# Question from a competition

1. Dec 5, 2009

### fawk3s

1. The problem statement, all variables and given/known data
What would be the last number of 1/52000 as a decimal fraction (or whatever it is for english)?

2. Relevant equations
3. 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 Im interested in how to solve it.

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

fawk3s

Last edited: Dec 5, 2009
2. Dec 5, 2009

### Staff: Mentor

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.

3. Dec 5, 2009

### fawk3s

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?

4. Dec 5, 2009

### Staff: Mentor

Read the rest of my other post. I laid out some things for you to do.

5. Dec 5, 2009

### HallsofIvy

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.