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I cheated a bit...

I used 24 and 25 to build the patterns:

24^n mod 7 is 3 2 6 4 5 1...

25^n mod 7 is 4 2 1...

5555 mod 6 is 5 so 2222^5555 mod 7 is the 5th position = 5

2222 mod 3 is 2 so 5555^2222 mod 7 is the 2nd position = 2

(5+2) mod 7 = 0

Edited by curr3nt
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any no of (x^y +y^x) is divisible by (x+y) in our case the (x+y) = 7777 which is divisible by 7. hence the remainder will be 0.

Edited by rupeshdubey
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any no of (x^y +y^x) is divisible by (x+y) in our case the (x+y) = 7777 which is divisible by 7. hence the remainder will be 0.

rupeshdubey, do you know of a proof of this?

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#	mod 7		#	mod 7		#	mod 7		#	mod 7		#	mod 7		#	mod 7

8	1		9	2		10	3		11	4		12	5		13	6

64	1		81	4		100	2		121	2		144	4		169	1

512	1		729	1		1000	6		1331	1		1728	6		2197	6

4096	1		6561	2		10000	4		14641	4		20736	2		28561	1

32768	1		59049	4		100000	5		161051	2		248832	3		371293	6

262144	1		531441	1		1000000	1		1771561	1		2985984	1		4826809	1

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10 mod 7 = 3 - Easy enough...

10 * 10 = 10 * 7 + 10 * 3 = 10 * 7 + 7 * 3 + 3 * 3 => (10 * 10) mod 7 equates to 9 mod 7 = 2

100 * 10 = 100 * 7 + 100 * 3 = 100 * 7 + 98 * 3 + 2 * 3 => (100 * 10) mod 7 equates to 6 mod 7 = 6

1000 * 10 = 1000 * 7 + 1000 * 3 = 1000 * 7 + 994 * 3 + 6 * 3 => (1000 * 10) mod 7 equates to 18 mod 7 = 4

each iteration is the previous (result * 3) mod 7

3

3 * 3 = 9 mod 7 = 2

2 * 3 = 6 mod 7 = 6

6 * 3 = 18 mod 7 = 4

4 * 3 = 12 mod 7 = 5

5 * 3 = 15 mod 7 = 1

1 * 3 = 3 mod 7 = 3

11 mod 7 = 4

11 * 11 = 11 * 7 + 11 * 4 = 11 * 7 + 7 * 4 + 4 * 4 => (11 * 11) mod 7 equates to 16 mod 7 = 2

121 * 11 = 121 * 7 + 121 * 4 = 121 * 7 + 119 * 4 + 2 * 4 => (121 * 11) mod 7 equates to 8 mod 7 = 1

same pattern but with a 4 (result * 4) mod 7

since 2222 mod 7 = 3 and 5555 mod 7 = 4 we know the pattern for each ^n

remainder of 3 follows a pattern of 6 while remainder of 4 follow a pattern of 3

5555 mod 6 = 5 so we want the 5th number in the pattern => 5

2222 mod 3 = 2 so we want the 2nd number in the pattern => 2

(5 + 2) mod 7 = 0

Does this count for not using a calculator?

edit - typo...

Edited by curr3nt
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