:06
emailmkarthik wrote: Bunuel wrote: 9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?
Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [1.5]=2, ...
(1) ab = 2. First of all this means that a and b are of the same sign.
If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b. So, this case is out.
If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or viseversa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or viseversa). But in this case ab cannot be equal to 2. So, this case is also out.
We have that the answer to the question is NO. Sufficient.
(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.
Answer: A. Hi Bunuel, Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = 2 for any negative a and b? i took the case of a=2 and b=1 and hence ab=2. In this case i am getting [a]+[b] = (2) + (1) = 3 Let me know if i am missing something. "If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b." Since, 3<2, then this statement holds for this set of numbers too. Does this make sense?
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Re: New Set: Number Properties!!!
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09 Sep 2013, 04:08
Bunuel wrote: emailmkarthik wrote: Bunuel wrote: 9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?
Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [1.5]=2, ...
(1) ab = 2. First of all this means that a and b are of the same sign.
If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b. So, this case is out.
If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or viseversa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or viseversa). But in this case ab cannot be equal to 2. So, this case is also out.
We have that the answer to the question is NO. Sufficient.
(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.
Answer: A. Hi Bunuel, Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = 2 for any negative a and b? i took the case of a=2 and b=1 and hence ab=2. In this case i am getting [a]+[b] = (2) + (1) = 3 Let me know if i am missing something. "If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b." Since, 3<2, then this statement holds for this set of numbers too. Does this make sense? Oh yes! It does make sense. I understand that the statement does not fall apart. But I was just interested in knowing if there is a case possible where we can get 2 as the maximum value.



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Re: New Set: Number Properties!!!
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09 Sep 2013, 04:20
emailmkarthik wrote: Bunuel wrote: emailmkarthik wrote: 9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?
Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [1.5]=2, ...
(1) ab = 2. First of all this means that a and b are of the same sign.
If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b. So, this case is out.
If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or viseversa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or viseversa). But in this case ab cannot be equal to 2. So, this case is also out.
We have that the answer to the question is NO. Sufficient.
(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.
Answer: A.
Hi Bunuel,
Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = 2 for any negative a and b?
i took the case of a=2 and b=1 and hence ab=2. In this case i am getting [a]+[b] = (2) + (1) = 3
Let me know if i am missing something. "If both are negative, then the maximum value of [a] + [b] is 2, for any negative a and b." Since, 3<2, then this statement holds for this set of numbers too. Does this make sense? Oh yes! It does make sense. I understand that the statement does not fall apart. But I was just interested in knowing if there is a case possible where we can get 2 as the maximum value. What I meant was that generally if a and b are negative, then the maximum value of [a] + [b] is 1+(1)=2. But you cannot get 2, if ab=2.
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Re: New Set: Number Properties!!!
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15 Sep 2013, 07:54
Quote: 6. Set S consists of more than two integers. Are all the integers in set S negative?
(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.
(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.
(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.
Answer: C. Dear Bunuel, First of all,i would like to thank you for all your sincere efforts that help innumerable gMAT takers like me. In the explanation provided for statement 2,i did not quite understand how all the numbers of the set would have the same sign,if the smallest and largest terms have the same sign.The set contains>2 terms and from statement,we can comment only about 2 terms.Can you help me understand? Thank you, Fido




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