Easy or Hard ..... I'll let u guys decide

Number Theory Level 3

Find the number of natural numbers N N such that there exists some positive integer x x that satisfies the following equation: 1 ! + 2 ! + 3 ! + ⋯ + x ! = N 2 . 1! + 2!+3! +\cdots+x! =N^{2}.

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The answer is 2.

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1 solution

Michael Ng
Dec 23, 2014

Excellent problem! We use the fact that x ! x! for x > 4 x>4 is divisible by 5 5 . So: 1 ! + 2 ! + 3 ! + 4 ! + ⋯ + x ! 1! + 2! + 3! + 4! + \cdots + x! for x > 4 x>4 is congruent to 3 m o d 5 3 \bmod 5 . But no squares are 3 m o d 5 3 \bmod 5 so no x > 4 x>4 satisfy the equation. Now we only have to consider the cases where 1 ≤ x ≤ 4 1\leq x \leq 4 .

Checking these cases shows that only x = 1 , 3 x=1, 3 satisfy the equation so the answer is 2 \boxed{2} .

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Sharks! Added 1 and 3 to get 4, not count. I can't even count properly...

Aloysius Ng - 6 years, 5 months ago

Surely that´s a Hard problem! It is a very non intuitive problem and think in the mod operation requires a bit of imagination. Like every Number Theory problem.

By the way, 1 ! + 2 ! + 3 ! + 4 ! ≡ 3 m o d 5 1!+2!+3!+4! \equiv 3 \bmod 5 .

Cleres Cupertino - 5 years, 9 months ago

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