massive non-Mersenne prime

[Sam Benry]

I want to solve this problem

The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form 2^6972593−1; it contains exactly 2,098,960 digits. Subsequently other Mersenne primes, of the form 2^p−1, have been found which contain more digits.

However, in 2004 there was found a massive non-Mersenne prime which contains 2,357,207 digits: 28433�2^7830457+1.

Find the last ten digits of this prime number.

I tried using BigInteger to compute the prime it self, I think it would take years to compute it, so I need another way around it. Can anyone suggest any method?
thanks

[fred rosenberger]

hmmm... couldn't you just calculate the bottom 10 digits? What i mean is you don't care about most of them so just ignore them. If i only wanted the last 2 digits of 2^10 + 1, i could do this:

2   2^1
4   2^2
8   2^3
16  2^4
32  2^5
64  2^6
28  2^7  //should be 128, but ignore the '1'
56  2^8  //found from 28 * 2, not 128 * 2
12  2^9  //found from 56 * 2, not 256 * 2
24  2^10 // found from 12 * 2

then add the one, giving me 25.
[ May 15, 2008: Message edited by: fred rosenberger ]

[Sam Benry]

didn't you calculate 2^10, then take the 2 last digits?
I didn't understand how you found the digits without calculating the number...

[Campbell Ritchie]

You can calculate the last few digits for much less effort than calculating the whole number.

[fred rosenberger]

i was not calculating the value. every time it rolled over 100 (in my case), i dropped whatever was in the hundreds place. i'm only ever dealing with 2 digits.

Heck, I think i could even get clever and when the 10's digit is greater than 4, i can pretty much ignore it and just double the 1's digit... i THINK.

this may not give me a performance boost (i think it would actually slow things down), but it is interesting...

[edit]nevermind... that's not correct.[/edit]

in your case, you'd have to account for the 10 digits, but that's going to process much faster than the hundreds or thousands of digits.
[ May 15, 2008: Message edited by: fred rosenberger ]

[Sam Benry]

ok lets try to find the last 10 digits of 2^7830457
whats wrong with my code?

 
import java.math.BigInteger;
public class Problem97 
{
public static void main(String[] args) 
{
BigInteger n = BigInteger.valueOf(2);
String number = n.toString();;
int pow = 1;
for(int i = 1; i <= 7830457; i++)
{
if (number.length() > 10)
{
number = number.substring(number.length()-10, number.length());
n = BigInteger.valueOf(Long.parseLong(number));
n = n.multiply(BigInteger.valueOf(2));
}
else
{
n = n.pow(pow);
number = n.toString();
pow++;
}
}
System.out.println(n);
}
}
 

[fred rosenberger]

could you give us a hint? does it compile? does it die when you run it? does it not give you the answer you are expecting? WHY do you think it's not right?

help us help you.

ok... i played with it a little. i made your for loop only go 5 times. and, i did this:

   else
   {  
      System.out.println("n is " + n);
      System.out.println("n.pow is " + pow);
      n = n.pow(pow);
      number = n.toString();
      pow++;
   }    

see if you can figure out the problem from here...
[ May 15, 2008: Message edited by: fred rosenberger ]

[Sam Benry]

it does compile put never displays an answer (loop never end or takes a long time to end)

ill try your code later and reply, thank you

[fred rosenberger]

My guess is that it will just take a LONG time. You have a logic error in there, that is ending up doing some MASSIVE computations.

another note... you really don't even need BigInteger for this. I think i have a solution that is 22 lines, uses only longs and ints. My answer is

8739992577

and my program runs in under a second.
[ May 15, 2008: Message edited by: fred rosenberger ]

[Sam Benry]

@fred rosenberger
please could you explain where I have logical errors? I can't figure out how to improve my program
I change BigInteger to Long (should it be double instead?)

 
public class Problem97 
{
public static void main(String[] args)
{
long n = 2;
String number = "" + n;
int pow = 1;
for(int i = 1; i <= 7830457; i++)
{
if (number.length() > 10)
{
number = number.substring(number.length()-10, number.length());
n = Long.parseLong(number);
n = 2 * n;
}
else
{
//System.out.println("n is " + n);
    System.out.println("n.pow is " + pow);
    n = (long)Math.pow((double)n, 2);
    number = "" + n;
    pow++;
}
}
System.out.println(n);
}
}

[fred rosenberger]

I changed your code to loop 5 times, and print out n each time. i get this:
java -cp . Problem97
n is 2
n.pow is 1
n is 4
n.pow is 2
n is 16
n.pow is 3
n is 256
n.pow is 4
n is 65536
n.pow is 5
n is 4294967296
n.pow is 6
n is 13709551614
n.pow is 7
n is 13709551614
n.pow is 8
9223372036854775807

What you are doing here each time is raising N to the power of 2. So, first you take 2, and raise it to the power of 2, giving 4.

you then take 4, and raise IT to the power of 2, giving you 16. you then raise 16 to the power of 2, giving 256, etc.

you don't want to raise it by the power of 2 each iteration through the loop... you want to MULTIPLY it by 2 each time through the loop.

Also, there is a much simpler way to chop of extra digits, using a basic operator. do you know what 'modulus' means? It basically means "give me the remainder after i divide". so, '23 mod 5' would be 3. '15 mod 7' would be 1.

so, '217 mod 100' is 17. there is an operator that will do this for you.
[ May 16, 2008: Message edited by: fred rosenberger ]