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Re: Good set of DS 3 [#permalink]
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25 Oct 2009, 08:38
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, 19:21
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, 19:50
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, 19:53
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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27 Oct 2009, 21:19
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, 04:18
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, 04:24



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Re: Good set of DS 3 [#permalink]
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26 Dec 2009, 02:47
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, 08:04
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, 18:23
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, 08:20
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, 08:24
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, 09:28
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, 10:58
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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Re: Good set of DS 3 [#permalink]
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03 Jan 2010, 02:14
Bunuel wrote: 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.
Not able to get why do you say this (marked in red)... I know that \(a\) would always give a positive value, but why cant we consider \(a\) as 2 or 3...? Any reason for this?
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Re: Good set of DS 3 [#permalink]
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03 Jan 2010, 08:23
jeeteshsingh wrote: Bunuel wrote: 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.
Not able to get why do you say this (marked in red)... I know that \(a\) would always give a positive value, but why cant we consider \(a\) as 2 or 3...? Any reason for this? OK. For statement (2): First of all I'm not saying that \(a\) can ONLY be 1. I'm saying that it MAY be 1 and in this case \(b\) can take any value statement (2): " \(a^b\) is a nonzero integer " to hold true. So \(b\) can take positive as well as negative values, hence \(ab > 0\) may or may not be true. That's why (2) is not sufficient. To elaborate further: we are told that \(a^b\) is A. not zero and B. it's an integer. A. Means that \(a\) is not zero, as the only way \(a^b\) to be zero, is \(a\) to be zero. B. \(a^b\) is an integer (non=zero). What does that mean? When \(a\) is any integer so that \(a>1\) (eg 2, 2, 3, 3, ...), \(b\) can take any value but negative. Example: \(a=3\) > \(3=3\), \(3\) in integer power (given \(a\) and \(b\) are integers) to be an integer, power (\(b\)), must be positive or zero, as when \(b\) is negative, let's say \(2\), we'll have: \(3^{2}=\frac{1}{3^2}\) which is not integer. But when \(a=1\) (\(1\) or \(1\)), then \(b\) can take ANY integer value as 1 in any power equals to 1 which is integer. So from statement (2) we'll have:\(a\) does not equals to 0. Plus, \(b\) can be ANY integer (positive or negative or zero) when \(a=1\) and \(b\) must be positive or zero when \(a>1\). Hope it's clear.
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Re: Good set of DS 3 [#permalink]
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18 Feb 2010, 12:38
jeeteshsingh,
To further clarify what Bunuel was saying. If b was negative and a was not 1 or 1, then a^b would be a fraction due to the properties of negative exponents. If a=1 (meaning a=1 or a=1) then the denominator of the fraction would be some power of 1 which is also one. In this case the fraction would be 1 over 1 which is again 1 and satisfies the conditions.
Also, Bunuel, for the second portion I noticed that you didn't include in your listed possibilities that for a^b is a non zero integer, b=0 is possible as it would create an answer of 1 for any value of a including 0.



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The sum of n consecutive positive integers is 45 [#permalink]
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18 Feb 2010, 13:07
theturk123 wrote: jeeteshsingh,
To further clarify what Bunuel was saying. If b was negative and a was not 1 or 1, then a^b would be a fraction due to the properties of negative exponents. If a=1 (meaning a=1 or a=1) then the denominator of the fraction would be some power of 1 which is also one. In this case the fraction would be 1 over 1 which is again 1 and satisfies the conditions.
Also, Bunuel, for the second portion I noticed that you didn't include in your listed possibilities that for a^b is a non zero integer, b=0 is possible as it would create an answer of 1 for any value of a including 0. Welcome to the Gmat Club. Your logic is correct. Though b=0 IS included as one of the possibilities. See the bolded part from the explanation: Bunuel wrote: So from statement (2) we'll have:
\(a\) does not equals to 0. Plus, \(b\) can be ANY integer (positive or negative or zero) when \(a=1\) and \(b\) must be positive or zero when \(a>1\). You also mention \(0^0\) issue, this not the case for statement 2 (as \(a\) cannot be zero), but still as you brought this up: \(0^0\), in some sources equals to 1, some mathematicians say it's undefined. Anyway you won't need this for GMAT: "During the past decade, mathematicians argued extensively about the value of 0^0. Some answer that 0^0 = 1, while others answer that 0^0 is undefined. In the unlikely event that this question appears in some format or is a required intermediary calculation, the correct answer is more likely that 0^0 = 1." http://www.platinumgmat.com/gmat_study_ ... ial_powersand: "Note: the case of 0^0 is not tested on the GMAT." http://www.manhattangmat.com/npexponents.cfm
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Re: Good set of DS 3 [#permalink]
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15 Sep 2010, 09:06
jax91 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
ab > 0? a is always +ve. So we need to know if b is +ve or ve. 1.) mod of any number is +ve. Insuff. 2.) a^b is an integer. we know a and b are integers. so a is a +ve integer. any +ve integer raised to a ve integer will give us a fraction. e.g. 4 ^ 3 = 1/ (4^3) which will never be an integer. so for a^b to be an integer b has to be +ve. So its suff. So B. Hi,Can anyone explain me how a is positive?We generally look at the sign of any number before deciding its abs value..Right?But how come that is not followed here? Generally we take x=x (x <0) x=x (x>0) but y is a taken positive directly?In the question,it is just mentioned as an integer Explanation plz..



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Re: Good set of DS 3 [#permalink]
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15 Sep 2010, 09:16
ravitejapandiri wrote: Hi,Can anyone explain me how a is positive?We generally look at the sign of any number before deciding its abs value..Right?But how come that is not followed here? Generally we take x=x (x <0) x=x (x>0) but y is a taken positive directly?In the question,it is just mentioned as an integer Explanation plz.. Seems that you need to brush up on absolute value. Please check Walker's post on Absolute Value at: mathabsolutevaluemodulus86462.html\(x=x\), when \(x<0\) and \(x=x\) when \(x>0\) > CORRECT. But when \(x=negative<0\) then \(x=x=(negative)=positive\) and when \(x=positive>0\) then \(x=x=positive\), so in any case when \(x\) is either positive or negative \(x\) is still positive. There is one more case though: when \(x=0\) then \(x=0\). So generally we can say that absolute value of an expression is alway nonnegative: \(some \ expression\geq{0}\) > \(x\geq{0}\). As for this particular question: see my posts on it on pages 1, 2, and 3. Hope it helps.
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Re: Good set of DS 3
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15 Sep 2010, 09:16



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