Fermat's little theorem (lemma)

I'm doing Number Theory, completed multiplication, division etc, but now moved on to Fermat's little theorem.
Having a bit of trouble seeing how to go about a couple of examples I have.
Generally
a^(p-1) _≡ 1 (mod p)

where p is a prime number
& a is an integer

i) 3^18 divided by 19
so p=19 & a=3
3^18 _≡ 1 (mod 19)
Hence remainder is 1

Fine

ii) 3^55 divided by 19
but 55 isn't a prime, so how do I do this, please?

Yes, the example solution tells me this, too & goes on two steps further which is
= 3 x 1^3
= 3 (mod19)

I'm still feeling a bit thick about this.

why is it 3x . . . . (3^18)^3
the 3x has me confused, sorry.

I'm not sure what you are confused about. Can you clarify?

Can you see that ?

the prime number p, in this case, is 19 (what goes in the modulus). the fact that 55 isn't prime isn't relevant.

what we DO know is that:

3¹⁸ = 1 (mod 19) this is FLT with p = 19, a = 3, and p-1 = 18.

THEN we use the division algorithm to write 55 = 18q + r (divide 55 by 18, and compute the remainder).

since 18*3 = 54, we get q = 3 and r = 1:

55 = 18(3) + 1. therefore:

3⁵⁵ (mod 19)

= 3^{18(3) + 1} (mod 19)

= (3¹⁸⁽³⁾)(3¹) (mod 19) (from the rules of exponents)

= (3¹⁸)³(3) (mod 19)

= (3¹⁸ (mod 19))³(3 (mod 19)) (the usual rules of multiplication still hold mod 19)

= (1 (mod 19))(3 (mod 19))

= (1)(3) (mod 19)

= 3 (mod 19)

Thanks for a brilliant explanation, however I have another similar question I don't seem to be able to finish.
Could someone help with this please?

43^43 divided by 17

FLT is a^p-1 ≡ 1 (mod p)
& I know that 43 = 2*16+11

I have used the steps as in above explanation,
43^43 = 2^16(2) + 11 (mod 17)
= 2^16(2) + 11 (mod 17)
= (2^16(2))(2^11) (mod 17)
= (2^16)^2 (2)^11 (mod 17)
= (2^16 (mod 17))^2 * (2 (mod 17))^11

er. . . not quite sure what to do next.

that's not the method i wrote.

we have, first of all, 43 = 9 (mod 17) (we reduce "a" mod p).

so 43⁴³ = 9⁴³ (mod 17).

now 43 = (16)(2) + 11, so

9⁴³ = 9^{(16)(2) + 11} (mod 17)

= (9¹⁶)²(9¹¹) (mod 17)

= (1)²(9¹¹) = 9¹¹ (mod 17) <---FLT used HERE (with a = 9, and p-1 = 16)

note we've done two things here: we've reduced the number being exponentiated from 43 to 9, and we've reduced the exponent from 43 to 11.

from here, we have to do things "the hard way":

9² = 81 = 13 (mod 17)
9⁴ = (9²)² = 13² = 169 = 16 = -1 (mod 17)
9⁸ = (9⁴)² = 1 (mod 17)

therefore, 9¹¹ = 9⁸⁺³ = (9⁸)(9³) = (1)(9³) = 9³ (mod 17)

finally, 9³ = (9²)(9) = (13)(9) = 117 = 15 (mod 17)

hence 43⁴³ = 15 (mod 17)