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Let $a = 31$. Consider the set of integers $T = \{a, 8a, 8^{2}a, 8^{3}a, \cdots \}$. Does $T$ contain the integer:

$999999999900000000000090909090000000000000000008$?

So far I've deduced that if we work mod $9$ that the set $T$ can be reduced to a reduced residue system modulo $9$ by using $8$ as a root. Additionally, because $(a, 9) = (31, 9) = 1$ we can eliminate $a$ from the elements of $T$.

Continuing, remark that $ord_{9}(2) = \phi(9) = 6$ and $ord_{9}(2^{3}) = \frac{6}{3} = 2$. Taking $8^{k} \mod 9$ for $k \in \mathbb{Z^{+}}$ we see that we get the set of reduced residues $\{8, 1\}$. $8$ is an element of this set so $T$ contains $999\ldots 0008$.

That's where I am so far but I have a feeling that I went wrong early on in this one.

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    I suppose...it was downvoted like 10 seconds after I uploaded it though so whoever downvoted it didn't even bother to think about it.2012-12-17

2 Answers 2

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Hint

$8 \equiv -1 \pmod 9$ $31 \equiv 4 \pmod 9$

Thus

$8^k a \equiv ??? \pmod 9$

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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6763/discussion-between-decave-and-n-s)2012-12-17
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The largest power of $8$ that divides our number is $8^1$. Our number is not $(8)(31)$.

Note that the only thing that was used is that our number ends in $008$.

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    But $31$ is further than I can comfortably count, even after taking my shoes off.2012-12-17