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Home » Now subtract the number of arrangements where R1 and R2 are consecutive. - GetMeFoodie

Now subtract the number of arrangements where R1 and R2 are consecutive. - GetMeFoodie

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 Solution: We are looking for the smallest two-digit positive integer $ L $ such that

📌 L \equiv 1 \pmod{13}

📌 and $ L $ is a multiple of 7 (since pipes are seven units long and total length is a sum of such pipes). However, note: the problem states the *total length of installed pipe* is one more than a multiple of 13âso $ L \equiv 1 \pmod{13} $, and $ L $ must be expressible as a sum of multiples of 7. But since individual pipes are 7 units, any valid total length must be a multiple of 7. So we seek the smallest two-digit number $ L $ such that:

📌 L \equiv 1 \pmod{13} \quad ext{and} \quad L \equiv 0 \pmod{7}

📌 L \equiv 0 \pmod{7} \

📌 Use the Chinese Remainder Theorem. Let $ L = 7k $. Substitute into the second congruence:

📌 7k \equiv 1 \pmod{13}

📌 We solve for $ k $: find the modular inverse of 7 modulo 13.

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