What's the remainder ?

$(2^{ 87 }+3) \mod 7= (2^{ 87 }\mod 7+3)\mod 7$

We know, $2^3=8\equiv 1\mod 7$

Therefore: $2^{87}\equiv(2^3)^{29}\equiv 1^{29}\equiv 1\mod 7$

Using this, the remainder needed is $1+3=\boxed{4}$

notice a sequence $2^{1} +3 = 5 (mod 7)$ $2^{2} +3 = 0 (mod 7)$ $2^{3} +3 =4 (mod 7)$ $2^{4} +3 = 5 (mod 7)$ so, every , 1= 5 2= 0 3= 4 since 87 is a multiple of 3 the answer will be 4

${ 2 }^{ 87 }+3\\ { (1+7) }^{ 29 }\quad +\quad 3\\ \quad using\quad the\quad expansion\quad of\quad { (1+a) }^{ n }\\ 29C0*7^{ 0 }+29C1*{ 7 }^{ 1 }.....29C29*7^{ 2 9}+3\\ 1+7*k+3\\ 7*k+4\quad .Thus,\quad remainder\quad is\quad 4.$

Fermat's Little Theorem