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Re: Good set of DS 3 [#permalink]
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17 Oct 2009, 19:30
GMAT TIGER wrote: Bunuel wrote: 10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n >= 9 1. n could be 2 or 6 or 10 n = 2: a + a+1 = 45 a = 22 n = 6: a + a+1 + a+2 + a+3 + a+4 + a+5 = 45 a = 5 2. n could be 9 or 10 or 14 or 15 or 18 & so on... 1&2: n could be 10 or 14 or 18. E. Did you say "consecutive positive integers"? not only "consecutive integers"? Agree with B if "consecutive positive integers". If only "consecutive integers", B is correct. Seems you said "consecutive positive integers"!
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Re: Good set of DS 3 [#permalink]
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17 Oct 2009, 20:43
GMAT TIGER wrote: Bunuel wrote: Yes and about the ZIP trap:
GMAT likes to act in the zone 1<=x<=1. So I always ask myself:
Did I assumed, with no ground for it, that variable can not be Zero? Check 0! Did I assumed, with no ground for it, that variable is an Integer? Check fractions! Did I assumed, with no ground for it, that variable is Positive? Check negative values!
I called it ZIP trap. Helps me a lot especially with number property problems. Thats cool. You can say PINZF (or better) trap as well: P = positive I = integer N = negative Z = zero F = fraction Sure you can call whatever suits you, no copyright on that term, for me ZIP sounds good.)))
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Re: Good set of DS 3 [#permalink]
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25 Oct 2009, 09:38
GMAT TIGER wrote: Bunuel wrote: 1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9 1. n could be 2 or 6 or 10 a + a+1 = 45 a = 22 n = 2 a + a+1 + a+2 + a+3 + a+4 + a+5 = 45 a = 5 n = 6 2. n could be 2, 3, 5 or 6 1&2: n could be 2 or 6. E. Bunuel wrote: 10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n >= 9 1. n could be 2 or 6 or 10 n = 2: a + a+1 = 45 a = 22 n = 6: a + a+1 + a+2 + a+3 + a+4 + a+5 = 45 a = 5 2. n could be 9 or 10 or 14 or 15 or 18 & so on...1&2: n could be 10 or 14 or 18. E. How can we find out 10, 14, 15 & 18??? I mean total is 45, how did tiger wrote 14, 15, 18 mark straight away?? I think its not possible!!! Pls help!!
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Re: Good set of DS 3 [#permalink]
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25 Oct 2009, 20:21
Bunuel wrote: ANSWERS:
10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is odd (2) n >= 9
Look at the Q 1 we changed even to odd and n<9 to n>=9
(1) not sufficient see Q1. (2) As we have consecutive positive integers max for n is 9: 1+2+3+...+9=45. (If n>9=10 first term must be zero. and we are given that all terms are positive) So only case n=9. Sufficient.
Answer: B. sum of n integers = (n*(n+1))/2 (n*(n+1))/2 = 45 which yields n = 9 could be the only answer Stmt 1 and 2 (D) are both sufficient. Does GMAT assume that we would not indulge in any formula? Just curious



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Re: Good set of DS 3 [#permalink]
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25 Oct 2009, 20:50
I like the concept of ZIP trap
check for 0,ve and fractions.



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Re: Good set of DS 3 [#permalink]
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25 Oct 2009, 20:53
goldgoldandgold wrote: Bunuel wrote: ANSWERS:
10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is odd (2) n >= 9
Look at the Q 1 we changed even to odd and n<9 to n>=9
(1) not sufficient see Q1. (2) As we have consecutive positive integers max for n is 9: 1+2+3+...+9=45. (If n>9=10 first term must be zero. and we are given that all terms are positive) So only case n=9. Sufficient.
