The minimum number of D flip-flops needed to design a mod-258 counter is

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Solution

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Official answer: A) 9

Combinational and Sequential Circuits

To design a modulo- $N$ counter, the minimum number of flip-flops $n$ required must satisfy:

2^n \ge N

For a mod-258 counter, $N = 258$ :

For $n = 8$ , $2^8 = 256 < 258$
For $n = 9$ , $2^9 = 512 \ge 258$

Therefore, the minimum number of D flip-flops needed is:

n = \lceil \log_2 258 \rceil = 9

Thus, (A) is the correct option.

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This Combinational and Sequential Circuits previous year question is Q15 of the GATE CSE 2011 paper. Find more GATE CSE PYQs on Combinational and Sequential Circuits in Digital Logic, or browse all GATE CSE PYQs.

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Combinational and Sequential Circuits in GATE CSE

Combinational and Sequential Circuits has 32 GATE CSE PYQs across 22 papers from 2011 to 2026. 8 of them are 1-mark MCQs like this one, and GATE CSE 2011 has 3.

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