What is the last digit of 1/5200?

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The discussion centers on finding the last nonzero digit of 1/5200, which is derived from the decimal representation of the fraction. The solution involves recognizing the pattern in the last digits of powers of 2, specifically noting that the last digits cycle every four powers. By analyzing these patterns, it is determined that the last digit of 1/5200 is 6. The approach emphasizes understanding the mathematical properties of fractions and powers rather than using calculators. Ultimately, the conclusion confirms that the last digit of 1/5200 is indeed 6.
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Homework Statement


What would be the last number of 1/52000 as a decimal fraction (or whatever it is for english)?

Homework Equations


The Attempt at a Solution



This is not a homework problem. Its a question in an old maths competition.
I already know the answer is supposed to be 6, but I am interested in how to solve it.

Dont tell me to think about it or demand for any attempted solutions because there arent any.

Thanks in advance,
fawk3s
 
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I don't understand what you are asking. Although you are using a subscript, I think that you might have meant it to be an exponent, so that what you want is the last nonzero digit of 1/(52000). If that's really the problem, start by looking at the decimal representations of 1/5, 1/25, 1/125, 1/625, and so on, and see if you recognize a pattern.
 
Mark44 said:
I don't understand what you are asking. Although you are using a subscript, I think that you might have meant it to be an exponent, so that what you want is the last nonzero digit of 1/(52000). If that's really the problem, start by looking at the decimal representations of 1/5, 1/25, 1/125, 1/625, and so on, and see if you recognize a pattern.

Yes, I made a little mistake in the OP. I edited it now.
But the thing is, I need to find the last number of this number at the state of
0,0000000000000000000534534... (<- these are random numbers inserted).
You are not allowed to use a calculator. How to do it?
 
Read the rest of my other post. I laid out some things for you to do.
 
The last digit of 1/5200. One way to do this would be to note that if you set x= 1/5^{200}, then (1/2^{200})x= 1/(10)^{200}= 10^{-200}. That means that x= 2^{200}(10^{-200}) and the last digit of x is the "ones" digit of 2200. Now look at powers of 2: 21= 2, 23= 8, 24= 16, 25= 32, 26= 64, 27= 128, 28= 256, 29= 512, 210= 1024, 211= 2048, 212= 4096, etc. The ones digits of those are 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, ... Do you see the pattern? Can you prove that pattern? Now that pattern clearly repeats every 4th time and 200/4= 50 with 0 remainder. If that pattern is correct, 1/5200 has last digit the same as the fourth digit in that sequence, 6.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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