dear friend, this question involves using the binomial theorem. on writing  4^87 as (3+1)^87 we can expand it as  C(87,0) * 3^87 + C (87,1) * 3^86  + C(87,2) * 3^85 + ............................. + C (87,87) we can easily see that all terms are divisible by 3 but the last one that is (87,87) that is 1 is not. so the remainder will be 1... rate me.....

see gal one person has already told one way to get to the ans easily

but for beginers i would sugest the folowing way

u hv got 4^87 / 3

= 4.4.4.    ..................4.4.4  87 times divided by 3

=4/3 4^86

=(1+ 1/3) 4^86

=4(1+1/3)4^85

=(4+4/3)4^85

=(4+1+1/3)4^85

=(5+1/3)4^85

=(20+4/3)4^84

=(21+1/3)4^84

...........................

..................

=(k+1/3) where k is a constant

hence remainder is 1

now try this one

its a bit tricky

5555^2222 + 2222^5555 when divided by seven gives remainder = ? ans is 0

apply modulo arithmatic ..

4^2=1mod(3)

raising both sides to the power of 43, we get

4^86=1mod(3)

so 4^87=4mod(3)=1mod(3)

Hence the remainder is 1