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Arithhmetic PS

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Arithhmetic PS

by Fokin » Mon Aug 09, 2010 9:35 am
Please, can any one explain the answer

How many bits of computer memory will be required to store the integer x , were x = - ((810,000)^(1/2)) - (minus sqr 810000), if each digit requires 4 bits of memory and the sign of x requires 1 bit?
(A) 25
(B) 24
(C) 17
(D) 13
(E) 12
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by selango » Mon Aug 09, 2010 9:42 am
Is this is a GMAT question?Can you post the source?

x = - ((810,000)^(1/2)) - (minus sqrt 810000)

x=-[900-(-900)]=-1800

1 digit require 4 bits.

-->4 digits require 16 bits

16+1[sign of x]=17 bits

Pick C
--Anand--
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by Fokin » Mon Aug 09, 2010 9:50 am
selango wrote:Is this is a GMAT question?Can you post the source?

x = - ((810,000)^(1/2)) - (minus sqrt 810000)

x=-[900-(-900)]=-1800

1 digit require 4 bits.

-->4 digits require 16 bits

16+1[sign of x]=17 bits

Pick C
The answer is D -13

It's a book- collection of GMAT math PS and DS problems . It's called GMAT PLUS. I have seen also this problem here
seen
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by selango » Mon Aug 09, 2010 9:55 am
Ok I think its each distinct digit.

0 is repeated twice.

So 3 digits-12 bits+1=13 bits
--Anand--
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