Competition problem — Number theory

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Number theory Difficulty 5.7 AIME, harder Prove it

Does there exist a number $h$ such that for no natural number $n$ the number $\left[h \cdot 1969^{n}\right]$ is divisible by $\left[h \cdot 1969^{n-1}\right]$ ?

Solution

Let $h=\frac{1969^{2}}{1968}$ . Then $h \cdot 1969^{n-1}=\frac{1969^{n+1}}{1968}=m+\frac{1}{1968} \quad$ and $h \cdot 1969^{n}=1969 m+1+\frac{1}{1968} \quad$ (where $m-$ is a natural number). For any natural $n$ , the number $m$ is greater than 1, so $1969 m+1$ does not divide by $m$ .

## Answer

It exists.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.