Problem 5 · Position 2026 in the sequence
Triangular numbers: where 2026 falls among $\tfrac{n(n+1)}{2}$.
Integer answer, at most 4 digitsWhich number occupies position 2026 in the sequence $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, \ldots$?
Copa Cangur · SCM
Medium
Closed answer
Reasoned solution
Key idea: the number $n$ appears $n$ times, so its last position is the triangular number $\tfrac{n(n+1)}{2}$.
We look for $n$ with $\tfrac{(n-1)n}{2} < 2026 \le \tfrac{n(n+1)}{2}$:
$$\frac{63 \cdot 64}{2} = 2016 < 2026 \le \frac{64 \cdot 65}{2} = 2080.$$
Position $2026$ falls inside the block of $64$s (which occupies positions $2017$ to $2080$).
Answer: 64