Find the number of natural numbers such that there exists some positive integer that satisfies the following equation:
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Excellent problem! We use the fact that x ! for x > 4 is divisible by 5 . So: 1 ! + 2 ! + 3 ! + 4 ! + ⋯ + x ! for x > 4 is congruent to 3 m o d 5 . But no squares are 3 m o d 5 so no x > 4 satisfy the equation. Now we only have to consider the cases where 1 ≤ x ≤ 4 .
Checking these cases shows that only x = 1 , 3 satisfy the equation so the answer is 2 .