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For integers a and b, if square root of (a^3 - a^2 - b) = 7,

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For integers a and b, if square root of (a^3 - a^2 - b) = 7, [#permalink] New post 22 Oct 2005, 21:22
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A
B
C
D
E

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For integers a and b, if square root of (a^3 - a^2 - b) = 7, what is the value of a?

A. a^2 - a = 12
B. b^2 - b = 2
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 [#permalink] New post 22 Oct 2005, 21:31
C

A) Insuff. Has 2 solutions. a=4, -3 (corresponding b=-1, -85)
B) Insuff. b=2, -1

Combining, we can say a=4
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 [#permalink] New post 22 Oct 2005, 21:49
from A,
a= -3 or 4
Substitute a to get b = 85, -1 ..both are integers. Thus, both values of a are ok==> still can't we determine what the value of a is.
--> insuff.


from B, b= -1 or 2
b=-1, a^3-a^2 = 49+(-1) = 48
a^3-a^2-48= > a =4
for b= 2, we find out other value of a
---> insuff

combine A and B, we find out a= 4
thus, C.

Sorry, edit: when b=2, we have a^3-a^2-51=0
There's a trick, the integer outcome of the equation can only be among factors of 51, that is to say +,- 1 ; +- 51, +- 3, +-17. As we notice that the LHS of equation involves all - , thus, negative factors of 51 can't be the solution. In fact ,there's ways for you to narrow down instead of trying these factors each by each.
Try these out and you get no integer results for a.
Thus, when b=2, there's no a.
the only value of a is 4 when b=-1
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 [#permalink] New post 22 Oct 2005, 22:21
Titleist wrote:
Ditto GSR

I go with C


Eagerly awaiting your next 'weekly' Avtar!!
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 [#permalink] New post 22 Oct 2005, 22:23
gsr wrote:
Titleist wrote:
Ditto GSR

I go with C


Eagerly awaiting your next 'weekly' Avtar!![/quote

What's wrong with the one I have now? :shock: Is it not "pretty" enough?
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 [#permalink] New post 22 Oct 2005, 22:28
Titleist wrote:
What's wrong with the one I have now? :shock: Is it not "pretty" enough?


Nothing is wrong with this one. Just that I want to see more awesome ones. Frankly, your avtars make this place a whole lot more lively! :cool

Infact, I thought your earlier avatar was your true picture! Now I know it isn't. Seeing this new one only makes me laugh everytime!! :lol:
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 [#permalink] New post 22 Oct 2005, 22:30
Titleist wrote:
Ditto GSR

I go with C


but when b=2, the equation a^3- a^2- 2= 49 or a^3-a^2-51= 0 has no integer solutions for a!
Thus, we get only a=4 when b=-1 -----> B is suff
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 [#permalink] New post 22 Oct 2005, 22:39
gsr wrote:
Titleist wrote:
What's wrong with the one I have now? :shock: Is it not "pretty" enough?


Nothing is wrong with this one. Just that I want to see more awesome ones. Frankly, your avtars make this place a whole lot more lively! :cool

Infact, I thought your earlier avatar was your true picture! Now I know it isn't. Seeing this new one only makes me laugh everytime!! :lol:


Well, for some reason I can't upload my own custom avatar - I think the admin might have blocked that feature. We can only use what's in the upload bank. When I found this picture on the upload bank, needless to say, I couldn't resist.
Obviously, it's definitely not advisable to put your personal mugshots on public forums because of hams like me. :wink:

Okay good night guys 2:32am my time! :sleep:
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"Wow! Brazil is big." —George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

http://www.nytimes.com/2005/11/21/inter ... prexy.html

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 [#permalink] New post 22 Oct 2005, 22:39
laxieqv wrote:
Titleist wrote:
Ditto GSR
I go with C

but when b=2, the equation a^3- a^2- 2= 49 or a^3-a^2-51= 0 has no integer solutions for a!
Thus, we get only a=4 when b=-1 -----> B is suff


:madd True! Good Catch!
Laxie, I suggest you take the GMAT tomorrow :lol:
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 [#permalink] New post 22 Oct 2005, 22:42
gsr wrote:
laxieqv wrote:
Titleist wrote:
Ditto GSR
I go with C

but when b=2, the equation a^3- a^2- 2= 49 or a^3-a^2-51= 0 has no integer solutions for a!
Thus, we get only a=4 when b=-1 -----> B is suff


:madd True! Good Catch!
Laxie, I suggest you take the GMAT tomorrow :lol:


Okay I'm definitely hitting the sack - good catch Laxi.

