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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 (32)*(52)*(72)*(112) 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: 10,471 Thanks: 693 
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 (21)*(31)*(51)*(71)*(111) = 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: 46 Thanks: 26 
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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