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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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Can whoever just downvoted my question explain the downvote? –  user39898 Dec 17 '12 at 19:22
    
I don't know, but maybe is because it looks too exaggerated to be an actual question and you show no own work on it? This can be solved sharing some of your own efforts, insights and self work in this problem and, perhaps, adding some background of where/how did this question come from. –  DonAntonio Dec 17 '12 at 19:25
    
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. –  user39898 Dec 17 '12 at 19:26
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2 Answers 2

up vote 3 down vote accepted

Hint

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

Thus

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

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I'm sure you're on the right track but could you explain where you're coming from? I'm not seeing your thought process here. –  user39898 Dec 17 '12 at 19:25
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@decave Your number is huge, but the "huge" part contains only 9's and 0's, so it is natural to look at it mod 9... Modulo 9, the question you ask implies "Can $8^ka \equiv 8 \pmod 9$?" –  N. S. Dec 17 '12 at 19:29
    
@N.S: Amusing! :) What if the number was 999999999900000000000090909090000000000000000004? –  Isomorphism Dec 17 '12 at 19:30
    
So we know that $9 \mid 99999\ldots00090909\ldots 0000$? I guess you could just factor out a 9....wow...I really need to learn to work better under pressure haha. –  user39898 Dec 17 '12 at 19:31
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@Isomorphism That's why I said implies... If the number was what you said, that one needs to find a differenta approach, because that is a DIFFERENT problem ;) –  N. S. Dec 17 '12 at 19:31
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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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Also, the number is not divisible by $31$. It is, in fact, $8 \times 220654023912941 \times 566497713347021146672084609906661$ where the last two factors are primes. –  Robert Israel Dec 17 '12 at 19:49
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But $31$ is further than I can comfortably count, even after taking my shoes off. –  André Nicolas Dec 17 '12 at 19:53
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