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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

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Bunuel
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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 [#permalink] New post  05 Nov 2015, 12:24
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If \(x = \frac{a}{2} + \frac{b}{2^3} + \frac{c}{2^4}\), 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
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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink] New post  05 Nov 2015, 12:59
Bunuel 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 = a/2 + b/2^3 + c/2^4 = 8a/16 + 2b/16 + c/16
(A) 1/16 when a=b=0, c=1
(B) 3/16 when a = 0, b=c=1
(D) 10/16 when a=b=1, c=0
(E) 11/16 when a=b=c=1

(C) 5/16 there is no value of a, b, c (0 or 1)

Ans: C
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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink] New post  05 Nov 2015, 18:57
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x= a/2 + b/2^3 + c/2^4
= a/2 + b/8 + c/16
= (8a + 2b + c)/16

Depending on whether a, b and c take 0 or 1 , we will have 1,3,10 and 11 as sum in the numerator of the expression .

Answer C
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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink] New post  05 Nov 2015, 22:54
x cannot be 5/16

Ans:c
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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink] New post  06 Nov 2015, 00:43
Nice post, tricky one.
IMO C
A,B,D,E can be easily eliminated as we have 0 or 1 to choose for values of a,b,c respectively.
A ->>> easy to get 1/16 (just setting 0 for a & b)
B - >>> setting 0 for a, now we have 2/16 (which is 1/8 reversed to 2/16) we get 3/16
D ->>> Getting (1/2 with 8/16 and b=1, c=0)
E - >>> a=b=c=1
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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink] New post  27 Jan 2018, 09:29
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Bunuel wrote:
If \(x = \frac{a}{2} + \frac{b}{2^3} + \frac{c}{2^4}\), 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



Something that may help you in getting to the answer..
All except D are odd/16, so C/16 =1/16....A
now work on other two to see if they add up to answer..
a/2+b/8+1/16....
a=b=1... 1/2+1/8+1/16=11/16 ...E
a=0 b=1...1/8+1/16 = 3/16...B
a=b=0...1/16...A

Ok now 10/16 is nothing but 5/8..
So c is 0... So 1/2+1/8=5/8...D

Ans C
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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 [#permalink] New post  24 Oct 2020, 14:35
If \(x = \frac{a}{2} + \frac{b}{2^3} + \frac{c}{2^4}\), then:

\(\frac{8a}{16} + \frac{2b}{16} + \frac{1c}{16}\)

If a, b, and c are each equal to 0 or 1, there will be on answer that is impossible. Lets look at the choices-

(A) 1/16 -- Possible with C

(B) 3/16 -- Possible with B and C

(C) 5/16 -- Not possible

(D) 10/16 -- Possible with A and B

(E) 11/16 -- Possible with A and B and C

Answer is C.
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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink] New post  25 Oct 2020, 01:45
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x = \(\frac{a}{2} + \frac{b}{2^3} + \frac{c}{2^4}\)

=> x = \(\frac{(2^3a + 2b + 1c)}{ 2^4} = \frac{(8a + 2b + c) }{ 16}\)

If a = b = c = 0, Then x = 0.

If a = b = c = 1, Then x = \(\frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^4}\)

=> x = \(\frac{(2^3 + 2 + 1)}{ 2^4} = \frac{11}{ 16}\)

So, \(0 ≤ x ≤ \frac{11}{16}\)

Various values for 8a + 2b + c:

=> All '0' = 0
=> All '1' = 11
=> 8 + 2 = 10 when c = 0
=> 8 + 1 = 9 when b = 0
=> 2 + 1 = 3 when when a = 0

Value not possible is '5'.

Answer C
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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink] New post  26 Mar 2021, 12:14
Correct me if i am wrong ,
If C=0 i.e as per option D how can the denominator be 16 ,
C / 2^4 = 0 / 2^4 = 0 ,
then the denominator becomes 8.
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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink] New post  20 May 2023, 01:04
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Re: If x = a/2 + b/2^3 + c/2^4, where a, b, and c are each equal to 0 or 1 [#permalink]
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