My Math Forum Using Totient to examine n = 4622

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 May 5th, 2017, 02:39 PM #1 Newbie   Joined: May 2017 From: England Posts: 2 Thanks: 0 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.....
 May 5th, 2017, 06:01 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,398 Thanks: 829 Did you google "Euler's Totient" ?
 May 5th, 2017, 09:26 PM #3 Newbie   Joined: May 2017 From: England Posts: 2 Thanks: 0 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 09:41 PM.
 May 7th, 2017, 07:19 AM #4 Member   Joined: Jan 2016 From: Athens, OH Posts: 88 Thanks: 47 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 07:26 AM.

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