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For integers a and b, if square root of (a^3 - a^2 - b) = 7, [#permalink]
 22 Oct 2005, 22:22
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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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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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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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Director
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Titleist wrote: Ditto GSR
I go with C
Eagerly awaiting your next 'weekly' Avtar!!
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Director
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gsr wrote: Titleist wrote: Ditto GSR
I go with C Eagerly awaiting your next 'weekly' Avtar! Is it not "pretty" enough?
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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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Director
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Titleist wrote: What's wrong with the one I have now? 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! 
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!! 
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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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Director
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gsr wrote: Titleist wrote: What's wrong with the one I have now? 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! 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!! 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. 
Okay good night guys 2:32am my time!  :
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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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Director
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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  True! Good Catch!
Laxie, I suggest you take the GMAT tomorrow
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Director
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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 True! Good Catch! Laxie, I suggest you take the GMAT tomorrow Okay I'm definitely hitting the sack - good catch Laxi.
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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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Director
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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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VP
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Nice Laxie...fei chang hao
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Director
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Antmavel wrote: Nice Laxie...fei chang hao Though i have no clue what you meant....il surely say this.....nee hao? 
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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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Titleist wrote: gsr wrote: Titleist wrote: What's wrong with the one I have now? 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! 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!! 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. Okay good night guys 2:32am my time! : 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.
awaiting eagerly.....
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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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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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Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
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