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For integers, a and b, if sqrt(a^3 - a^2 - b) = 7, what is

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For integers, a and b, if sqrt(a^3 - a^2 - b) = 7, what is [#permalink] New post 05 Oct 2004, 17:52
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For integers, a and b, if sqrt(a^3 - a^2 - b) = 7, what is the value of a?
I) a^2 - a = 12
II) b^2 - b = 2
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 [#permalink] New post 05 Oct 2004, 18:37
1.a=4 or-3. Two values for a and hence insuff.
2. b can be 2 or -1. Applying the values of b in a^3-a^2-b=49, we have a as either some fraction or a=4. Since a need to be an integer, a is 4.

hence B.
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 [#permalink] New post 06 Oct 2004, 12:55
Quote:
Applying the values of b in a^3-a^2-b=49, we have a as either some fraction or a=4. Since a need to be an integer, a is 4.


Please explain your method in solving this. My calc with polynomial equations is hazy. Thanks.
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 [#permalink] New post 06 Oct 2004, 13:05
I think it is 49 = a.(a^2-a)-b so a = (49+b)/12

knowing that (stat. 2) b = 2 or -1, a being an integer, b = -1 and a = 4
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 [#permalink] New post 06 Oct 2004, 14:18
I think C
1 gives 12-b=49
2 gives quadratic b=-1 and 2
Only b=-1, gives a as integer=4
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 [#permalink] New post 06 Oct 2004, 15:01
I think you just prove statement 2 is suff !
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 [#permalink] New post 06 Oct 2004, 15:23
how do you solve
a^3-a^2-48=0 to find the value of a to conclude B as answer?
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 [#permalink] New post 06 Oct 2004, 21:49
venksune wrote:
1.a=4 or-3. Two values for a and hence insuff.
2. b can be 2 or -1. Applying the values of b in a^3-a^2-b=49, we have a as either some fraction or a=4. Since a need to be an integer, a is 4.

hence B.


Hi Venk, I see that your solution for a in statement 1 neglects the original statement which states that if sqrt(a^3 - a^2 - b) = 7, what is the value of a, You solve statement 1 independ of the original equation and that does not seem right.
At best, from the original equation we can say that
a^3 - a^2 - b = 49
Now given 1, a2-a=12 implies that 12a -b =49 from (a(a^2-a) -b =49, making statement 1 insufficient.
Statement 2 now gives you two values of b (2, -1) which when substitude into the original equation yield 2 values, one an integer of 4 and the other a non integer. Thus for a to be an integer it has to be 4. B is sufficient
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 [#permalink] New post 06 Oct 2004, 21:53
saurya_s wrote:
how do you solve
a^3-a^2-48=0 to find the value of a to conclude B as answer?
S


It is posible, try picking numbers adnd you will land at a = 4 or a = 3.25.
It is easy to fall into the trap that
statement 1 combine with the original statement yields
12a -b = 49 and thus the moment one gets b, one can find a making C the dangerous answer. But if you substitue the value of B calculated and substitude into the original equation you might get only one integer for a and that is 4.
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 [#permalink] New post 06 Oct 2004, 21:57
sdanquah,
I am also saying only B is sufficient. In stem 1, we have two values of 'a' - question is what is the value of 'a'. We cant conclusively say the value of 'a'. Also, when applied to the original equation, we still need 'b'. I didnt neglect it, I thought the added data is not necessary to prove insufficiency of stem 1.
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 [#permalink] New post 06 Oct 2004, 22:27
for the second we have
b=-1,2

for b=-1 , sqrt(a^3-a^2+1)=7. implies a^3-a^2+1=49

or a^3 - a^2 -48 = 0
a = 4 is a solution
divide a^3-a^2-48 by a-4 gives the polynomial a^2+3a+12

so (a^2+3a+12)(a-4)=0
a^2+3a+12 is never zero since discriminant is negative (no roots for a) :-D
giving a =4


for b = 2

we get a^3-a^2=51
a^2(a-1)=51

from this we cannot find any value of A


hope this helps.
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 [#permalink] New post 07 Oct 2004, 06:04
OA is B. From Princeton.
OE = "Statement 2 gives b = 2, -1. Since a cant be an interger if b = 2, it must be true that b = -1 and a = 4."
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 [#permalink] New post 07 Oct 2004, 07:12
venksune wrote:
sdanquah,
I am also saying only B is sufficient. In stem 1, we have two values of 'a' - question is what is the value of 'a'. We cant conclusively say the value of 'a'. Also, when applied to the original equation, we still need 'b'. I didnt neglect it, I thought the added data is not necessary to prove insufficiency of stem 1.


Agreed with your answer venksume, Just that I found that you neglected orignial statement in your earlier solution.
  [#permalink] 07 Oct 2004, 07:12
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