How many AND gates would be required to completely decode ALL the states of a MOD-64 counter, and how many inputs must each AND gate have?

ANS:C - 64 gates, 6 inputs to each gate

MOD = 64;
MOD = 2^6;

Because,
STARTING MOD = 000000.
.
.
.
ENDING MOD = 111111 FOR UNDERSTANDING(32 16 8 4 2 1).
Total state = 64(0 to 63).

For every state need of 6 input gate then,
Total no.of AND gate = 64;(every 6 input AND GATE).