+16 using powers of 2, for finding the number of 64-bit words that can be held in a main memory of size 2 TiB.
+36916 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
+9466 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 = \(\dfrac{2^{44}}{64}\ =\ \dfrac{2^{44}}{2^{6}}\ =\ 2^{44-6}\ =\ 2^{38}\) .
+36916 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 !
