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Topic: Last 1000 digits (Read 735 times) |
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gkwal
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Last 1000 digits
« on: May 3rd, 2007, 11:29am » |
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Find the last 1000 digits of the number N, where
N = 1 + 50 + 50^2 + 50^3 +..........+ 50^999
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towr
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Re: Last 1000 digits
« Reply #1 on: May 3rd, 2007, 1:55pm » |
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I'd have a look at (50N - N)/49 i.e. (50^1000-1)/49
That's not a final answer, I know. But it's an easier question 
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Aryabhatta
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Re: Last 1000 digits
« Reply #2 on: May 3rd, 2007, 5:19pm » |
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on May 3rd, 2007, 1:55pm, towr wrote:I'd have a look at (50N - N)/49 i.e. (50^1000-1)/49
That's not a final answer, I know. But it's an easier question |
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Actually, that approach might not work.
I get the answer as:
3265306122448979591836734693877551020408163265306122448979591836734693877551
Where the underline means, repeat till you hit 1000 digits.
The approach used:
Instead of N = 1 + 50 + 50^2 + .... + 50^999
consider
N' = 1 + 50 + 50^2 + .... + 50^999 + 50^1000 + ... + 50^1007
Notice that N and N' have the same last 1000 digits.
Now 49N' = (50^1008 - 1)
Rewrite it as (5^1008 - 1)*10^1008 + 10^1008 - 1
Now we can show that 5^1008-1 is divisible by 49 (using the fact that 5^42 = 1 modulo 49, 42 = phi(49), where phi is euler-totient function)
Thus the last 1008 digits of N' are given by
(10^1008 - 1)/49 = [10^1008/49] (integral part of that)
Now 1/49 is a recurring decimal with period 42 (and repeating part 020408163265306122448979591836734693877551).
The rest follows.
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| « Last Edit: May 3rd, 2007, 5:38pm by Aryabhatta » |
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Eigenray
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Re: Last 1000 digits
« Reply #3 on: May 3rd, 2007, 5:55pm » |
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This problem was posted before here.
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rmsgrey
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Re: Last 1000 digits
« Reply #4 on: May 4th, 2007, 4:12am » |
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on May 3rd, 2007, 5:19pm, Aryabhatta wrote:I get the answer as:
3265306122448979591836734693877551020408163265306122448979591836734693877551
Where the underline means, repeat till you hit 1000 digits. |
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Also known as (using the same notation):
326530612244897959183673469387755102040816
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ThudnBlunder
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Re: Last 1000 digits
« Reply #5 on: May 4th, 2007, 4:40am » |
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on May 4th, 2007, 4:12am, rmsgrey wrote:
Also known as (using the same notation):
326530612244897959183673469387755102040816 |
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But this number seems to end with 102040816
The required number does not.
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rmsgrey
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Re: Last 1000 digits
« Reply #6 on: May 5th, 2007, 6:01am » |
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on May 4th, 2007, 4:40am, ThudanBlunder wrote:
But this number seems to end with 102040816
The required number does not. |
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That depends how you handle the truncation
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