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x = a/3 + b/3^2 + c/3^3

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x = a/3 + b/3^2 + c/3^3

by sanju09 » Wed Mar 23, 2011 4:31 am
If x = a/3 + b/3^2 + c/3^3, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT:
A. 1/27
B. 1/9
C. 4/27
D. 2/9
E. 4/9



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by HSPA » Wed Mar 23, 2011 4:39 am
1/27, a=0,b=0 and C=1
1/9 ,a=0,b=1,c=0
4/27 a=0,b=1,c=1
4/9 a=1,b=1,c=0

left is 2/9 by POE
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by Night reader » Wed Mar 23, 2011 12:01 pm
I would think and rewrite before moving on ...
a/3 + b/9 + c/27 --> (9a+3b+c) divided by 27

Quick prompt: three numbers can be 0 or 1 --> 2*2*2 ways to form number x too long :(

let's work with answer choices and use plugging in techniques by noting the denominators 9 and 27 for (9a+3b+c) divided by 27
A. possible
B. possible
C. possible
D. not possible
E. possible

answer D
sanju09 wrote:If x = a/3 + b/3^2 + c/3^3, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT:
A. 1/27
B. 1/9
C. 4/27
D. 2/9
E. 4/9



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