Answer: B. sum of n integers = (n*(n+1))/2 (n*(n+1))/2 = 45 which yields n = 9 could be the only answer Stmt 1 and 2 (D) are both sufficient. Does GMAT assume that we would not indulge in any formula? Just curious This is not correct. First of all the formula you referring n(1+n)/2 is the formula for counting the sum of n FIRST integers (meaning that starting from 1). Question stem does not give us that information: "The sum of n consecutive positive integers is 45." You can not assume anything. The formula for counting the sum of consecutive integers (not necessarily first n integers), which is in fact sum of AP (arithmetic progression) is: Sn=n*(a1+an)/2 or Sn=n*(2a1+d(n1))/2, you can substitute d=1, as the numbers are consecutive. If you further substitute a1 (the first term of the progression) with 1 you'll get exactly the formula you wrote: Sn=n(2+n1)/2=n(n+1)/2 Second, you can find the n consecutive odd positive integers (n=odd) to total 45: n=3 > 14, 15, 16 n=5 > 7, 8, 9, 10, 11. Hope it's clear.
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Re: Good set of DS 3 [#permalink]
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26 Oct 2009, 01:12
Quote: 3. Multiplication of the two digit numbers wx and cx, where w,x and c are unique nonzero digits, the product is a three digit number. What is w+cx? (1) The three digits of the product are all the same and different from w c and x. (2) x and w+c are odd numbers.
(1) wx+cx=aaa (111, 222, ... 999=37*k) > As x is the units digit in both numbers, a can be 1,4,6 or 9 (2,3,7 out because x^2 can not end with 2,3, or 7. >>What about 8?5 is out because in that case x also should be 5 and we know that x and a are distinct numbers). 1 is also out because 111=37*3 and we need 2 two digit numbers. 444=37*12 no good we need units digit to be the same. 666=37*18 no good we need units digit to be the same. 999=37*27 is the only possibility all digits are distinct except the unit digits of multiples. Sufficient (2) x and w+c are odd numbers. Number of choices: 13 and 23 or 19 and 29 and w+cx is the different even number.
Answer: A.
Could somebody please elaborate the solution..i don't understand:(
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Re: Good set of DS 3 [#permalink]
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27 Oct 2009, 22:19
tejal777 wrote: Quote: 3. Multiplication of the two digit numbers wx and cx, where w,x and c are unique nonzero digits, the product is a three digit number. What is w+cx?
(1) The three digits of the product are all the same and different from w c and x. (2) x and w+c are odd numbers.
(1) wx+cx=aaa (111, 222, ... 999=37*k) > As x is the units digit in both numbers, a can be 1,4,6 or 9 (2,3,7 out because x^2 can not end with 2,3, or 7. >>What about 8?5 is out because in that case x also should be 5 and we know that x and a are distinct numbers). 1 is also out because 111=37*3 and we need 2 two digit numbers. 444=37*12 no good we need units digit to be the same. 666=37*18 no good we need units digit to be the same. 999=37*27 is the only possibility all digits are distinct except the unit digits of multiples. Sufficient
(2) x and w+c are odd numbers. Number of choices: 13 and 23 or 19 and 29 and w+cx is the different even number.
Answer: A.
Could somebody please elaborate the solution..i don't understand:( 8 is also not possible. ........................................................................ 1. wx * cx = aaa (aaa can only be 111 (for 9), 444 (for 2), 666 (for 8), and 999 (for 3 and 7)). 222, 333, 555, 777, and 888 are not possible as noneo f the unit difgits of the 3 digit numbers is square of an integer. wx * cx = 111 = 3x37. Not possible as there is only one 2 digit number. Needs 2 two digit numbers. wx * cx = 444 = 2x3x37 = 6x37. Not possible as there is only one 2 digit number. Needs 2 two digit numbers. wx * cx = 666 = 2x3x3x37 = 18x37. Not possible as the unit digit of these 2 digit numbers are not the same. wx * cx = 999 = 3x3x3x37 = 27x37. Possible as the unit digit of these 2 digit numbers are same. 2. is self explanatory..
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Re: Good set of DS 3 [#permalink]
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08 Nov 2009, 00:46
GMAT TIGER wrote: tejal777 wrote: Quote: 3. Multiplication of the two digit numbers wx and cx, where w,x and c are unique nonzero digits, the product is a three digit number. What is w+cx?
