# find the remainder of (5+5^1 + 5^2+.....5^55)...

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Incognito

find the remainder of (5+5^1 + 5^2+.....5^55) divide by 13

Incognito
5^1 = 5 leavs reminder 5 5^2

5^1 = 5 leavs reminder 5
5^2 = 25 leaves reminder 12
5^3 = 125 leaves reminder 8
5^4 = 625 leaves reminder 1
this cycle of 5,12,8,1  countinues

Incognito

What is the answer of this question?

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Joined: 2011-03-13

the Q can be simplified as follows-
5 + 5^1 + 5^ 2 + 5 ^ 3........5^55 = 5 + 5 ( 5^55 - 1 )/ ( 5 - 1 )

= 5 + ( 5 ^ 56  - 5 )/4 = ( 5 ^ 56 + 15 ) / 4
now by fermats littlle theorem-

5 ^ 13 - 5 = 13k

5 ^ 13 = 13k + 5
raising to power 4-
5 ^ 52 = (13k + 5) ^ 4 = 13k + 625 = 13k + 1
multiplying both sides by 5 ^ 4-
5 ^ 56 = 13k + 625  =  13k + 1
5 ^ 56 + 15 = 13k + 16
left side is divisible by 4, 16 is div. by 4, hence 13k is also and so is k.
dividing by 4 we get 4 in place of 16 and the required eqn on left

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Incognito
fermats theorem

what is fermats theorem.. i dint hear till now.. can you explain?