Divisibility by 7 (Posted on 2005-05-22)

	Submitted by pcbouhid
Rating: 2.0000 (3 votes)
Solution:	(Hide)
The answer is "IT IS", and we can prove it by several means. Here, a different way (a little insight and some knowledge of divisibility) from the correct answers given : We have : (2222^5555 + 5555^2222) = (2222^5555 + 4^5555) + (5555^2222 - 4^2222) - (4^5555 - 4^2222) Let's analyse each sum enclosed by parentheses. The first (2222^5555 + 4^5555) is divisible by (2222 + 4) = 2226 = 7 x 318, since (a^n + b^n) is divisible by (a + b) if n is odd, and so divisible by 7. The second (5555^2222 - 4^2222) is divisible by (5555 - 4) = 5551 = 7 x 793, since (a^n - b^n) is always divisible by (a - b), and so divisible by 7. The third (4^5555 - 4^2222) may be written : 4^2222(4^3333 - 1) = 4^2222(64^1111 - 1) Since (64^1111 - 1) is divisible by (64 - 1) = 63 = 7 x 9 (see the second sum), so it is also divisible by 7. So, (2222^5555 + 5555^2222) is divisible by 7. Good work, everybody !!

	Subject	Author	Date
	Alternative Methodology	K Sengupta	2007-11-14 04:12:54
	re: Big numbers are NO PROBLEM for divisibility	phi	2006-03-09 04:43:36
	re(2): Big numbers are NO PROBLEM for divisibility	Rupesh Khandelwal	2005-05-25 15:43:38
	re: Big numbers are NO PROBLEM for divisibility	Ravi Raja	2005-05-25 06:05:19
	Big numbers are NO PROBLEM for divisibility	Rupesh Khandelwal	2005-05-25 05:52:18
	divisible by 7?	heather	2005-05-23 14:02:42
	re: Divisible by 7	Justin	2005-05-22 17:02:50
	Divisible by 7	Ravi Raja	2005-05-22 16:40:36