ITPEC IP October 2020 Question 6
ITPEC IP October 2020 — Question 6 of 100
After two executions the state is 4. Starting at 1, 1 × 11 leaves remainder 2 when divided by 3, so move two states clockwise: 1 → 2 → 3. Next, 3 × 11 = 33 leaves remainder 0, so move one state clockwise: 3 → 4.
Answer (d)
Why not others:
- state 1 is only the initial state
- state 2 is passed during the first two-step move
- state 3 is the result after only the first execution
Key rule: Recalculate the remainder from the new current state on every execution, then apply exactly the transition specified for that remainder.
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