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Find all natural numbers \(n\) such that \(n! | a^n+1 \) for all \(0\) <\( a \le n! \) with natural number \(a\).

Note by Surya Prakash Bugatha 2 years, 1 month ago

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Ehh, \(2^n+1\) is less than \(n!\) for \(n<4\), so you got only 3 cases to check... – Sreejato Bhattacharya · 2 years, 1 month ago

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Hint: What happens when \( a = 0 \)? – Calvin Lin Staff · 2 years, 1 month ago

@Calvin Lin – Ooops!! Typo fixed – Surya Prakash Bugatha · 2 years, 1 month ago

@Surya Prakash Bugatha – Hint: What happens when \( a = 1 \)? What happens when \( a = n \)? – Calvin Lin Staff · 2 years, 1 month ago

@Sreejato Bhattacharya @Surya Prakash Bugatha – Surya Prakash Bugatha · 2 years, 1 month ago

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## Comments

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TopNewestEhh, \(2^n+1\) is less than \(n!\) for \(n<4\), so you got only 3 cases to check... – Sreejato Bhattacharya · 2 years, 1 month ago

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Hint:What happens when \( a = 0 \)? – Calvin Lin Staff · 2 years, 1 month agoLog in to reply

– Surya Prakash Bugatha · 2 years, 1 month ago

Ooops!! Typo fixedLog in to reply

Hint:What happens when \( a = 1 \)? What happens when \( a = n \)? – Calvin Lin Staff · 2 years, 1 month agoLog in to reply

@Sreejato Bhattacharya @Surya Prakash Bugatha – Surya Prakash Bugatha · 2 years, 1 month ago

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