Question

find the remainder when x = 5^5^5^5^5^5 ........24 times 5 is divided by 24

Answer 1

since 5'5=25=odd no.

5'25=5*5*5*....=odd no.

5'oddno.=odd no.

so x=5'5'5'5....24times =5*5*5*5*....oddtimes

and since a no. 5'a can be written as

5*5*5*5......a times

if a is to be odd no.

then. 5'a/24=25*5*5*5...a-12times /24

=(1+1/24)*5*5*5*......a-2 times

=5*5*.....+5*5*5.......a-2 times/24

=5*5*.....+25*5*5*5...a-4times/24

=5*5*...+(1+1/24)*5*5...a-4times.

so if we keep doing this we will keep getting multiplication of fives a-2,a-4,a-6,....

5,3,finally 1 times

so x/24= some integer +(1+1/24)*5

x/24 =integer+5/24

hence the ans is 5.

the answer may seem a bit of annoying but very simple arithmatic has been used. so if you find any difficulty you may ask.