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?

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.

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

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.

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.

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.

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...

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.

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

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

Re: Answer
 

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.

Re: Answer
 

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.

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]

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]

No you are not an ameteur ...
 

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

Answer: 7

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 ?

Re: 99,193^4131= ...
 

If you want to see all the digits, follow the attached link.
99193^4131

My Answer
 

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.

My answer is 7.

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!