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200^200*40^40/(20^20*400^400) =

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200^200*40^40/(20^20*400^400) =  [#permalink]

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New post 05 Jul 2018, 04:43
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

72% (03:08) correct 28% (02:47) wrong based on 127 sessions

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200^200*40^40/(20^20*400^400) =  [#permalink]

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New post 05 Jul 2018, 05:06
Bunuel wrote:
\(\frac{200^{200} 40^{40}}{20^{20}400^{400}}=\)


A. \(\frac{1}{2^{920}5^{380}}\)

B. \(\frac{1}{2^{240}5^{240}}\)

C. \(\frac{1}{2^{180}5^{180}}\)

D. \(1\)

E. \(2^{180}5^{180}\)



separate the 10s and convert 4 also in terms of 2..

\(\frac{200^{200} 40^{40}}{20^{20}400^{400}}=\frac{(2*10^2)^{200} (2^2*10)^{40}}{(2*10)^{20}(2^2*10^2)^{400}}=\frac{2^{200}*10^{400}*2^{80}*10^{40}}{2^{20}*10^{20}*2^{800}*10^{800}}=\frac{2^{200+80} 10^{400+40}}{2^{820}10^{820}}=\frac{1}{2^{(820-280)}*10^{(820-440)}}=\frac{1}{2^{(540+380)}5^{380}}\)

A
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Re: 200^200*40^40/(20^20*400^400) =  [#permalink]

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New post 05 Jul 2018, 08:27
Bunuel wrote:
\(\frac{200^{200} 40^{40}}{20^{20}400^{400}}=\)


A. \(\frac{1}{2^{920}5^{380}}\)

B. \(\frac{1}{2^{240}5^{240}}\)

C. \(\frac{1}{2^{180}5^{180}}\)

D. \(1\)

E. \(2^{180}5^{180}\)


\(\frac{200^{200} 40^{40}}{20^{20}400^{400}}\)
= \(\frac{(2^3*5^2)^{200}*(2^3*5)^{40}}{(2^2*5)^{20}(2^4*5^2)^{400}}\)
=\(\frac{2^{600}*5^{400}*2^{120}*5^{40}}{2^{40}*5^{20}*2^{1600}*5^{800}}\)
=\(2^{600+120-40-1600}*5^{400+40-20-800}\)
=\(2^{-920}*5^{-380}\)

Ans. A
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Re: 200^200*40^40/(20^20*400^400) =  [#permalink]

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New post 07 Jul 2018, 18:27
Bunuel wrote:
\(\frac{200^{200} 40^{40}}{20^{20}400^{400}}=\)


A. \(\frac{1}{2^{920}5^{380}}\)

B. \(\frac{1}{2^{240}5^{240}}\)

C. \(\frac{1}{2^{180}5^{180}}\)

D. \(1\)

E. \(2^{180}5^{180}\)


If we re-express all of the factors to have bases of 10, 20, or 40, we can cancel and simplify the expression.

[20^200 x 10^200 x 40^40]/[20^20 x 40^400 x 10^400]

[20^180]/[10^200 x 40^360]

We can re-express 40^360, obtaining:

[20^180]/[10^200 x 20^360 x 2^360]

1/[10^200 x 20^180 x 2^360]

1/[10^200 x 10^180 x 2^180 x 2^360]

1/[10^380 x 2^540]

1/[5^380 x 2^380 x 2^540]

1/[5^380 x 2^920]

Answer: A
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200^200*40^40/(20^20*400^400) =  [#permalink]

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New post 06 Jan 2019, 19:25
What a great question! The key here is not to get scared by high powers. By looking at the answer choices you can most likely figure out that it is asking you to break the big numbers down into powers of 2 and powers of 5!
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Re: 200^200*40^40/(20^20*400^400) =  [#permalink]

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New post 06 Jan 2019, 20:33
Bunuel wrote:
\(\frac{200^{200} 40^{40}}{20^{20}400^{400}}=\)


A. \(\frac{1}{2^{920}5^{380}}\)

B. \(\frac{1}{2^{240}5^{240}}\)

C. \(\frac{1}{2^{180}5^{180}}\)

D. \(1\)

E. \(2^{180}5^{180}\)


Just remember \((m*n)^x\) => \(m^x * n^x\)
and continue expanding the term to the simplest form and voila you will get the answer

Will solve each part individually, so that the solution can be understood clearly.
\(200^{200} 40^{40}\) -(x)
=> \((2^3 * 5^2)^{200} * 2^{120} * 5^{40}\)
=> \(2^{600} * 5^{400} * 2^{120} * 5^{40}\)
=> \(2^{720} * 5^{440}\)

\(20^{20}400^{400}\) -(y)
=> \((2^2 * 5)^{20} * (2^4 * 5^2)^{400}\)
=> \(2^{40} * 5^{20} * 2^{1600} * 5^{800}\)
=> \(2^{1640} * 5^{820}\)


\(\frac{x}{y}\) = \(\frac{1}{2^{1640-720} * 5^{820-440}}\)
\(\frac{x}{y}\) = \(\frac{1}{2^{920}5^{380}}\)

Correct answer A.
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Re: 200^200*40^40/(20^20*400^400) =   [#permalink] 06 Jan 2019, 20:33
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