# Are there non-trivial integer $a$ and $c$ such that, for some natural values of $b_i$, the collection of $a^{b_i}+c$ contain every digit once?

Do there exist positive integers $$a$$ and $$c$$ such that for some nonnegative values $$b_i$$, there exists a set of numbers of the form $$a^{b_i}+c$$ that, when written in base 10, contain all the digits from $$0$$ to $$9$$?

The answer to this question is not only "yes", but in fact that there are infinitely many cases if you want to be cheeky about it. If you choose arbitrary integers $$a$$ and $$b$$ such that $$a^b<9876543210$$, and then allow $$P=9876543210-a^b$$, then by simple, grade-school arithmetic, $$a^b+P$$ will contain every digit from $$0$$ to $$9$$ exactly once.

So, let's tighten up our restrictions. Let's say that you can't cheese it by cramming all the digits into one number. Say instead that you need multiple values of $$b_i$$ and multiple numbers.

For an example that's almost right, consider the case where $$a=2$$ and $$c=3$$. Then for $$b_0=0, b_1=8$$, and $$b_2=14$$,

$$2^0+3=4$$ $$2^8+3=259$$ $$2^{14}+3=16387$$ In these three numbers, every digit from $$1$$ to $$9$$ appears exactly once. Another near miss: $$2^{11}+10$$ and $$2^{14}+10$$ contain every digit from 0 to 9 except for 7 between them. Can the full set ever be attained?

Some notes:

• If there is a valid value of $$a$$ it is not the case that $$a\equiv 1,5,6$$, or $$0\ \text{mod}\ 10$$, since then $$a^{b_0}$$ and $$a^{b_1}$$ would end in the same digit.
• If $$a\equiv4$$ or $$9\ \text{mod}\ 10$$, then they have to be split between two different numbers such that $$b_0\not\equiv b_1\ \text{mod}\ 2$$.
• Finally, for any other value of $$a$$, there can be no more than four different numbers which make up the full set of digits, or else there will be a case where $$b_i\equiv b_j\ \text{mod}\ 4$$, leading to the two numbers having the same end digit.

Try $$a = 2$$, $$c = 0$$: $$2^{2} = 4$$, $$2^{29} = 536870912$$.
Or $$a = 2$$, $$c = 1$$: $$2^0+1 = 2$$, $$2^{12}+1 = 4097$$, $$2^{14}+1 = 16385$$. Of course this could also be done as $$a=4$$, with $$4^0+1$$, $$4^6+1$$ and $$4^7+1$$.
Or $$a = 18$$, $$c = 0$$: $$18^3 = 5832$$, $$18^4 = 104976$$.