(1) The three digits of the product are all the same and different from w c and x. (2) x and w+c are odd numbers.
(1) wx+cx=aaa (111, 222, ... 999=37*k) > As x is the units digit in both numbers, a can be 1,4,6 or 9 (2,3,7 out because x^2 can not end with 2,3, or 7. >>What about 8?5 is out because in that case x also should be 5 and we know that x and a are distinct numbers). 1 is also out because 111=37*3 and we need 2 two digit numbers. 444=37*12 no good we need units digit to be the same. 666=37*18 no good we need units digit to be the same. 999=37*27 is the only possibility all digits are distinct except the unit digits of multiples. Sufficient
(2) x and w+c are odd numbers. Number of choices: 13 and 23 or 19 and 29 and w+cx is the different even number.
Answer: A.
Could somebody please elaborate the solution..i don't understand:( 8 is also not possible. ........................................................................ 1. wx * cx = aaa (aaa can only be 111 (for 9), 444 (for 2), 666 (for 8), and 999 (for 3 and 7)). 222, 333, 555, 777, and 888 are not possible as noneo f the unit difgits of the 3 digit numbers is square of an integer. wx * cx = 111 = 3x37. Not possible as there is only one 2 digit number. Needs 2 two digit numbers. wx * cx = 444 = 2x3x37 = 6x37. Not possible as there is only one 2 digit number. Needs 2 two digit numbers. wx * cx = 666 = 2x3x3x37 = 18x37. Not possible as the unit digit of these 2 digit numbers are not the same. wx * cx = 999 = 3x3x3x37 = 27x37. Possible as the unit digit of these 2 digit numbers are same. 2. is self explanatory.. I am sorry.. but I still dont understand how did you arrive at that 37..



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Re: Good set of DS 3 [#permalink]
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08 Nov 2009, 05:18
Hey Bunuel,
Thanks for the great set of DS problems. They are always the hardest for me. I just cant get used to how open ended some of those problems can be.
Just out of curiousity, where are these problems coming from? If they are from a specific book, I definitely need to get it!
Thanks again!!!
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Re: Good set of DS 3 [#permalink]
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08 Nov 2009, 05:24



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Re: Good set of DS 3 [#permalink]
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25 Dec 2009, 18:18
thanks for the questions
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Re: Good set of DS 3 [#permalink]
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26 Dec 2009, 03:47
please explain 2nd Q.
i want an example where XYZ can be prime using STATEMENT 1 ALONE



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Re: Good set of DS 3 [#permalink]
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26 Dec 2009, 09:04
jan4dday wrote: please explain 2nd Q.
i want an example where XYZ can be prime using STATEMENT 1 ALONE First note prime numbers are only positive. (Also note that \(x\), \(y\) and \(z\) are integers) Q: \(xyz=p\), is \(p\) prime? (1) \(x=y\) > \(p=x^2z\). Let's check when this expression gives a prime number: Well first of all \(p\) to be prime \(z\) MUST be negative, as \(p\) MUST be positive to be a prime. Next if \(x>1\), (eg \(2\), \(3\), ...) OR equals to zero, \(p\) won't be prime. So \(x\) must be equal to \(1\). But it's not enough. We'll have \(p=x^2z=z\), so \(p\) to be a prime number \(z\) must be equal to \(prime\). You are asking how using statement (1) \(p\) could be a prime: according to above, when \(x=1\) and \(z=p\). eg.: \(x=1\) > \(y=1\) > z\(=7\) > \(p=(1)*1*(7)=7\), which is prime. Statement (1) may or may not give the prime number for \(xyz\). Not sufficient. (2) \(z=1\) > \(p=xy\). Again for \(p\) to be a prime number \(xy\) must be \(>0\) (both positive or both negative). Then if \(x=prime\) and \(y=1\), OR \(y=prime\) and \(x=1\), so that \(xy>0\), then \(xy\) is a prime number. For any other values or combinations of \(x\) and \(y\), \(p\) won't be a prime. Not sufficient. (1)+(2) \(p=xyz=x^2\) (as \(x=y\) and \(z=1\)). \(x^2\) is never positive, hence \(p\) is not a prime. Sufficient. Answer: C. Hope it's clear.
