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#11
11-22-2002, 01:11 AM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636

Therefore X is prime should read "Therefore X is prime or X is divisible by a prime larger than N". Either way there is a prime larger than our assumed largest prime N, so there is no largest prime N.
#12
11-22-2002, 01:18 AM
 Mano Senior Member Join Date: Sep 2002 Location: Salt Lake City, Utah Posts: 265

I think we are saying the same thing - I probably could have worded the end of the proof better, as you did - I held onto the assumption of the primes being bounded for one more step. Assuming there is a largest prime leads to a contradiction either way through my reasoning above.
#13
11-22-2002, 01:30 AM
 BB King's Senior Member Join Date: Sep 2002 Posts: 244
You\'re full of crap ...

<font color="red">In the first place, you changed your last digit to 7 after I posted.</font color> NO - I didn't !!! You were only faster than me by ONE minut !!! Actually it's a pretty simple prolem.

<font color="red">In the second place, your number doesn't have anywhere near enough digits. If this was just (10,000)^4131 it would have 16,525 digits.</font color> Close enough !?! [img]/forums/images/icons/grin.gif[/img]
#14
11-22-2002, 01:48 AM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: You\'re full of crap ...

I saw your first answer, it had a 1 at the end. Who do you think you're dealing with some amateur? [img]/forums/images/icons/shocked.gif[/img]
#15
11-22-2002, 02:11 AM
 BB King's Senior Member Join Date: Sep 2002 Posts: 244
No you are not an ameteur ...

I'm reposting my original post - ONE minut later than your post :

99,193^4131= 5683287653286836532810926345876......7610981435761 09281435761092634587610981

I forgot to some 7's in on the end - my mistake ;-)

No you are not an ameteur - You are the King of this forum - like Dynasty is it in small/stakes - like Greg in tourney/no/limit - like ...

Do you have a PMS-problem ? Do you have a sense of humour ?
#16
11-22-2002, 10:38 AM
 irchans Senior Member Join Date: Sep 2002 Posts: 157
Re: 99,193^4131= ...

If you want to see all the digits, follow the attached link.
99193^4131
#17
11-22-2002, 02:23 PM
 Jim Brier Senior Member Join Date: Sep 2002 Location: Las Vegas, NV Posts: 189

The key to this problem is realizing that when positive integers are raised to powers, the one's digit follows identifiable patterns. Note the pattern with 3 (or any number ending with 3)

3 to the zero power is 1
3 to the first power is 3
3 to the second power is 9
3 to the third power is 27 (one's digit is 7)
3 to the fourth power is 81 (one's digit is 1 again)

If you were to keep going you notice the pattern is 1, 3,9, and 7. It then repeats. Every fourth power has the one's digit as 1.

The number 4131 is divisible by 4 with a remainder of 3. Therefore, the number raised to the 4128th power would have a "1" as its one's digit. Following the pattern, the next power (4129) would have a "3" in the one's digit. The next power (4130) would have a "9" in the one's digit. Finally, raising a number ending in 3 to the 4131 power would have a "7" in the one's digit.

#18
11-26-2002, 09:05 PM
 Guest Posts: n/a
Re: Math Problem

I'm pleased to say that my 11 year old daughter solved this in about 3 minutes. We're ready for the next one!

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