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What is the remainder when 129^(34)+96^(38) is divided by 11?

Guest Aug 12, 2018
 #1
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0

(129^34+96^38) =[212043 3921703930 3573395047 5179191787 1974830700 5020117631 0367952329 7321308417] mod 11 = 9 - the remainder.

Guest Aug 12, 2018
 #2
avatar+93894 
+3

Heureka, could you please check if what I have done is valid    indecision

I am not sure about the rules for modulus but this is my attempt

 

 

 

What is the remainder when 129^(34)+96^(38) is divided by 11?

 

129^(34)+96^(38) [mod11]

= (-3)^34+(-3)^38 [mod 11]

=(-3)^34(1+(-3)^4) [mod11]

= 9^17 (1+81) [mod11]

= (-2)^17(82) [mod11]

=( -32*32*32*4)(5) [mod11]

= (-1*-1*-1*-1 *4)(5) [mod11]

=20 mod11

=9  mod 11

Melody  Aug 12, 2018
 #3
avatar+20153 
+2

excellent

 

Heureka

laugh

heureka  Aug 13, 2018
 #4
avatar+93894 
+2

Thanks Heureka   laugh

Melody  Aug 13, 2018

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