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The sum of n consecutive positive integers is 45

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The sum of n consecutive positive integers is 45 [#permalink] New post 16 Oct 2009, 18:59
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Please find below new set of DS problems:

TIP: many of these problems act in GMAT zone, so beware of ZIP trap.

1. The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is even
(2) n < 9

2. Is a product of three integers XYZ a prime?
(1) X=-Y
(2) Z=1

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.

4. Is y – x positive?
(1) y > 0
(2) x = 1 – y

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

6. If M and N are integers, is (10^M + N)/3 an integer?
1. N = 5
2. MN is even

7. If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value of c?
(1) d = 3
(2) b = 6

8. If x and y are non-zero integers and |x| + |y| = 32, what is xy?
(1) -4x - 12y = 0
(2) |x| - |y| = 16

9. Is the integer n odd
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

10. The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is odd
(2) n >= 9

Please share your way of thinking, not only post the answers.

OA and explanations to follow.


Also you can check new set of PS problems: good-set-of-ps-85414.html
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Last edited by Bunuel on 16 Oct 2009, 21:33, edited 1 time in total.
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Re: Good set of DS 3 [#permalink] New post 16 Oct 2009, 21:14
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.
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Re: Good set of DS 3 [#permalink] New post 16 Oct 2009, 21:55
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Bunuel wrote:
TIP: many of these problems act in GMAT zone, so beware of ZIP trap.

1. The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is even
(2) n < 9

2. Is a product of three integers XYZ a prime?
(1) X=-Y
(2) Z=1


What is the ZIP trap?

Q1)
Statement 1) n = 2,4,6 etc
n = 2 => x+(x+1)=45 => x=22 (works)
n = 4 => x+(x+1)+(x+2)+(x+3)=45 => 4x+6=45 => x=39/4 (doesn't work)
n = 6 => Take above equation+(x+4)+(x+5) => 6x+15=45 => x=5 (works)
Not suff.
Statement 2) n < 9. This is proven insufficient from the working above since both n=2 and n=6 n<9.
1 and 2 together still prove insufficient due to above working.

ANS = E.

Q2)
Statement 1) X=-Y
This means Z needs to be negative and for XYZ to have a chance of being prime. Z can be anything.
Insufficient.
Statement 2) Z=1
X and Y could be anything such as 2 and 3 (non prime multiple) or 1 and 2 (prime).
Insufficient.
1 and 2 Together) Z = 1. X=-Y
1*Y*(-Y) = -Y^2 which cannot be prime as it is negative.

ANS = C

Edited: Got the right working but wrote E instead of C. I gotta stop doing that :(

Last edited by yangsta8 on 16 Oct 2009, 22:03, edited 1 time in total.
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Re: Good set of DS 3 [#permalink] New post 16 Oct 2009, 22:01
Bunuel wrote:
2. Is a product of three integers XYZ a prime?

(1) X=-Y
(2) Z=1


(1) If x=-y = 2 and z = -1, yes. Otherwise, no..
(2) If z = 1, x could be 2 and y = 1. xyz is a price. If something else, no.

From 1 and 2: x = -y and z = 1, xyz is always a -ve integer, which cannot be a prime....C.
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Re: Good set of DS 3 [#permalink] New post 16 Oct 2009, 22:03
Bunuel wrote:
4. Is y – x positive?
(1) y > 0
(2) x = 1 – y


Statement 1) y>0 Not suff, X could be anything larger or smaller than X.
Statement 2) x=1-y
x+y=1
Let x=3 and y=-2 then y-x < 0.
But if x=1/4 and y=3/4 then y-x >0
Not suff.

1 and 2 together)
From the example above we have:
if x=1/4 and y=3/4 then y-x >0
but if we flip it around:
if x=3/4 and y=1/4 then y-x <0
not suff.

ANS = E
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Re: Good set of DS 3 [#permalink] New post 16 Oct 2009, 22:13
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Bunuel wrote:
6. If M and N are integers, is (10^M + N)/3 an integer?
1. N = 5
2. MN is even


Statement 1) N=5
If M>=0 then it is always divisible by 3. Since the number will always consist of 1, trailing 0's and a 5. Of which the sum of digits =6 which is the rule for divisibility by 3.
If M<0 then the equation is not divisble by 3. For example if M=-1.
Insufficient

Statement 2) MN is even. Again this means M could still be negative so insufficient. For example M could be -1 and N could be 2 which is not divisible by 3. Or n=5 but m=-2 which is not.

Statements together) Still insuff. m=2 n=5 works. But m=-2 n=5 doesn't work.

ANS = E
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Re: Good set of DS 3 [#permalink] New post 16 Oct 2009, 22:17
Bunuel wrote:
7. If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value of c?
(1) d = 3
(2) b = 6


Expanding it out we get :
x^2 + bx + c = x^2 + 2dx + d^2

Statement 1) d = 3
d^2 = c = 9
Suff.
Statement 2) b=6
b=6=2d
d=3
d^2=c=9
Suff

ANS = D
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Re: Good set of DS 3 [#permalink] New post 16 Oct 2009, 22:41
Bunuel wrote:
9. Is the integer n odd
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

10. The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is odd
(2) n >= 9



Q9)
Statement 1)
N is a multiple of 3. N could be 3 or 6.
Insufficient.
Statement 2)
I am not sure how to prove this except by examples:
Example 1:n=9 factors={1,3,9}, 2n=18 factors={1,2,3,6,9,18}
N is odd is true.
Example 2:n=6 factors={1,2,3,6} 2n=12 factors={1,2,3,4,6,12} Does not have twice as many factors.
Example 3: n=3 factors={1,3} 2n=6 factors={1,2,3,6}
N is odd is true.

