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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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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.

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, 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.

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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Re: Good set of DS 3 [#permalink]

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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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Re: Good set of DS 3 [#permalink]

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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 $$a$$ can not 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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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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Re: Good set of DS 3 [#permalink]

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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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Re: Good set of DS 3 [#permalink]

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16 Sep 2010, 11:30
Thanx Bunuel!Some good questions
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Re: Good set of DS 3 [#permalink]

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19 Dec 2010, 17:08
yangsta8 wrote:
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

I selected that both stmts were sufficient together because the examples I chose worked both ways so if x=-2 y=3 you get 3-(-2)=5 or if y=2 than x = (-1) so 2-(-1)=3. Can the insufficiency only be seen with fractional numbers? Thanks.
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Re: Good set of DS 3 [#permalink]

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19 Dec 2010, 17:33
gettinit wrote:
yangsta8 wrote:
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

I selected that both stmts were sufficient together because the examples I chose worked both ways so if x=-2 y=3 you get 3-(-2)=5 or if y=2 than x = (-1) so 2-(-1)=3. Can the insufficiency only be seen with fractional numbers? Thanks.

Yes, in order to get NO answer you should choose values of x and y from (0, 1), note that this range can give you the YES answer as well.

Is $$y-x>0$$? --> is $$y>x$$?

(1) y > 0, not sufficient as no info about $$x$$.
(2) x = 1 - y --> $$x+y=1$$ --> the sum of 2 numbers equal to 1 --> we can not say which one is greater. Not sufficient.

(1)+(2) $$x+y=1$$ and $$y>0$$, still can not determine which one is greater: if $$y=0.1>0$$ and $$x=0.9$$ then $$y<x$$ but if $$y=0.9>0$$ and $$x=0.1$$ then $$y>x$$. Not sufficient.

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Re: Good set of DS 3 [#permalink]

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23 Dec 2010, 19:07
thanks Bunuel should have simplified stmt 2 better to understand the fractional solutions. Thank you sir.
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Re: Good set of DS 3 [#permalink]

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31 Jan 2011, 06:33
Hi Bunuel

For question # 8, please explain how this is true :

either 4y=32 y=8 x=-3, xy=-24 OR -4y=32

Regards,
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Re: Good set of DS 3 [#permalink]

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31 Jan 2011, 06:48
subhashghosh wrote:
Hi Bunuel

For question # 8, please explain how this is true :

either 4y=32 y=8 x=-3, xy=-24 OR -4y=32

Regards,
Subhash

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.

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Re: Good set of DS 3 [#permalink]

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31 Jan 2011, 19:10
Thanks a lot, I can understand it fully now.
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Re: Good set of DS 3 [#permalink]

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23 Feb 2011, 01:13
For question number 3, consider this:

1st number = 10w+x and 2nd number = 10c+x so three digit number = 100*(wc)+10*(w+c)+x^2

In 1, We are given, wc=x*(w+c)=x^2

or x*(w+c-x) = 0 implying either x=0 or w+c-x = 0
Since it is given that x is non-zero, we have w+c-x = 0 so 1 is sufficient
w+c and x being odd just says that w+c-x is even number, so insufficient
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Re: Good set of DS 3 [#permalink]

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07 Aug 2011, 11:53
Quote:
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.

is true when and does not equal to zero.

(1) --> does not equal to zero, but we don't know about , it can be any value, positive or negative. Not sufficient.

(2) is a non-zero integer --> can be 1 and 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.

Thanks for the explanation, Bunuel..
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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20 Apr 2012, 21:10
Bunuel wrote:
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

Bunuel, Is this approach correct?

sum of n consecutive +ve integers = $$\frac{n(n+1)}{2}$$

so, $$\frac{n(n+1)}{2}$$ = 45
n(n+1) = 90
$$n^{2}+ n - 90 = 0$$
solving this equation gives n= -10 or 9

Both A and B doesn't help. Hence E.
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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20 Apr 2012, 21:16
ENAFEX wrote:
Bunuel wrote:
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

Bunuel, Is this approach correct?

sum of n consecutive +ve integers = $$\frac{n(n+1)}{2}$$

so, $$\frac{n(n+1)}{2}$$ = 45
n(n+1) = 90
$$n^{2}+ n - 90 = 0$$
solving this equation gives n= -10 or 9

Both A and B doesn't help. Hence E.

Bunuel, Ignore my request. I found out the mistake. The question does not say sum of first consecutive integers, so, I cant use this formula.
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Re: Good set of DS 3 [#permalink]

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02 May 2012, 08:11
Hi Bunuel/Karishma,
Request you to clarify the below colored part. What difference does this makes in the question?
Thanks
H
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

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

.

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Re: The sum of n consecutive positive integers is 45 [#permalink]

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29 Jul 2012, 04:31
1. The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is even
(2) n < 9

Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45
n^2 +n -90 = 0
n=9 or -10
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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29 Jul 2012, 05:02
dvinoth86 wrote:
1. The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is even
(2) n < 9

Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45.

[/color]
n^2 +n -90 = 0
n=9 or -10

Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45.

This formula gives the sum of the first n consecutive positive integers: 1,2 3, ..., n.
Nowhere is stated that our sequence starts with 1.
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Re: The sum of n consecutive positive integers is 45   [#permalink] 29 Jul 2012, 05:02

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