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OSN SMA 2017 No 6

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Tentukan banyaknya bilangan asli $n$ yang tidak lebih besar dari $2017$ sedemikian sehingga $n$ habis membagi $20^n+17k$ untuk suatu bilangan asli $k$.

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Kalau n bukan kelipatan 17 maka 20^n pasti punya inverse modulo, kalau n kelipatan 17 maka $17 | n | 20^n + 17k$ sehingga $17 | 20^{n}$ tidak mungkin. Jadi $n$ memenuhi jika dan hanya jika kelipatan 17

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