ANS = B

Q10)
Statement 1) N is odd.
N could be 1. 45
N could also be 3. x+(x+1)+(x+2)=45 => 3x=42 x=14
Insufficient.
Statement 2) N>=9
Let n=9.
9x+8+7+6+5+4+3+2+1=45 => 9x+36=45 => 9x=9 x=1
we cannot use n>10 because adding anymore positive integers means sum > 45.
Sufficient.

ANS = B
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 00:09
Hi Bunuel, Awesome questions...keep them coming esp DS :)
+1 to you.
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 01:50
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Bunuel wrote:
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.



WX x CX = IJK

1.) I,J,K are the same and not equal to W,C or X.

so 3 digit numbers with all digit same are 111,222,...., 999.

basically multiples of 111 (37x3).

so we get 1 number = 37

conditions the second number has to meet = last digit = 7, multiple of 3, double digit.

so we get 27.

27 x 37 = 999

So suff.

2.) x and w+c are odd.

this gives multiple values.

So A.
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 01:57
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.
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 02:06
Bunuel wrote:
8. If x and y are non-zero integers and |x| + |y| = 32, what is xy?
(1) -4x - 12y = 0
(2) |x| - |y| = 16


1.) -4x = 12y
or -x = 3y

so we get x and y to be = 24,-8 or -24,8

xy = -192 in both cases

so suff.

2.) |x| - |y| = 16

we can get |x| and |y|.

but the signs of x and y cannot be determined. So insuff.

IMO A.
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 04:38
Great questions, keep em coming ..
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 09:24
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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.



This is not correct.
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 09:29
Expert's post
jax91 wrote:
Bunuel wrote:
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.



WX x CX = IJK

1.) I,J,K are the same and not equal to W,C or X.

so 3 digit numbers with all digit same are 111,222,...., 999.

basically multiples of 111 (37x3).

so we get 1 number = 37

conditions the second number has to meet = last digit = 7, multiple of 3, double digit.

so we get 27.

27 x 37 = 999

So suff.

2.) x and w+c are odd.

this gives multiple values.

So A.


Though answer is right and one number indeed is 37, but it was not the only possibility for it. So you got the right answer with right numbers, but missed one case to consider.
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 09:33
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

Basically the question asks, is b>0?

stmt1: b can be -ve or +ve or 0. Insuff.
stmt2: b can be +ve or 0. Insuff.
Combining, b can be +ve or 0. Insuff.

E.
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 16:29
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ANSWERS:

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=2 --> 22+23=45, n=4 --> n=6 x1+(x1+1)+(x1+2)+(x1+3)+(x1+4)+(x1+5)=45 x1=5. At least two options for n. Not sufficient.
(2) n<9 same thing not sufficient.
(1)+(2) No new info. Not sufficient.

Answer: E.


2. Is a product of three integers XYZ a prime?
(1) X=-Y
(2) Z=1

(1) x=-y --> for xyz to be a prime z must be -p AND x=-y shouldn't be zero. Not sufficient.
(2) z=1 --> Not sufficient.
(1)+(2) x=-y and z=1 --> x and y can be zero, xyz=0 not prime OR xyz is negative, so not prime. In either case we know xyz not prime.

Answer: C


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. 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.


4. Is y – x positive?
(1) y > 0
(2) x = 1 – y

Easy one even if y>0 and x+y=1, we can find the x,y when y-x>0 and y-x<0
Answer: E.


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.


6. If M and N are integers, is (10^M + N)/3 an integer?
(1) N = 5
(2) MN is even

Note: it's not given that M and N are positive.
(1) N=5 --> if M>0 (10^M + N)/3 is an integer ((1+5)/3), if M<0 (10^M + N)/3 is a fraction ((1/10^|M|+5)/3). Not sufficient.
(2) MN is even --> one of them or both positive/negative AND one of them or both even. Not sufficient
(1)+(2) N=5 MN even --> still M can be negative or positive. Not sufficient.

Answer: E.


7. If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value of c?
(1) d = 3
(2) b = 6

Note this part: "for all values of x"
So, it must be true for x=0 --> c=d^2 --> b=2d
(1) d = 3 --> c=9 Sufficient
(2) b = 6 --> b=2d, d=3 --> c=9 Sufficient

Answer: D.


8. If x and y are non-zero integers and |x| + |y| = 32, what is xy?

(1) -4x-12y=0 --> x=-3y --> x and y have opposite signs.

So either: |x|=x and |y|=-y --> in this case |x|+|y|=x-y=-3y-y=-4y=32: y=-8, x=24, xy=-24*8;

OR: |x|=-x and |y|=y --> |x|+|y|=-x+y=3y+y=4y=32 --> y=8 and x=-24 --> xy=-24*8, the same answer.

Sufficient.

(2) |x| - |y| = 16. Sum this one with th equations given in the stem --> 2|x|=48 --> |x|=24, |y|=8. xy=-24*8 (x and y have opposite sign) or xy=24*8 (x and y have the same sign). Multiple choices. Not sufficient.

Answer: A.


9. Is the integer n odd
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

(1) 3 or 6. Clearly not sufficient.
(2) TIP:
When odd number n is doubled, 2n has twice as many factors as n.
Thats because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2.

Sufficient.

Answer: B.


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.
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 16:44
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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.
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 17:48
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Last edited by Bunuel on 17 Oct 2009, 19:40, edited 1 time in total.
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Re: Good set of DS 3 [#permalink] New post 17 Oct 2009, 18:14
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. 8-)

You can say PINZF (or better) trap as well:

P = positive
I = integer
N = negative
Z = zero
F = fraction
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Re: Good set of DS 3   [#permalink] 17 Oct 2009, 18:14
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