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Topic: The digit in the hundreds place (Read 453 times) |
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pcbouhid
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The digit in the hundreds place
« on: Nov 26th, 2005, 3:35am » |
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Let N be an even number not divisible by 10. What digit will be in the hundreds place of N^200?
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Barukh
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Re: The digit in the hundreds place
« Reply #1 on: Nov 26th, 2005, 10:04am » |
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The way to figure this out is to find N200 mod 125.
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Icarus
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Re: The digit in the hundreds place
« Reply #2 on: Nov 26th, 2005, 10:23am » |
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(10k+x)200 = x200 + 200*10kx199 + (200*199/2)*100k2x198 +M*1000 for some integer M, so (10k+x)200 = x200 mod 1000.
Therefore it is only necessary to check the values N=2, 4, 6, 8. However 6200 = 996200 = (-4)200 = 4200 mod 1000, and similarly, 8200 = 2200 mod 1000. So it is only needful to check N=2 and N=4.
Working always mod 1000, 210 = 24, 310 = 49 = 72.
so 2200 = 2420 = 260330 = 246320 = 218326 = 28(24)326 = 211327 = (2)(24)327 = (16)328 = (16)3874 = 16*561*401 = 376.
4200 = 2400 = 2440 = 2120340 = ... 2836710.
28 = 256, 36 = 729, 710 = 49*(401)2 = 49*801= 249.
So 4200 = 256*729*249 = 376.
So the last 3 digits of N200 are 376, if N is even and not divisible by 10.
Probably some little trick was available to show that 2200 = 4200 mod 1000, which would have avoided the need for double calculation if I had spotted it.
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| « Last Edit: Nov 27th, 2005, 9:37am by Icarus » |
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Eigenray
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Re: The digit in the hundreds place
« Reply #3 on: Nov 26th, 2005, 1:14pm » |
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Well, here's a trick.
The multiplicative group (Z/125)* has order
phi(53) = 524 = 100,
so if n is relatively prime to 125, i.e., n not divisble by 5, then
n200=(n100)2 = 1 mod 53.
And if n is even, then clearly
n200 = 0 mod 23.
Then just note 376 (mod 103) is the unique solution to these two equations.
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Icarus
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Re: The digit in the hundreds place
« Reply #4 on: Nov 27th, 2005, 9:34am » |
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Okay, except I don't believe that noting "376 mod 103 is the unique solution to these two equations" is any simpler than the calculations for 2200 and 4200 that I performed.
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| « Last Edit: Nov 27th, 2005, 9:38am by Icarus » |
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Barukh
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Re: The digit in the hundreds place
« Reply #5 on: Nov 27th, 2005, 10:46am » |
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on Nov 27th, 2005, 9:34am, Icarus wrote:| Okay, except I don't believe that noting "376 mod 103 is the unique solution to these two equations" is any simpler than the calculations for 2200 and 4200 that I performed. |
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I think it's easier: the first equation has solutions 126, 251, 376, 501, 626, 751, 876, and only one of these numbers is divisible by 8.
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Eigenray
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Re: The digit in the hundreds place
« Reply #6 on: Nov 27th, 2005, 1:36pm » |
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Indeed, and you can stop after 376, because the solution is unique by the Chinese Remainder Theorem. (And you need only consider the sequence 1+5k mod 8.)
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