(a) 
since 13 is prime.
(b) 
, and there are 81/3 = 27 multiples of 3 between 1 and 81, which leaves 81 - 27 = 54 numbers between 1 and 81 that are coprime to 81, so 
.
(c) 
; there are 50 multiples of 2, and 20 multiples of 5, between 1 and 100; 10 of these are counted twice (the multiples of 2*5=10), so a total of 50 + 20 - 10 = 60 distinct numbers not coprime to 100, leaving us with 
.
(d) 
; there are 51 multiples of 2, 34 multiples of 3, and 6 multiples of 17, between 1 and 102. Among these, we double-count 17 multiples of 2*3=6, 3 multiples of 2*17=34, and 2 multiples of 3*17=51; we also triple-count 1 number, 2*3*17=102. There are then 51 + 34 + 6 - (17 + 3 + 2) + 1 = 70 numbers between 1 and 102 that are not coprime to 102, and so 
.
(x^2 - 4^2)^2 = x^4 - 4^4
(x^4 - 4^4) (x^2 + 4^2) = x^2 - 4^6
The Angela's solution is wrong cause when you multiply you get a different answer.
I think using a computer software would be the best answer because it is then fair
