To solve this problem, we recognize it as the classic Josephus problem where we need to find the last remaining student in a circle after eliminating every 3rd student.

Key Recurrence Relation

The Josephus function (J(n)) (1-based position of the last remaining person) follows this recurrence:

• (J(1) = 1) (only one person left)
• (J(n) = (J(n-1) + k) \mod n), where (k=3) (every 3rd person is eliminated). If the result is 0, it is replaced by (n).

Calculation Steps

We compute (J(n)) step-by-step up to (n=40):

[ \begin{align} J(1)&=1 \ J(2)&=(1+3)\mod2=0→2 \ J(3)&=(2+3)\mod3=2 \ J(4)&=(2+3)\mod4=1 \ J(5)&=(1+3)\mod5=4 \ &\vdots \ J(40)&=(J(39)+3)\mod40=28 \ \end{align} ]

Answer: (\boxed{28})

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