EXPLANATION: Both 61 and 17 are prime numbers and since (17,61)=1, by Fermat’s Little Theorem, 61^16=1 (Mod 17) giving, 61^2000=1 (Mod 17). We observe that, 61=10 (Mod 17), so that 61^6=(-7)^6 (Mod 17)= 7^6 (Mod 17). Now, 7^2= -2 (Mod 17); so that 7^6= - 8 (Mod 17). Hence, 61^6= -8 (Mod 17), giving 61^2006= (61^2000)(61^6)= 1*(-8) (Mod 17)= - 8 Mod(17))= 9 Mod(17).

Since, 61 is odd, it follows that 61^2006 is also odd. odd.Accordingly:

61^2006 = 27 Mod (59) = 9(Mod 17)= 1 Mod(2).---------(#)

Let, us denote 61^2006 by x, and the required remainder by r.
So, x-r is divisible by 2006 = 2*17*59; and since 2,17 and 19 are pairwise coprime, it follows that (x-r) is divisible by each of 2,17 and 59.

From the first congruence, we obtain,

x = 59*L + 27 = 9(Mod 17)
or, r = 9 (mod 17) giving, 59L + 18 = 0 (Mod 17); or, 8L + 1 = 0 (Mod 17); giving L = 2 (Mod 17) or, L = 17*M + 2, for some M.

Consequently, r = 59(17*M + 2)+ 27 = 1003*M + 145.
For M=0, we obtain r = 145.

For, M greater than or equal to 2,we obtain r > = 2151 > 2006, which is not feasible.

For M=1, we obtain r = 1148, which is even and thus contradicts (#)

Consequently, it follows that the required remainder must be 145.