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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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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:

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.

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:

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.

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(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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26 Oct 2009, 01:12
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.

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 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.

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 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.

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!!!

-h2polo
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Re: Good set of DS 3 [#permalink]

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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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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

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
2
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jan4dday wrote:

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.

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 |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.

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 |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.

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 |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.

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] 02 Jan 2010, 11:58

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