Question

Consider a linear arrangement of seven bulbs, each of which can be in the ON or OFF state. The initial configuration of the bulbs is shown below. In every step, the states of the bulbs change based on the following rules:

• Any OFF bulb with exactly one ON neighbor at the end of the previous step is turned ON.
• Any ON bulb with both neighbors ON at the end of the previous step is turned OFF.
• The state of any bulb not meeting the conditions above is left unchanged.

The states of the bulbs at the end of Step 1 and Step 2 are also shown below.

Stage	Bulb 1	Bulb 2	Bulb 3	Bulb 4	Bulb 5	Bulb 6	Bulb 7
Initial	OFF	OFF	OFF	ON	OFF	OFF	OFF
Step 1	OFF	OFF	ON	ON	ON	OFF	OFF
Step 2	OFF	ON	ON	OFF	ON	ON	OFF

The number of bulbs which are ON at the end of Step 8 is ______

• A. 5
• B. 4
• C. 3
• D. 0

Hint:

Simulate the ON/OFF rule step by step; the pattern reaches a fixed alternating state well before Step 8, so counting the ON bulbs there gives the answer directly.

Answer 1

The Correct Option is B

Instead of drawing every bulb row, we can track just the set of positions that are ON at each stage and use the same two rules directly on that set.

Let $S_t$ be the set of ON bulb positions after stage $t$, numbering the bulbs $1$ to $7$.

• $S_0=\{4\}$ (given initial state).
• $S_1=\{3,4,5\}$ (given Step 1): bulbs 3 and 5 gain exactly one ON neighbor (bulb 4) and switch on; bulb 4 keeps both neighbors off so it is untouched.
• $S_2=\{2,3,5,6\}$ (given Step 2): bulb 2 gains one ON neighbor (bulb 3) and turns on, bulb 6 gains one ON neighbor (bulb 5) and turns on, while bulb 4 has two ON neighbors (3 and 5) which is not "exactly one", so it turns off, and bulbs 3, 5 are untouched.

Continue the same set update for the next steps:

• $S_3=\{1,2,3,5,6,7\}$: bulb 1 and bulb 7 each have exactly one ON neighbor (2 and 6 respectively) and switch on; bulb 4 again sees two ON neighbors and stays off; bulbs 2, 3, 5, 6 each have only one ON neighbor so they remain on.
• $S_4=\{1,3,5,7\}$: bulb 2 now has both neighbors (1 and 3) ON, so it turns off; bulb 6 similarly turns off since both neighbors (5 and 7) are ON; bulbs 3 and 5 each have only one ON neighbor left, so they stay on; bulbs 1 and 7 stay on since, as end bulbs, they can never see two ON neighbors.

Check $S_4$ for stability: with ON positions $\{1,3,5,7\}$, every ON bulb (3 and 5) has both its neighbors OFF (an even position on each side), so neither condition to turn off is met and they stay on; bulbs 1 and 7 stay on as end bulbs. Every OFF bulb (2, 4, 6) has both its neighbors ON (odd positions on each side), which is two ON neighbors, not exactly one, so none of them switch on. Therefore $S_5=S_6=S_7=S_8=S_4=\{1,3,5,7\}$.

So the number of bulbs ON at the end of Step 8 equals $|S_4|=4$.

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