Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Prove it United States

Problem:

Define a sequence $\{a_{n}\}$ by $a_{1}=1$ and $a_{n}=(a_{n-1})!+1$ for every $n>1$ . Find the least $n$ for which $a_{n}>10^{10}$ .

Solution

Solution:

We have $a_{2}=2$ , $a_{3}=3$ , $a_{4}=7$ , $a_{5}=7!+1=5041$ , and $a_{6}=5041!+1$ . But
$5041!+1 \gg 5041 \cdot 5040 \cdot 5039 > 10^{10}$
Hence, the answer is 6.

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