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The sum of n consecutive positive integers is 45

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Re: Good set of DS 3 [#permalink]

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New post 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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New post 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 :lol:

Stmt 1 and 2 (D) are both sufficient.

Does GMAT assume that we would not indulge in any formula? Just curious :roll:

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Re: Good set of DS 3 [#permalink]

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New post 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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New post 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 :lol:

Stmt 1 and 2 (D) are both sufficient.

Does GMAT assume that we would not indulge in any formula? Just curious :roll:


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(n-1))/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+n-1)/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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New post 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 non-zero digits, the product is a three digit number. What is w+c-x?

(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+c-x 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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New post 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!!!

-h2polo

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Re: Good set of DS 3 [#permalink]

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New post 08 Nov 2009, 05:24
h2polo wrote:
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!!!

-h2polo


These questions are from my collection of questions which I collected from the web. So, no specific book...
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Re: Good set of DS 3 [#permalink]

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New post 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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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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New post 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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New post 02 Jan 2010, 09:20
Bunuel wrote:
5. If a and b are integers, and a not= b, is |a|b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer

This is tricky |a|b > 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 non-zero 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? :oops:
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Re: Good set of DS 3 [#permalink]

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New post 02 Jan 2010, 09:24
jeeteshsingh wrote:
Bunuel wrote:
5. If a and b are integers, and a not= b, is |a|b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer

This is tricky |a|b > 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 non-zero 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? :oops:


\(a\) does not equal to \(b\). Sorry for confusion.
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New post 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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New post 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 |a|b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer.

\(|a|b > 0\) 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 non-zero integer --> \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient.

(1)+(2) If a=1 and b=2, then |a|b > 0, but if a=1 and b=-2, then |a|b <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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New post 03 Jan 2010, 03:14
Bunuel wrote:

If a and b are integers, and a not= b, is |a|b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer.

\(|a|b > 0\) 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 non-zero integer --> \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient.

(1)+(2) If a=1 and b=2, then |a|b > 0, but if a=1 and b=-2, then |a|b <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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New post 03 Jan 2010, 09:23
jeeteshsingh wrote:
Bunuel wrote:

If a and b are integers, and a not= b, is |a|b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer.

\(|a|b > 0\) 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 non-zero integer --> \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient.

(1)+(2) If a=1 and b=2, then |a|b > 0, but if a=1 and b=-2, then |a|b <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 non-zero integer " to hold true. So \(b\) can take positive as well as negative values, hence \(|a|b > 0\) 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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New post 18 Feb 2010, 13:38
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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New post 18 Feb 2010, 14:07
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_powers

and:
"Note: the case of 0^0 is not tested on the GMAT."
http://www.manhattangmat.com/np-exponents.cfm
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Re: Good set of DS 3 [#permalink]

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New post 15 Sep 2010, 10:06
jax91 wrote:
Bunuel wrote:
5. If a and b are integers, and a not= b, is |a|b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer


|a|b > 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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New post 15 Sep 2010, 10:16
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: math-absolute-value-modulus-86462.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 non-negative: \(|some \ expression|\geq{0}\) --> \(|x|\geq{0}\).

As for this particular question: see my posts on it on pages 1, 2, and 3.

Hope it helps.
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