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galactus · Sep 16, 2007, 04:24 PM

I've been wrestling with an induction. I think this'll work:

We must show $10^{2k+1}$ is divisible by 11.

In other words, $\frac{10^{2k+1}+1}{11}=n$

Let k=1 and we get $\frac{10^{3}+1}{11}=91$ .

Base case is true. Now, we must show the induction step.

Let:

$\frac{10^{2(k+1)+1}+1}{11}=s$

$\frac{10\cdot{10^{2k+2}+1}}{11}$

= $\frac{10^{2}(10^{2k+1})+1}{11}$

Add and subtract 10^2:

= $\frac{10^{2}(10^{2k+1})+1+10^{2}-10^{2}}{11}$

= $\frac{10^{2}(10^{2k+1}+1)-99}{11}$

$10^{2}(10^{2k+1}+1)$ is divisible by 11 and so is 99.

Therefore, QED and all that.

pj_martins · Sep 17, 2007, 06:34 AM

This proof works if we have an even number of 0s, since 10^(2k+1) + 1 will always give you an even number of 0s, e.g.. 1001, 100001...
What about if there's an odd number of 0s, i.e. Prove the formula
(10^(k+1))^2 + 1 is not prime for k > 1

galactus · Sep 17, 2007, 07:57 AM

Look again. 2k+1 is an odd integer. The proof is sound.

asterisk_man · Sep 21, 2007, 01:06 PM

I'm really curious about why my months old post about only 101 being prime matters to anyone.

problemsolving · Oct 1, 2007, 05:25 PM

Originally Posted by Wolfie1963

I need to find 2 prime numbers that, if multiplied, would generate a 400-digit number. Any idea how to go about this. I have all the prime numbers but no matter what I do I do not get 400. Am I looking at this question wrong maybe?
Thanks ::confused:

Do you go to itt tech?? I have this same question and there's 4 or 5 of us working on it and have yet to find an answer myself..

trinaity · Jun 18, 2008, 08:11 PM

Did anyone ever find the answer? I'm going to itt tech, and I have to solve this also. Any help would be greatly appreiciated

jkersey76 · Jul 3, 2008, 09:49 PM

I have found one, there is a program you can down load free at:Mersenne Primes: History, Theorems and Lists you can visit:Math Forum - Ask Dr. Math for instructions on use...