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

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

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24 Oct 2006, 08:30
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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 ff EXCEPT:

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

... i got the answer by solving the long way... anyone can give me shortcut of solving this? thanks
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Joined: 10 Oct 2005
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24 Oct 2006, 08:36
Hermione 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 ff EXCEPT:

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

... i got the answer by solving the long way... anyone can give me shortcut of solving this? thanks

(8a/16)+(2b/16)+c/16=
(8a+2b+c)/16 plugging 1 and 0 instead a, b and c only C is out
so C is the answer total 40 secs
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Re: PS: exponents   [#permalink] 24 Oct 2006, 08:36
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