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505-555 Level|   Arithmetic|   Exponents|      
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Bunuel
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1/2^12 + 2/2^13 + 4/2^14 + 8/2^15 = 01 May 2019, 23:28
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\(\frac{1}{2^{12}} + \frac{2}{2^{13}} + \frac{4}{2^{14}} + \frac{8}{2^{15}} =\)


A. 1/2^10

B. 1/2^12

C. 15/2^15

D. 2/2^10

E. 23/2^16
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A
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1/2^12 + 2/2^13 + 4/2^14 + 8/2^15 = 01 May 2019, 23:42
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Lcm is 2^15. So it will become like

8 + 8 + 8 + 8/2^15 = 2^5/2^15
= 1/2^10, so option A.

Please give me Kudos if you like my explanation.

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1/2^12 + 2/2^13 + 4/2^14 + 8/2^15 = 02 May 2019, 00:26
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We can reduce the fractions
\(\frac{1}{2^12}\)=\(\frac{1}{2^12}\) :)
\(\frac{2}{2^13}\)=\(\frac{1}{2^12}\)
\(\frac{4}{2^14}\)=\(\frac{1}{2^12}\)
\(\frac{8}{2^15}\)=\(\frac{1}{2^12}\)

So we get 4*\(\frac{1}{2^12}\)=\(\frac{1}{2^10}\)

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Ans: A
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1/2^12 + 2/2^13 + 4/2^14 + 8/2^15 = 02 May 2019, 00:30
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After reducing each term we get 4*1/(2^12)= 1/(2^10)
IMO A

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1/2^12 + 2/2^13 + 4/2^14 + 8/2^15 = 02 May 2019, 02:52
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Bunuel
\(\frac{1}{2^{12}} + \frac{2}{2^{13}} + \frac{4}{2^{14}} + \frac{8}{2^{15}} =\)


A. 1/2^10

B. 1/2^12

C. 15/2^15

D. 2/2^10

E. 23/2^16

simplify expression
we get ; 4/2^12 ; 1/2^10
IMO A
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1/2^12 + 2/2^13 + 4/2^14 + 8/2^15 = 02 May 2019, 10:38
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Bunuel
\(\frac{1}{2^{12}} + \frac{2}{2^{13}} + \frac{4}{2^{14}} + \frac{8}{2^{15}} =\)


A. 1/2^10

B. 1/2^12

C. 15/2^15

D. 2/2^10

E. 23/2^16

\(\frac{1}{2^{12}} + \frac{2}{2^{13}} + \frac{4}{2^{14}} + \frac{8}{2^{15}}\)

\(=\frac{1}{2^{12}} + \frac{1}{2^{12}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)

\(=\frac{4}{2^{12}}\)

\(=\frac{1}{2^{10}}\); Answer must be (A)
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1/2^12 + 2/2^13 + 4/2^14 + 8/2^15 = 02 May 2019, 12:49
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Bunuel
\(\frac{1}{2^{12}} + \frac{2}{2^{13}} + \frac{4}{2^{14}} + \frac{8}{2^{15}} =\)


A. 1/2^10

B. 1/2^12

C. 15/2^15

D. 2/2^10

E. 23/2^16

I would say, simply take 1/2^12 common.

--> 1/2^12 [1+2/2+4/4+8/8] = 1/2^12[4] = 1/2^2 * 2^10 [4] = 1/2^10 cheers
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1/2^12 + 2/2^13 + 4/2^14 + 8/2^15 = 11 Jan 2025, 22:57
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