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Bunuel
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Bunuel
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DJ1986
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DJ1986
In order to have a terminating decimal, you need a power of ten on the bottom.

I re-wrote original equation as:

3^4/(10^3)*5^3

(3^4)*(10^-3) / 5^3

.081/125

And then you don't even need to finish the long division to see that there will be three zeros.

You don't need a power of ten on the bottom. It has to be a power of 2, 5, or both.

E.g
1/2 = .5, which terminates
1/5 = .2, which terminates
1/20 = .1, which terminates

Anything other than a power of 2, 5, or a a combination of those two powers will not terminate. E.g. 1/3, 1/6, 1/7, 1/9, etc.
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DJ1986
In order to have a terminating decimal, you need a power of ten on the bottom.

I re-wrote original equation as:

3^4/(10^3)*5^3

(3^4)*(10^-3) / 5^3

.081/125

And then you don't even need to finish the long division to see that there will be three zeros.

Question asks for the number of non-zeros. Brunels explanation is the easiest way to sort this and make it "visual", unless I'm missing something in your explanation?
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Sure so, here is a trick that I came up with by running a few cases (Experts pls point to any mistakes, if present):

If you have 2 or 5 power n, the number of non-zero decimals will be n.

If you have a combination of 2 and 5, say 2^a and 5^b: Then the no of decimal digits will be a or b, depending on which is greater. I would like to caveat that some digits here could be zero so if you want non-zero digits might just be easier to divide by separating powers of 10 out


Example:
81/2 = 40.5 (2^1)
81/5 = 16.2 (5^1)
81/50 = 1.62 (5^2, 2^1)
81/400 = 0.2025 (5^2, 2^4)
81/4000 = 0.02025 (I would have divided here so just captured this case)

Hope this helps for anyone who visits this post in the future:)
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Bunuel
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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This is a great question that’s helpful for learning and I like the solution - it’s helpful.
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