Answer:
1. n ≥ 4
2. N ≥ nlog(2)÷log3
Step-by-step explanation:
1
1 byte = 8 bits
each byte = 8 trits
a 6 bit number = [tex]2^{6} = 64[/tex]
3n ≥ 64
divide through by3
n ≥ 64/3
we take the logarithm
n ≥ [tex]\frac{log64}{log3}[/tex]
n ≥ 1.8062 ÷ 0.4771
n≥3.785
when approximated, n ≥ 4. So the number of trits needed in the storage of 6 bit is atleast 4
2.
to hold n bits, we can do something similar.
3N ≥ 2n
we take log of both sides
such that;
N ≥ [tex]\frac{nlog(2)}{log(3)}[/tex]
