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# maths computer question

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using powers of 2, for finding the number of 64-bit words that can be held in a main memory of size 2 TiB.

Mar 22, 2020

#1
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264  = 1.844674 x 1019  different  64 bit words are possible

64 is  26

2 terabytes is 2 000 000 000 000 bytes

each byte is 8 bits    = 16 000 000 000 000 bits         log(16000000000000,2) = 43.86313713864835

243.863137 / 26       =  237.863137      64 bit words

Mar 22, 2020
#2
+8963
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TiB  specifically means  Tebibyte

2 TiB   =   2 Tebibytes   =   2 * 240  bytes   =   2 * 240 * 8  bits   =   2 * 240 * 23  bits   =   244  bits

There are  244  bits total. We want to know how many  64-bit words  will  "fit into"  that.

number of 64-bit words that can be held  =  \(\dfrac{2^{44}}{64}\ =\ \dfrac{2^{44}}{2^{6}}\ =\ 2^{44-6}\ =\ 2^{38}\)  .

Mar 22, 2020
#3
+23710
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Cool.....never heard of Tebibyte before ....thought it was a typo in the question .....

I found that Tib and TB   they are close enough in value they are often used as synonyms, though they are not EXACTLY the same value...as your answer shows

same methodology to solution though.....  THANX !

ElectricPavlov  Mar 22, 2020
edited by ElectricPavlov  Mar 22, 2020