"Triple Palindromic" Date CalculationsUK Format only  dd/mm/yyyy

The approach for evaluating palindromes for the US date
format  mm/dd/yyyy  follows along pretty much the same lines. This is not intended to be a rigorous mathematical proof  it is more a layman's way of understanding what's going on. If you have any comments or disagreements, please contact me. 
For the sake of both sanity and brevity, we'll consider only 4 digit years (i.e. from the year 1000 AD to 9999 AD). I think this is reasonable  apart from the fact that in his days Julius C didn't use a 24 hour clock, I'm pretty certain they did not insert a gratuitous zero in front of their 1, 2 and 3 digit years. At the other end of the scale  beyond the year 9999  who really gives a damn anyway. So we are considering very simply 3 groups of 4 digits each for genuine "triple palindromicity"  ie they must take the form nmmn nmmn nmmn. (See "normal palindromicity" for corresponding calculations/estimations)
We start by defining the two digits of each element:
H1H2  Hours 
Mi1Mi2  Minutes 
D1D2  Day 
Mo1Mo2  Month 
C1C2  Century 
Y1Y2  Years 
Using this notation we can express the time and date as H1H2:Mi1Mi2 D1D2/Mo1Mo2 C1C2Y1Y2
Now we know that each of these elements is subject to certain constraints:
1) we define triple palindromicity by the relationships:
a) H1 = Mi2 =
D1 = Mo2 = C1 = Y2
and
b) H2 = Mi1 = D2 =
Mo1 = C2 = Y1
2) Looking at the constraints of each element in turn:
a)  The hours (H1H2) must lie between 00 and 23:  H1 = 0, 1 or 2 and H2 = 0...9 and 00 <= H1H2 <= 23 
b)  The minutes (Mi1Mi2) must lie between 00 and 59  Mi1 = 0...5 and Mi2 = 0...9 
c)  The days (D1D2) must lie between 01 and 31 (but watch out for 28, 29 and 30 as well) 
D1 = 0, 1, 2, or 3 and D2 = 0...9 and 01 <= D1D2 <= 31 
d)  The months (Mo1Mo2) must lie between 01 and 12  Mo1 = 0 or 1 and Mo2 = 0...9 and 01 <= Mo1Mo2 <= 12 
e)  The century (C1C2) must lie between 10 and 99 (we're looking only at the years 1000 to 9999) 
C1 = 1...9 and C2 = 0...9 
f)  The years (Y1Y2) are totally unconstrained  Y1 = 0...9 and Y2 = 0...9 
Table 1  combining 1a above with 2a2f
H1 =  0,1,2 
Mi2 =  0,1,2,3,4,5,6,7,8,9 
D1 =  0,1,2,3 
Mo2 =  0,1,2,3,4,5,6,7,8,9 
C1 =  1,2,3,4,5,6,7,8,9 
Y2 =  0,1,2,3,4,5,6,7,8,9 
This leads directly to the conclusion that if H1=Mi2=D1=Mo2=C1=Y2 then they can each be only either 1 or 2
Table 2  combining 1b above with 2a2f
H2 =  0,1,2,3,4,5,6,7,8,9 
Mi1 =  0,1,2,3,4,5 
D2 =  0,1,2,3,4,5,6,7,8,9 
Mo1 =  0,1 
C2 =  0,1,2,3,4,5,6,7,8,9 
Y1 =  0,1,2,3,4,5,6,7,8,9 
This leads directly to the conclusion that if H2=Mi1=D2=Mo1=C2=Y1 then they can each be only either 0 or 1
From there it can be demonstrated that there are only four possible combinations of palindromic UK dates and times:
Combination  Time  Date  Year 
1001  10:01  10 January  1001 
1111  11:11  11 November  1111 
2002  20:02  20 February  2002 
2112  21:12  21 December  2112 
QED
