# Base $b$ such that as many $x_b$ divide $x$ in set $\mathcal{Q}$

Assume $$d|n$$ means that integer $$d$$ divides integer $$n$$. Define $$\chi^k$$ to be the $$k$$-th element in a set. Consider the set $$\mathcal{Q} =\{1, 2, 3, \dots, 2019\}$$. For a base $$b$$ such that $$1, the amount of elements $$\chi^k$$ in the set $$\mathcal{Q}$$ such that $$\chi^k | {\chi^k}_b$$, is maximum. If $$d_b$$ means $$d$$ in base $$b$$, find $$b$$.

This is a problem I made.

I started by thinking about how we can turn something in base $$10$$ into base $$b$$. We have $$b^{e_1} + b^{e_2} + b^{e_3} +\cdots +b^{e_n} = \chi$$, where $$e_n$$ is an exponent. We know that $$b>10$$, as it is necessary. I have no idea where to go from here, as there is nothing I can do. Someone help me out?

• The last two sentences in the highlighted problem does not make sense. Can you reply a translation? – Alberto Takase Mar 8 at 2:35
• I wanted to say that there is a base $b$ such that the number of elements in set Q that divide itself in base b is maximum. sorry for my poor english. – weareallin Mar 8 at 2:40
• I see. So if zero was in the list, then that would be the number (since every number divides zero). – Alberto Takase Mar 8 at 2:41
• Yeah, but the base should be nonzero. I should add that. – weareallin Mar 8 at 2:43