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:

I'm guessing you need 2 200 digit numbers...

How about a 399 digit prime with the left most digit greater than 4 and 2?

At any rate, how you're going to find 1 prime with that many digits is a pretty big problem to begin with not to mention how you're going to multiply it with something else.

Also, you clearly don't have "all the prime numbers". There are infinitely many prime numbers.

NEWS FLASH: look here for the same question with partial answer http://mathforum.org/library/drmath/view/61527.html

OK. I did the math on the page I just linked to.
(2^1279 - 1) * 100000000000031 = 10407932194667625540905586291079482363116745113783 67553661576502889325698450198541651073818342470797 90693586344515821584917093933607387938136768868797 98844133971889478825726825921094007968773237311354 51144833642863206346098902410390687824141288701866 71615247460015978566928630448851025138075835911283 65331544954407031713112175581409050251234543879558 08672513909648433892849492473214935926798230601697

which, I think has 400 digits

So you were trying to multiply two primes to equal 400? Clearly this is not possible because the prime factors of 400 are (2,2,2,2,5,5). I guess I don't understand how you were initially viewing the problem. Maybe if you tell us your initial understanding of the problem we can try to look at it from that point of view in case you were right and we were misunderstanding. Thanks!

I am going to have to clarify the question with my instructor. I posted it as it is in the problem. Thanks for all the input though. It made me think in a bigger view than I originally was looking.

Ok it seems Capuchin had the right idea. Thanks for all the help

Yeah thanks for the negative rating for my correct idea :p

My ONLY negative rating, by the way! Shame!

So what was the final clarification from the instructor?

I am already considering dropping out of college before my aid kicks in. This question was on the first home work assignment ON THE FIRST DAY OF COLLEGE. We did not receive books to go along with the workbooks. We only received a resource cd that has a pdf of a book that in my opinion is poorly written. The question refers to a section in the ebook that is 3 short paragaphs and has nothing really to do with the question. After 5 days of working on the problem.. I haven't even touched the pencil to the paper yet! I know that this is excited for some, but I have been out of school for 12 years, and I am just trying to get an education so I can provide for my family. I can dropout up to 2 weeks after the start of college and still not owe anything. I am going to do that. How can you ask such a question to a new (old) student. I have been to hundreds of sites trying to figure this out and all I get is some sort of an explanation that I can not understand. I am trying to go through with a 90-100% so I can get more money for school. But when it starts like this... I am just giving up. I am to old now to try to be young and understand what is going on in the mathmatical world.

What I have accomplished (or not):

From what I understand, I will need to find 2 prime number that when multiplied will create a 400 digit number. Now after reading many many many many MANY pages... I have noticed that it will take more power than what my Dell B130 can do in my lifetime. So if that is the case how can you do it? I tried that gp thing and I acctually got 2 different 200 digit prime numbers and actually multiplied them and got a 400 digit number. Now, Do I write that down as my answer and explain out I got it? How do I even check to see if that is even right? Or does the question just wan to know how to do it and not really a result? Colloge is to hard.

What course is this question for? What is the exact question? Maybe there are instructions that somewhat nullify the question or cause it to be a trick, like "explain the research steps you would use to answer the following" or "using the internet, answer the following", or "answer any 2 of the following 3 questions". Please supply more info. I agree that without using the internet I would never have found the solution. And I also agree that even if you have a guess of the 2 primes it's certainly a feat to perform the multiplication. At any rate, don't make any hasty decisions about college based on one homework assignment!

I agree with Asterisk Man. You're dropping out of college because of one math problem? That's just aggravation talking. This is something that needs to be done with computers. I do know that factoring huge primes is how RSA Encryption works.

Here's some help pertaining to your query:

Finding Primes (http://www.perlmonks.org/index.pl?node_id=283725)

Math Forum - Ask Dr. Math (http://mathforum.org/library/drmath/view/69152.html)


Also, consider other math forums. Here are a few:

Freemathhelp.com
SOSmath.com
Mathnerds.com

Relax. That's a horse you got to git back on. No one would be in college if they quit as soon as the going got rough.
I, personlly, hated College Geometry. Proving all that silly, intuitive stuff made me ill.

Yes it was aggravation talking. The question reads exactly "What two prime numbers when multiplied equal a 400 digit numbe? Show your strategy and results in the space provided" The space provided is about 1"x7". The class is Strategies for Problem Solving. I pretty much just gave up and answered the question like this;

I explained how I researched to try to complete the question, then I just guesed on an answer. It didn't say how long the original 2 primes had to be so I multiplied a single digit prime by a 400 digit prime (I think) and got a 400 digt answer. I am not sure but wrote in theory if 101, 1001, 10001... are all prime then it doesn't matter how many zeros you throw in there if it will still be prime.


I hope this works, and I am sorry for the long, possibly annoying posts.

If I were grading your answer I'd give nearly full credit. Your plan of multiplying a 1 digit prime by a 400 digit prime is pretty good. Obviously you'd want to avoid the situation that makes 2*50 result in 3 digits instead of 2 but you're on the right track. Unfortunately, your answer is wrong. You assumed that (10^n)+1 n>1 is always prime which is not true.
101 is prime
1001 is divisible by 7,11,13
10001 is divisible by 73,137
100001 is divisible by 11,9091
essentially it may be the case that the only prime of that form is 101.
I'm really interested in finding out what your professor says the "correct" answer is for this one. Please post an update when you get it!

Ok, I see what you are saying. Now ciphering down Sieve of Eratosthenes, I see that most number ending in 1 is prime, or at least more than any other number that is why I choose that number. I also noticed that every other _91 was prime. So I made an excel sheet that had 200 digits and ran it down to so how many 100 number sieves it would take. I came up with 102 and multiplied that by 2. and came to the conclusion that 1... (398 0's)... 91 may be prime, but that idea failed when I got to 391. (17, 23).

It's just that there is only three questions and getting 1 wrong means that I will get a ~66%.

I will let you know Thursday night what the "answer" is.

Thanks. You guys.

Did you check out those links I posted. They may prove to be helpful. They address your particular problem.

OK. My final conclusion is that this exercise is intended to force you to use some computer software and some thought to produce your answer. Something like mathematica would work great for example. I have used the free software Maxima (http://maxima.sourceforge.net/) for a while now. In it's language you can determine your solution thusly:
p1:next_prime(7*10^199);
p2:next_prime(2*10^199);
prod: p1*p2;
float(ceiling(log(prod)/log(10)));
?print(p1);
?print(p2);
?print(prod);


the line with the float(celing(... just verifies that the result is 400 digits. For this example I got

70000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000039

times

20000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000001019

equals

14000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000721 10000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000039741

Actually, now that I have been getting help from you guys, this IS kind of interesting. I'll let you know what the proffessor says is the right answer Thursday night.

Yes, Asterisk Man is right on. That's probably what the professor wants. Answers will vary.

Here's one I got with Maple 10.

20000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000132 50000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000155907

101 is prime
1001 is divisible by 7,11,13
10001 is divisible by 73,137
100001 is divisible by 11,9091
essentially it may be the case that the only prime of that form is 101.


How can you prove that this is true, that every number of the form 1000... 01 where there is an odd number of 0s in between is not prime except for 101?

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.

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

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

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

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..

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

I have found one, there is a program you can down load free at:Mersenne Primes: History, Theorems and Lists (http://primes.utm.edu/mersenne/index.html) you can visit:Math Forum - Ask Dr. Math (http://mathforum.org/library/drmath/view/61527.html) for instructions on use...