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May 5th, 2017, 03:39 PM   #1
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Using Totient to examine n = 4622

For the number of pairs (a,b) with a+b= 4622 and neither a nor b divisible by 2,3,5,7,11.

I was told be a friend that the number of such pairs is found using Euler's Totient resulting in (3-2)*(5-2)*(7-2)*(11-2)

Is this the case and can anyone explain how Euler's Totient is used to determine this?

Be gentle with me.....
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May 5th, 2017, 07:01 PM   #2
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Did you google "Euler's Totient" ?
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May 5th, 2017, 10:26 PM   #3
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Yes......I can see that the Totient applies to the first n integers but how does this help when I am examining a set of pairs of integers ?

So for 2310 integers....the totient would be (2-1)*(3-1)*(5-1)*(7-1)*(11-1) = 480

However,again....how can I apply this to 2311 pairs?

Last edited by Elaine; May 5th, 2017 at 10:41 PM.
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May 7th, 2017, 08:19 AM   #4
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Elaine,
Unless your friend's name is Socrates, I should have thought your friend would have explained the answer to you. Here's an explanation that I hope you can follow.



If you're into computer programming, here's a little Java program that verifies the above proposition:

Code:
   public static void main(String[] args) {
      int[] primes = {2, 3, 5, 7, 11};
      int total = 0;
      int i, j, n = 2310;
      for (i = 1; i < n; i++) {
         for (j = 0; j < 5; j++) {
            if (i % primes[j] == 0) {
               break;
            }
         }
         if (j == 5) {
            for (j = 0; j < 5; j++) {
               if ((2 * n + 2 - i) % primes[j] == 0) {
                  break;
               }
            }
            if (j == 5) {
               total++;
            }
         }
      }
      System.out.println(total);
   }

Last edited by johng40; May 7th, 2017 at 08:26 AM.
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