Maths Problem

p4radox

p4radox

p4radox

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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
 
123456789
10111213141516171819
20
30
40
50
60
70
80
90 2x9x10+9figures = 189
100 199
200 299
90 999 3x9x100 + 99 = 2799
1000 1999
9000 9999 4 x 9 x 1000 + 2799 = 38799
10000 19999
90000 99999 5 x 9 x 10000 + 38799 = 488799
100000 199999 6 x 100000 + 488799 = 1,088,799
88799/6 = 14799.83 199999 - 14799.83 = 185199.16 thats my approximation
The actual answer is 185185
 
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... :(
 
Yeah that's it, what is the millionth digit.

So the first one is 1, second is 2, etc

I get the answer as being 1, if you guys are still stuck in a bit, I will post my way, I don't want to spoil the fun too early.
 
:eek: CHAMBERS (p4radox) YOU CHEAT!!!!

That was the special homework Mr O'neil gave you for not doing your other homework!!!

This is very naughty Chambers... i will be printing this thread off an handing it to Mr O'neil when i see him next :p mwahahaha i'm so mean... i like it :cool:

edit: and you watch neighbours?! don't seem the type tbh
 
Last edited:
Right, I have it. I believe the 1,000,000th digit is "8", and that the "number" it is supposed to be a part of (because the sequence is nothing more than a number line with the numbers end on end) is 185185. The "8" I am referring to would be the second "8" in "185185".

Am I correct?
 
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?
 
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