:P
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"Wow! Brazil is big." —George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

http://www.nytimes.com/2005/11/21/inter ... prexy.html

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 [#permalink] New post 22 Oct 2005, 22:42
True, Titleist! Don't take this one off unless you find a better one! I guess we don't have any better ones on the forum!
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 [#permalink] New post 23 Oct 2005, 00:13
Nice Laxie...fei chang hao :lol:
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 [#permalink] New post 23 Oct 2005, 02:33
Antmavel wrote:
Nice Laxie...fei chang hao :lol:


Though i have no clue what you meant....il surely say this.....nee hao? :)
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 [#permalink] New post 23 Oct 2005, 10:51
hmm...I am not sure about B cause

if we just look at b and ignore anything we know about a, then we still dont know if -2 or 1 for B will work or not...

It only makes sense that b=1 when we know what a^3-a^2 simplifies too?

laxieqv wrote:
Titleist wrote:
Ditto GSR

I go with C


but when b=2, the equation a^3- a^2- 2= 49 or a^3-a^2-51= 0 has no integer solutions for a!
Thus, we get only a=4 when b=-1 -----> B is suff
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 [#permalink] New post 23 Oct 2005, 11:23
Titleist wrote:
gsr wrote:
Titleist wrote:
What's wrong with the one I have now? :shock: Is it not "pretty" enough?

Nothing is wrong with this one. Just that I want to see more awesome ones. Frankly, your avtars make this place a whole lot more lively! :cool
Infact, I thought your earlier avatar was your true picture! Now I know it isn't. Seeing this new one only makes me laugh everytime!! :lol:

Well, for some reason I can't upload my own custom avatar - I think the admin might have blocked that feature. We can only use what's in the upload bank. When I found this picture on the upload bank, needless to say, I couldn't resist.
Obviously, it's definitely not advisable to put your personal mugshots on public forums because of hams like me. :wink:
Okay good night guys 2:32am my time! :sleep:

mr./ms. Titleist,

after going through the posts above, i also couldnot stop myself from asking you for your awsome and "old is gold" avatar. :wink:

awaiting eagerly..... :P
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 [#permalink] New post 23 Oct 2005, 18:11
laxieqv wrote:
but when b=2, the equation a^3- a^2- 2= 49 or a^3-a^2-51= 0 has no integer solutions for a!
Thus, we get only a=4 when b=-1 -----> B is suff


a^3-a^2-b=49
a^2(a-1)=49+b
We know that a(a-1) is even, so 49+b has to be even. That's why we can determine that b=-1. Then we write 49+(-1) = 48 = 4*4*3 and determine a=4.

You could stop here. But if you want to think further, you can.
This is a polynomial equation of the third order, so we have to be able to decide whether the other two solutions are integers. We know a-1>0, in other words a>1. Also a has to be even because if a is odd then a^2 is odd and a^2(a-1)/2 is odd, which cannot be 48. Obviously a has to be smaller than 7 because 7^2=49>48. So we only need to try a=2 and a=6, which can be easily shown that they are not one of the solutions.
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 [#permalink] New post 23 Oct 2005, 18:15
HongHu wrote:
Also a has to be even because if a is odd then a^2 is odd and a^2(a-1)/2 is odd, which cannot be 48.


Actually this is not right, but is ok to use for a<7. Also better change it to be a^2(a-1)/4 is odd.
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  [#permalink] 23 Oct 2005, 18:15
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