Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer

Let $N$ be the largest positive integer that can be expressed as a 2013-digit base -4 number. What is the remainder when $N$ is divided by 210?

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

The largest is $\sum_{i=0}^{1006} 3 \cdot 4^{2 i}=3 \frac{16^{1007}-1}{16-1}=\frac{16^{1007}-1}{5}$ . This is $1(\bmod 2), 0(\bmod 3), 3 \cdot 1007 \equiv 21 \equiv 1(\bmod 5)$ , and $3\left(2^{1007}-1\right) \equiv 3\left(2^{8}-1\right) \equiv 3\left(2^{2}-1\right) \equiv 2$ $(\bmod 7)$ , so we need $1(\bmod 10)$ and $9(\bmod 21)$ , which is $9+2 \cdot 21=51(\bmod 210)$ .

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