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I know that it is a terminating one. But I need help in the part of finding the digit where it terminates.

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    Total lack of context (unsurprisingly, since this is posted 20 minutes after another question). – 2017-01-04
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    So what do u mean to tell – 2017-01-04
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    Did implies to tell that it seems you have opened up your homework sheet and firing questions here for us to solve it for you. Your questions shows no effort that you have put on your part in trying to solve it and where exactly you got stuck or what part of question you could not tackle. – 2017-01-04
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    @naveendankal well.....It's not my homework sheet fyi. These are sums from a reference book's exercise problem.....I have solved most of them but was not getting some. So I marked them and now I am asking them one by one over here coz this site seems to be very good. But if u have a problem with this then sorry. But I repeat, this is not my *homework* sheet coz I am smart enough to do my school HW sheet. – 2017-01-04
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    Kindly see the reason why your question has been put on hold and please try to work upon the points mentioned in there to improve the question so that it does not give that "do my homework " impression to others . – 2017-01-04
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    Ok....But I have written what ever was there in the book....Even I felt the question was incomplete and so I asked over here – 2017-01-04
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    That's exactly what is discouraged here. To type exact question from the textbook. Maybe you can show on your part as to how you arrived that it is a terminating decimal . – 2017-01-04
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    Oooo.... Well if the denominator is in the form of 2^m or 5^n is a terminating decimal – 2017-01-04
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    That should be made part of your question instead of the comment.You may edit your question to include this – 2017-01-05

1 Answers 1

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A decimal terminates if it can be written as $x = \frac{k}{10^{n}}$ where $k$ is an integer.

We have that: $$\frac{3^{2017}}{5^{7102}} = \frac{3^{2017}2^{7102}}{2^{7102}5^{7102}} = \frac{3^{2017}2^{7102}}{10^{7102}}$$ The numerator is some positive integer, and the denominator is of the form $10^{7102}$. I'm unsure how to interpret where it precisely terminates, but I'd guess after 7102 places (this would make sense at least).

Edit: The last nonzero digit.

The last nonzero digit should be the lowest digit of $3^{2017}2^{7102}$. Funnily enough, this is "small" enough that I can compute it on my computer, so the answer is $2$. But, we could get the same result by looking at $3^{2017}2^{7102}\pmod{10}$. For this, we need that $3^{4}\pmod{10} = 1$, so $3^{2017}\pmod{10} = 3$. For $2$ it's a little more annoying because $2$ and $10$ aren't coprime, but we can instead use the Chinese Remainder theorem. To do this, we need to solve $2^{7102}\pmod{2}$ and $2^{7102}\pmod{5}$. Clearly, the first answer is $0$. Note that $2^{4}\pmod{5}\equiv 1$. So, we get that $2^{7102}\pmod{5} \equiv 2^2 \times 1 \equiv 4\pmod{5}$. So, the solution to this is $x\equiv 4\pmod{5}$ and $x\equiv 0\pmod{2}$. Note that $5\times 1-2\times 2 = 1$. We get that: $$x = 0+2\times (-2)\times 4 = -16\pmod{10}\equiv 4\pmod{10}$$ (note that here I just used the formula from the case of two moduli section of the chinese remainder theorem).

We get that the last digit is $3\times 4\pmod{10} = 2$, which agrees with our computation.