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Re: Good set of DS 3 [#permalink]
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26 Dec 2009, 19:23
Hey thanks a lot! I forgot to take a negative value for z in 1st statement. And i am writing GMAT in one week .damn it!!!!



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Re: Good set of DS 3 [#permalink]
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02 Jan 2010, 09:07
Great Questions and Great Tips Bunuel!! +1 Kudos! for this Cheers! JT
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Re: Good set of DS 3 [#permalink]
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02 Jan 2010, 09:20
Bunuel wrote: 5. If a and b are integers, and a not= b, is ab > 0? (1) a^b > 0 (2) a^b is a nonzero integer
This is tricky ab > 0 to hold true: a#0 and b>0.
(1) a^b>0 only says that a#0, because only way a^b not to be positive is when a=0. Not sufficient. NOTE having absolute value of variable a, doesn't mean it's positive. It's not negative > a>=0
(2) a^b is a nonzero integer. What is the difference between (1) and (2)? Well this is the tricky part: (2) says that a#0 and plus to this gives us two possibilities as it states that it's integer: A. 1>a>1 (a>1), on this case b can be any positive integer: because if b is negative a^b can not be integer. OR B. a=1 (a=1 or 1) and b can be any integer, positive or negative. So (2) also gives us two options for b. Not sufficient.
(1)+(2) nothing new: a#0 and two options for b depending on a. Not sufficient.
Answer: E.
Bunuel... apologize... but am I little lost with this explanation. I am not able to understand what exactly you mean by a#0  Does this mean a = 0?
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Re: Good set of DS 3 [#permalink]
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02 Jan 2010, 09:24
jeeteshsingh wrote: Bunuel wrote: 5. If a and b are integers, and a not= b, is ab > 0? (1) a^b > 0 (2) a^b is a nonzero integer
This is tricky ab > 0 to hold true: a#0 and b>0.
(1) a^b>0 only says that a#0, because only way a^b not to be positive is when a=0. Not sufficient. NOTE having absolute value of variable a, doesn't mean it's positive. It's not negative > a>=0
(2) a^b is a nonzero integer. What is the difference between (1) and (2)? Well this is the tricky part: (2) says that a#0 and plus to this gives us two possibilities as it states that it's integer: A. 1>a>1 (a>1), on this case b can be any positive integer: because if b is negative a^b can not be integer. OR B. a=1 (a=1 or 1) and b can be any integer, positive or negative. So (2) also gives us two options for b. Not sufficient.
(1)+(2) nothing new: a#0 and two options for b depending on a. Not sufficient.
Answer: E.
Bunuel... apologize... but am I little lost with this explanation. I am not able to understand what exactly you mean by a#0  Does this mean a = 0? \(a\) does not equal to \(b\). Sorry for confusion.
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Re: Good set of DS 3 [#permalink]
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02 Jan 2010, 10:28
Hi Bunuel..... Would appreciate if you could explain this in detail as I am confused... Thanks! JT
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Re: Good set of DS 3 [#permalink]
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02 Jan 2010, 11:58
jeeteshsingh wrote: Hi Bunuel..... Would appreciate if you could explain this in detail as I am confused... Thanks! JT If a and b are integers, and a not= b, is ab > 0? (1) a^b > 0 (2) a^b is a nonzero integer. \(ab > 0\) is true when \(b>0\) and \(a\) does not equal to zero. (1) \(a^b > 0\) > \(a\) does not equal to zero, but we don't know about \(b\), it can be any value, positive or negative. Not sufficient. (2) \(a^b\) is a nonzero integer > \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient. (1)+(2) If a=1 and b=2, then ab > 0, but if a=1 and b=2, then ab <0. Not sufficient. Answer: E. This is the easiest way to solve this problem. In my previous solution I just tried to show what each statement means algebraically. Please tell which part needs more clarification.
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