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Jim Brier 11-21-2002 11:44 PM

Math Problem
 
Answer the following problem without using computers or logarithms. If you were to take the number 99,193 and raise it to the 4131 power, what would the one's digit be?

BruceZ 11-22-2002 12:32 AM

Answer
 
You asked one like this before. Each time we muliply by 99,193, the one's digit gets multiplied by 3, so it goes (starting from the 1st power) 3,9,7,1,3,9,7,1,3,9,7,1... Since 4131/4 has a remainder of 3, the result ends with 7.

Now, find the largest prime number there is, or prove that they go on forever.

BB King's 11-22-2002 12:33 AM

99,193^4131= ...
 
Answer: 7

99,193^4131= 56832876532868365328109263458761098143576109263458 76109814357610926345876109814357610926345876109814 35761092634587610981435765876109814357610926345876 10981587610981435761092634587610981587610981435761 09263458761098158761098143576109263458761098158761 09814357610926345876109815683287653286836532810926 34587610981435761092634587610981435761092634587610 98143576109263458761098143576109263458761098143576 58761098143576109263458761098158761098143576109263 45876109815876109814357610926345876109815876109814 35761092634587610981587610981435761092634587610981 56832876532868365328109263458761098143576109263458 76109814357610926345876109814357610926345876109814 35761092634587610981435765876109814357610926345876 10981587610981435761092634587610981587610981435761 09263458761098158761098143576109263458761098158761 09814357610926345876109815683287653286836532810926 34587610981435761092634587610981435761092634587610 98143576109263458761098143576109263458761098143576 58761098143576109263458761098158761098143576109263 45876109815876109814357610926345876109815876109814 35761092634587610981587610981435761092634587610981 56832876532868365328109263458761098143576109263458 76109814357610926345876109814357610926345876109814 35761092634587610981435765876109814357610926345876 10981587610981435761092634587610981587610981435761 09263458761098158761098143576109263458761098158761 09814357610926345876109815683287653286836532810926 34587610981435761092634587610981435761092634587610 98143576109263458761098143576109263458761098143576 58761098143576109263458761098158761098143576109263 45876109815876109814357610926345876109815876109814 35761092634587610981587610981435761092634587610981 56832876532868365328109263458761098143576109263458 76109814357610926345876109814357610926345876109814 35761092634587610981435765876109814357610926345876 10981587610981435761092634587610981587610981435761 09263458761098158761098143576109263458761098158761 09814357610926345876109815876109814357610926345876 10981587610981435761092634587610981587610981435761 09263458761098158761098143576109263458761098158761 09814357610926345876109815683287653286836532810926 34587610981435761092634587610981435761092634587610 98143576109263458761098143576109263458761098143576 58761098143576109263458761098158761098143576109263 45876109815876109814357610926345876109815876109814 35761092634587610981587610981435761092634587610981 58761098143576109263458761098158761098143576109263 45876109815876109814357610926345876109815876109814 35761092634587610981587610981435761092634587610981 56832876532868365328109263458761098143576109263458 76109814357610926345876109814357610926345876109814 35761092634587610981435765876109814357610926345876 10981587610981435761092634587610981587610981435761 09263458761098158761098143576109263458761098158761 09814357610926345876109815876109814357610926345876 10981587610981435761092634587610981587610981435761 09263458761098158761098143576109263458761098158761 09814357610926345876109815683287653286836532810926 34587610981435761092634587610981435761092634587610 98143576109263458761098143576109263458761098143576 58761098143576109263458761098158761098143576109263 45876109815876109814357610926345876109815876109814 35761092634587610981587610981435761092634587610981 77777777777777777777777777777777777777777777777777 7777777777777777777777777777









Mano 11-22-2002 12:37 AM

Re: Math Problem
 
The ones digit for powers of any number ending in 3 follows the pattern 3,9,7,1 so if you take 4131mod4 =3, so we choose the third number in this sequence, which is 7.

Mano 11-22-2002 12:43 AM

Re: Answer
 
They go on forever. Proof:

Suppose there were a finite # of primes, and suppose N is the largest. Multiply them all together and add 1 to the result and call this X. Then X can not be divisible by any of the primes, since X-1 is divisible by all of them. Therefore X is prime, but it must be larger than N which is a contradiction, therefore there is not a finite number of primes.

BruceZ 11-22-2002 12:47 AM

Re: 99,193^4131= ...
 
You're full of crap [img]/forums/images/icons/smirk.gif[/img] In the first place, you changed your last digit to 7 after I posted. 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.

BruceZ 11-22-2002 12:50 AM

Re: Answer
 
So you are claiming to have an algorithm for generating larger and larger primes X. Better publish it, because nobody else knows how to do this, and it's a really important problem. I think you are missing something...

Mano 11-22-2002 12:59 AM

Re: Answer
 
Not my claim at all. Assuming that there are a finite # of primes leads to these conclusions, which leads to a contradiction. In practice, since the set of primes is not bounded, if we multiply the first N of them together and add one to the result, the result will often be divisible by a prime larger than the Nth prime - but my initial assumptions did not allow for such primes.

RocketManJames 11-22-2002 01:01 AM

Re: Answer
 
They go on forever... there is no largest prime.

I could spit out the very elegant proof by Erdos, but what's the point in that? In any case, if you haven't seen it, look up the proof by Erdos... amazingly simple. At the time of his proof discovery, the proof that primes went on forever was rather complicated.

Cheers all,
RMJ

RocketManJames 11-22-2002 01:09 AM

Re: Answer ADDENDUM
 
My mistake... it's been a while since I've thought about this stuff. Erdos' elegant proof was that there exists at least one prime between n and 2n for n >= 2. By this proof of his, you can deduce the proof BruceZ is looking for.

Now, for a simple proof on the infinity of primes. Let me think about that.

Sorry for the possible confusion,
RMJ


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