find the remainder when (5+5^1+5^2+... +5^55)...
find the remainder when (5+5^1+5^2+... +5^55) divided by 13?
n/a
solve it
5+5^1+5^2+5^3..........5^55
5 to the pwer anything will result 5 in the last digit
eg- 5^2=25
5^3=125
in all places the last digit is 5 , so
the sum of all , last digit is
= 5+5+5............55(terms)
=5+5*55
=56*5=280%13=7
Hence ans is 7.............
the answer is 16
its converts to 5/13 + (5^1 + 5^2 + .......5^55)/13
applying GP formation in 2nd part it comes out to be 5/13 + 5(5^55 - 1)/4*13
5/13 + 5(5^55-1)/52
taking lcm its
(20 + 5^56 - 5)/52
(15 + 5^56)/52
special method
to find the remainer of 5^56/52
divide 5 by 52 it gives remainder 5
SQUARE THE REMAINER that is 25 again divide by 52 remainer is 25
again multiply by 5 its 125 divide by 52 its 21 DO THIS UNTIL REMAINDER = 1
again multiply by 21 by 5 its 105 again divide by 52 the remainder is 1 STOP NOW !
count the number of cycles its 4
now divede power thats 56 by 4 its 0
if 0 comes go to the last cycle its 1
so remainder of 5^56/52 = 1
and remainder of 15/52 is 15
ADD the remainers = 15 + 1 = 16
i wrote a program to calculate this
it gives answer as 0
you guys find out the short method
ans is 8