Dear NumberTheory Scholar-Apprentices.
Question Q1:
`Compute the last digit of 37^{471} and its last TWO digits.'
Myself, I would rather write than compute, so I will organize my
thoughts so that I hardly need compute, by making a lot of bitsy steps.
Firstly, since B := 37 is coprime to 10 and to 100, the EFT tells us that
P := Per_{10}(B) <| phi(10) =note= 4, and
Q := Per_{100}(B) <| phi(100) =note= 40.
Also, ten divides one-hundred, so
1: P <| Q <| 40.
Now let == mean cong.mod.10. Our P is either 1,2, or 4, but it is
neither 1 nor 2 since, B^1 == -3 which is not ==1, and
B^2 == [-3]^2 = 3^2 == -1,
which is not 1. Thus the mod-10 period of B is 4.
The mod-10 cycle that we see for B is
n: 0 1 2 3 4
B^n: 1 -> -3 -> 9==-1 -> 3 -> 1
where "->" means multiplication by B, mod 10.
Reducing the exponent, 471, modulo 4 yields 3, so
B^{471} == B^3 == 3.
LAST TWO DIGITs: From (1), the mod-100 period, Q, is either
4, 8, 20 or 40.
NOW let == mean cong.mod.100. I now compute mod-100, writing my
results to end in digit "1", so that I can do all the computations
trivially in my head [as opposed to out of my mind]. I put "NO" next
to each POTENTIAL value of Q, which turned out NOT to be Q.
n: **_{100} == Std.Form
-----------------------------------------------
1: 37 == 37
2: 69 == -31
NO 4: [-31]^2 = [31]^2 == 61
NO 8: [61]^2 == 21
16: [21]^2 == 41
YES 20: B^4 * B^{16} == 61*41 == 01
Then 471 =20= 71 =20= 11. (You guessed what "=20=" means, yes?)
Now note that eleven equals 8+2+1. So our table continues as
11: 21*[-31]*37 == 21*31*[-37] == 51*[63] == 13 .
Consequently,
LastTwoDigits(37^{471}) = "13" .
--Best Wishes, "Prof. Jonathan"
**