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what is the quikest way to solve this

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by crackinggmat » Mon Aug 01, 2011 8:17 am
the cube root of what integer power of 2 is closest to 50?

1)16 2) 17 3)18 4 ) 19 5) 20

THOUGH I COULD EASILY SOLVE THIS PROBLEM BUT IT TOOK ME 3 MINUTES ....SO PLS TELL ME A BETTER WAY.
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by Frankenstein » Mon Aug 01, 2011 8:27 am
Hi,
2^n = 50^3 = 125*10^3
125 -> closest to 128(2^7),
1000 -> 1024(2^10).
So, 2^n is close to 2^7*2^10 = 2^17
So, n = 17

Hence, 2
Cheers!

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by MBA.Aspirant » Mon Aug 01, 2011 11:22 am
@ Frankenstein

by closest they mean < or =, or it may surpass the number?

because 2^(17/3) is > 50


anyway this was my approach:

2^5 < 50 < 2^6

(2^5)^3 < (50)^3 < (2^6)^3

2^15 < 50^3 < 2^18

if closest means < or = then 2^16. otherwise, 2^(17/3) = ~50.8
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by Frankenstein » Mon Aug 01, 2011 8:57 pm
MBA.Aspirant wrote:@ Frankenstein

by closest they mean < or =, or it may surpass the number?

because 2^(17/3) is > 50


anyway this was my approach:

2^5 < 50 < 2^6

(2^5)^3 < (50)^3 < (2^6)^3

2^15 < 50^3 < 2^18

if closest means < or = then 2^16. otherwise, 2^(17/3) = ~50.8
Hey,
closest can be either greater than or less than but the difference should be least.
For example, 2^5<50<64
50-32 = 18
64-50 = 14.
So, 50 is closest to 2^6.
Cheers!

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