Maths Problem

p4radox

p4radox

p4radox

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Joined
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Hello,

I've just encountered this maths problem (been given it by my teacher - it's not homework, honest!):

What digit holds the millionth decimal place of the following number:
0.123456789101112131415161718192021222324252627282930 ...

I don't even know where to start! Maybe finding out how many digits it takes to count to 100, then how many more digits are used up for each hundred after that... argh... my head hurts... :(


edit: There's not supposed to be a space in the number after 28, i just can't get rid of it :confused:.
 
Rich_L said:
When I did this I was given the hint that you are essentially counting up from 1 to a number which has the millionth digit so ignore the decimal place, makes it easier :)

:p

*deleted spoiler*
Click to expand...

Gawd, damn you Rich, I half-read that before hitting F5. I think I'm getting somewhere now actually. I'll be back in 25min, Neighbours is on... :p
 
eXSBass said:
Just use Sequences and Series.

Un = a+(n-1)d
a is the first term. n is the nth term. And d is the common difference

:cool:
Click to expand...

I thought about it, but it just won't work, it's not an arithmetic sequence, more of a periodic one that's difficult to work out!
edit: in fact it's not even periodic i don't think...

So far I've got:
1->9 requires 9 digits
10->99 requires 90*2digits=180 digits
100->999 requires 900*3digits=2700 digits
1000->9999 requires 9000*4 digits=36000 digits
10000->99999 requires 90,000*5 digits = 450,000 digits

9+180+2700+27000+450,000 = 488,889 digits used in total.

This leaves 1,000,000 digits-488,889=511111 digits remaining. The millionth digit is obviously going to be somewhere within a 6 digit number.

511,111/6 = 85185.16666666r . Does this mean that it's one of the digits in the 85185th 6 digit number?



Brain is still aching... :(
 
KillerKebab: I've got pretty much the same:

1->9 requires 9 digits
10->99 requires 90*2digits=180 digits
100->999 requires 900*3digits=2700 digits
1000->9999 requires 9000*4 digits=36000 digits
10000->99999 requires 90,000*5 digits = 450,000 digits

9+180+2700+27000+450,000 = 488,889 digits used in total.

This leaves 1,000,000 digits-488,889=511111 digits remaining. The millionth digit is obviously going to be somewhere within a 6 digit number.

511,111/6 = 85185.1666666

So the 85185.166th 6digit number contains the millionth digit.

The first 6digit number is 100,000 so it's 99,999+85185 = 185184. So the 1millionth digit is 1. How do you get it as 8?
 
Killerkebab said:
Because there are 511,111 digits, which means 85185 numbers remaining, since your sequence is nothing more than a number line.

So what you need is the 85185th number, which you quite rightly said is 185184 (I got it wrong, I guess).

This number will have six digits.
The fraction, 0.16r is one sixth.

So the digit you are looking for will be the 85185th number (185184), and then you go along one sixth of this number, to find your digit, which is still 8.

:)
Click to expand...

In a way I see what you're saying, but I can't quite get it into my head. Why do you need to go along one sixth of the number to find the 1 millionth digit?

I can see that having 511,111 digits remaining, and dividing this by 6 gives 85185.16r, but why does this mean you have to take the 2nd digit of 185184? Isn't the 1 one sixth of the way along the number? :)
 
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