Guest Posted April 4, 2011 Report Share Posted April 4, 2011 find out remainder for (2222^5555+5555^2222)\7 without using any mathematical tool and do explain it.. All the best.. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 4, 2011 Report Share Posted April 4, 2011 (edited) Define mathematical tool my brain? Edited April 4, 2011 by maurice Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted April 4, 2011 Report Share Posted April 4, 2011 (edited) 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 April 4, 2011 by curr3nt Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 4, 2011 Report Share Posted April 4, 2011 using modular arithmetic it is the same remindar of 3^2222 + 4^5555. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 4, 2011 Report Share Posted April 4, 2011 (edited) 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 April 4, 2011 by rupeshdubey Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 4, 2011 Report Share Posted April 4, 2011 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? Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted April 4, 2011 Report Share Posted April 4, 2011 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. (2^3 + 3^2) = 8 + 9 = 17 which is not divisible by 5 or (2 + 3) Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted April 4, 2011 Report Share Posted April 4, 2011 # 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 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 4, 2011 Report Share Posted April 4, 2011 Is the same as 3^5555 + 4^ 2222. (mod 7) is the same as 3^5 + 4^2 Is the same as 0 Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted April 4, 2011 Report Share Posted April 4, 2011 (edited) 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 April 4, 2011 by curr3nt Quote Link to comment Share on other sites More sharing options...
0 Aaryan Posted April 5, 2011 Report Share Posted April 5, 2011 Going... to faint... so... much... math!!! My brain's swimming! Quote Link to comment Share on other sites More sharing options...
0 k-man Posted April 5, 2011 Report Share Posted April 5, 2011 Going... to faint... so... much... math!!! My brain's swimming! Judging by that X-Ray... you're probably right Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 7, 2011 Report Share Posted April 7, 2011 6 Quote Link to comment Share on other sites More sharing options...
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find out remainder for
(2222^5555+5555^2222)\7
without using any mathematical tool and do explain it..
All the best..
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