A binary ripple counter is designed to count (0)₁₀ to (64)₁₀. Which of the following is/are the number of flip-flops required to design the counter?

The correct option is 7.

To understand why 7 flip-flops are required, let us analyze the counter's requirements step-by-step.

A binary ripple counter needs to count from (0)₁₀ to (64)₁₀.
This means the counter must represent a total number of states. The range of count is from 0 to 64, which contains a total of 65 distinct states (0, 1, 2, ..., 64).

For a counter designed using $n$ flip-flops, the maximum number of states it can represent is given by:
$N=2^n$
where $n$ is the number of flip-flops.

To successfully count up to (64)₁₀ (which requires representing 65 states, or at least being able to represent the number 64 in binary), the capacity of the counter $2^n$ must be greater than or equal to the total number of states, or more simply, the maximum value to be represented must be less than $2^n$.
Specifically, the binary representation of the decimal number 64 is:
${\left(64\right)}_{10}={\left(1000000\right)}_2$
Counting the number of bits in the binary representation of 64:
The number 1000000₂ has 1 one followed by 6 zeros, which is a total of 7 bits.

Let us check the capacity of counters with different numbers of flip-flops:
If we use $n=6$ flip-flops, the maximum value we can represent is:
$2^6-1=64-1=63$
Since 63 is less than 64, a 6-bit counter can only count from 0 to 63. It cannot represent the state 64.

If we use $n=7$ flip-flops, the maximum value we can represent is:
$2^7-1=128-1=127$
Since 127 is greater than or equal to 64, a 7-bit counter can easily count from 0 to 64 (with states to spare).

Therefore, the minimum number of flip-flops required to design a ripple counter that counts from 0 to 64 is 7.