2004 Pan African MO Problems/Problem 3

Problem

One writes 268 numbers around a circle, such that the sum of 20 consecutive numbers is always equal to 75. The number 3, 4 and 9 are written in positions 17, 83 and 144 respectively. Find the number in position 210.

Solution

Let $n_1, n_2, n_3, \cdots n_{267}, n_0$ be the 268 numbers. First, we will prove that a number twenty positions after said number would be the same number.

Lemma: $n_{i \pmod{268}} = n_{i+20 \pmod{268}}$ for integers $1 \le i \le 268$

We know that $\sum_{k=1}^{20} n_{a_k} = 75$ , where $a_k = i+k-1 \pmod{268}$ . By subtracting $n_{a_1}$ and adding $n_{a_{21}}$ from both sides, we must have $75 - n_{a_1} + n_{a_{21}} = 75$ . Therefore, $n_{a_1} = n_{a_{21}}$ , so $n_{i \pmod{268}} = n_{i+20 \pmod{268}}$ . $\blacktriangleright$

Thus, we know that $n_{i \pmod{268}} = n_{i+20m \pmod{268}}$ , where $m$ is an integer. If $m = 14m_1$ , then $i + 20m \equiv i + 280m_1 \equiv i + 12m_i \pmod{268}$ . If $m_1 = 22m_2$ , then $i + 12m_1 \equiv i + 264m_2 \equiv i - 4m_2 \pmod{268}$ . Note that $4$ is a factor of $268$ , so we can know that if $a \equiv b \pmod{4}$ , then $n_a = n_b$ .

With this in mind, $\begin{align*} \sum_{k=1}^{20} n_k &= 75 \\ 5(n_1 + n_2 + n_3 + n_4) &= 75 \\ 5(3 + n_2 + 4 + 9) &= 75 \\ 5(16 + n_2) &= 75 \\ 16 + n_2 &= 15 \\ n_2 &= -1. \end{align*}$ Thus, since $n_2 = n_{210}$ , we must have $n_{210} = \boxed{-1}$ .

2004 Pan African MO (Problems)
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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2004 Pan African MO Problems/Problem 3

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