Barely an Integer!

Number Theory Level 5

$\large{n = \left \lfloor m \right \rfloor + \left \lfloor \dfrac{m}{2!} \right \rfloor + \left \lfloor \dfrac{m}{3!} \right \rfloor + \ldots + \left \lfloor \dfrac{m}{k!} \right \rfloor + \ldots }$

For every natural number $n$ , define $S(n)$ to be the unique integer $m$ (if it exists) which satisfies the equation above. Submit the value of $S(3438)$ as your answer.

Bonus: Does there exist a number $k$ such that, for any non-negative integer $n$ , at least one of $S(n+1), \ S(n+2),\ \ldots \ , \ S(n+k)$ exists?

The answer is 2003.

1 solution

Patrick Corn
Sep 25, 2015

If we take the floor functions out, then the right side is $m(e-1)$ . So $m$ should be about $3438/(e-1)$ , which is between $2000$ and $2001$ . It's easy to compute $S(3434) = 2000$ , and then $S(3438) = 2003$ .

The extra credit answer is no. Given any positive integer $k$ , pick $n$ such that $S(n) = (k+1)!-1$ . Then $S(n+1), \ldots, S(n+k)$ all don't exist, and $S(n+k+1) = (k+1)!$ .

at m =3, $[3(e-1)]=5,[3]+[3/2]+[3/6]+...=4$ and taking out ceiling is wrong, for example $[\frac{1}{2}]+[\frac{1}{2}]\neq[\frac{1}{2}+\frac{1}{2}]$

Aareyan Manzoor - 5 years, 8 months ago

Yes, but you can take the floors out to get a rough estimate of how $n$ and $m$ are related. Then you can do the exact computations once you have a ballpark for $m$ .

Patrick Corn - 5 years, 8 months ago

Barely an Integer!

The answer is 2003.

1 solution

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