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(sqrt)abc=504, is b divisible by 2?

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(sqrt)abc=504, is b divisible by 2? [#permalink] New post 19 Sep 2007, 12:46
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A
B
C
D
E

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(sqrt)abc=504, is b divisible by 2?

(1) c = 168

(2) a is a perfect square
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Re: DS [#permalink] New post 19 Sep 2007, 13:02
12345678 wrote:
(sqrt)abc=504, is b divisible by 2?

(1) c = 168

(2) a is a perfect square


I get E.

c=168 insuff because we dont know about a .

a being perfect sqaure could be again anything .
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Re: DS [#permalink] New post 19 Sep 2007, 15:53
12345678 wrote:
(sqrt)abc=504, is b divisible by 2?

(1) c = 168

(2) a is a perfect square


I get C, combined 1 and 2 sufficient

sqrt:abc=504
or
abc=504*504

now 1: c=168 <=> ab=3*504 so not suff. we need info about a,
now2: a is perfect square, doesnt give anything,
then combine

abc=504^2
168ab=504*504
ab=3*504
ab=9*168 a can be 9, then b=168 or 6 so divisible by 2
or
ab=4*378 so a can be 4 then b is divisible by 2
or
ab=36*42 a can be 36 then b=42 so divisible by 2

C
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Re: DS [#permalink] New post 20 Sep 2007, 07:09
Ravshonbek wrote:
12345678 wrote:
(sqrt)abc=504, is b divisible by 2?

(1) c = 168

(2) a is a perfect square


I get C, combined 1 and 2 sufficient

sqrt:abc=504
or
abc=504*504

now 1: c=168 <=> ab=3*504 so not suff. we need info about a,
now2: a is perfect square, doesnt give anything,
then combine

abc=504^2
168ab=504*504
ab=3*504
ab=9*168 a can be 9, then b=168 or 6 so divisible by 2
or
ab=4*378 so a can be 4 then b is divisible by 2
or
ab=36*42 a can be 36 then b=42 so divisible by 2

C


I think this fine if you assume a b c are all integers. What if a = 81 or 64 ?
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 [#permalink] New post 20 Sep 2007, 07:35
I do agree with Ravshonbek, C

ba=504*3 = 7*2^3*3^3

If I perfect square means just squared number, than even if you take 3^2 or 2^2 there is always left at least one 2 which will be enough to conclude that b is divisible by 2.

Ans: C

And one more thing I think 81 and 64 is not possible to derive.
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 [#permalink] New post 20 Sep 2007, 10:21
Ferihere wrote:
I do agree with Ravshonbek, C

ba=504*3 = 7*2^3*3^3

If I perfect square means just squared number, than even if you take 3^2 or 2^2 there is always left at least one 2 which will be enough to conclude that b is divisible by 2.

Ans: C

And one more thing I think 81 and 64 is not possible to derive.


I dont understand what you mean by 81 and 64 is not possible to derive.

Stmt 1: c= 168
Stmt2: a is a perfect square

So if abc=504^2 or 254,016
a=81 b=18 2/3 and c=168 would be a valid soloution where b is not divisible by 2.
  [#permalink] 20 Sep 2007, 10:21
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