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If x= (a/2) + (b/2^3) + (c/2^4) , where a,b, and c are each

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If x= (a/2) + (b/2^3) + (c/2^4) , where a,b, and c are each [#permalink]

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New post 24 Jun 2005, 05:29
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If x= (a/2) + (b/2^3) + (c/2^4) , where a,b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT:

a) 1/16
b) 3/16
c) 5/16
d) 10/16
e) 11/16


fastest way to solve this?
.
.
.
.
.
OA is C

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Re: PS [#permalink]

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New post 24 Jun 2005, 06:50
considering the denomenator, c/2^4 cannot be 0. so the possible numerator value for any combination of a, b, and c are: 0, 1, 3, 9, 10, 11.

5 as numerator cannot be possible so this is the answer..

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New post 24 Jun 2005, 07:41
I don't get it. :oops: why can't c/16 =0? and how did you get the possible numerators?

thanks!

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New post 24 Jun 2005, 08:43
I dont know of a short way, but I started plugging in...eventually realize 5/16 cannot be possible..you have 5/8, which can 10/16 but not 5/16!

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New post 24 Jun 2005, 14:07
break them into 1/2 + 1/8 + 1/16
=> ( 8 + 2 + 1) / 16

u cannot get 5 by adding the numerators. so u can figure it out just by looking at the answers. More formally, you can do this:

a,b,c
0,0,0 -> 0
0,0,1 -> 1/16
0,1,0 -> 2/16
0,1,1 -> 3/16
1,0,0 -> so forth
1,0,1
1,1,0
1,1,1


--Andrew

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Re: PS [#permalink]

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New post 26 Jun 2005, 06:09
gumbico wrote:
If x= (a/2) + (b/2^3) + (c/2^4) , where a,b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT:

a) 1/16
b) 3/16
c) 5/16
d) 10/16
e) 11/16



x = (8a + 2b + c)/16,
and since a,b,c are either 0 or 1.
the numerator can never be equal to 5.
Hence C.

HMTG.

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 [#permalink]

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New post 26 Jun 2005, 06:50
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If x= (a/2) + (b/2^3) + (c/2^4) , where a,b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT:

a) 1/16
b) 3/16
c) 5/16
d) 10/16
e) 11/16

Fastest way is to get all three fractions together. This is easily done : x = (8a + 2b + c)/16

We can quickly look for the possibilites. 1/16 is possible if a==b=0, c=1; 3/16 is possible if a=0, b=c=1; 5/16 is not possible. 10/16 is possible if a=b=1, c=0, 11/16 is possible if a=b=c=1.

So the answer is C

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 [#permalink]

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New post 26 Jun 2005, 18:45
same approach

(8a+2b+c)/16 with a,b,c = 1 or 0

then you check all numerators : 1,3,5,10,11

5 is impossible to reach

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Re: If x= (a/2) + (b/2^3) + (c/2^4) , where a,b, and c are each [#permalink]

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Re: If x= (a/2) + (b/2^3) + (c/2^4) , where a,b, and c are each   [#permalink] 16 Aug 2017, 00:20
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