I got it through observations purely.
Checksum alphabet letters used:
A
B
C
D
E
G
H
J
K
L
M
P
R
S
T
U
X
Y
Z
Easy?
I observed that there was a pattern between checksum letters,
through observing SMRT buses on the run.
An example: TIB 425 R.
Next letters:
TIB 426 M
TIB 427 K
TIB 428 H
TIB 429 E
TIB 430 A
Now, take a look at the letters in an upward direction, starting
from R.
A
B
C
D
E
G
H
J
K
L
M
P
R
Notice that they are in alternatives with the exception of "C".
This is because at a start of a new tens (in this case, 430.), the
next alternative letter is skipped and replaced by the alternative
to the next alternative.
Next, let me give you another example.
TIB 589 B
Next on hand, according to the list of checksum letters:
TIB 590 X
TIB 591 T
TIB 592 R
TIB 593 M
TIB 594 K
TIB 595 H
TIB 596 E
TIB 597 C
TIB 598 A
TIB 599 Y
TIB 600 B
Now you would ask, why isn't it "U", and "B" instead?
Here's a pretty reasonable rationale I found.
TIB 500 G
TIB 501 D
TIB 502 B
TIB 800 R
Going backwards...
TIB 700 X
TIB 701 T
TIB 702 R
TIB 703 M
TIB 704 K
Going forward again...
TIB 801 M
TIB 802 K
TIB 900 K
What do you notice?
The checksum letter of any second number of a hundred always
appears as the first checksum letter of any consecutive hundred.
This applies to ALL registration plates.
Also, any checksum letter with the numbers "XX89" always appears as
the first hundred of "X(X+1)00".
Now, you would ask... why? I don't really know, except that it
makes it easy to track when you've got a easy-to-understand pattern
on hand.
Now, something interesting for you to find out...
TIB 600 B
TIB 700 X
TIB 800 R
TIB 900 K
Slowing down the pace...
TIB 902 E
TIB 989 E
TIB 990 A
TIB 991 Y
TIB 992 U
TIB 993 S
TIB 994 P
TIB 995 L
TIB 996 J
TIB 997 G
TIB 998 D
TIB 999 B
I shall stop here...
Shouldn't 1000 have checksum letter of "E", according to the
rationales I've proven? Why is it that...
TIB 1000 B?
It corresponds to TIB 600 B.
TIB 1100 X corresponds to TIB 700 X
TIB 1200 R corresponds to TIB 800 R
TIB 1300 K corresponds to TIB 900 K
Will 1400 be B again? That's for you to find out.
In summary,
Let the checksum letter of any number be "N".
XX00 N = XX[00 + 18] N = XX(18 + 17) N = XX(18 + 17 + 17) N = XX[18
+ 17 + 17 + (18 + 17)] N.
XX02 N = XX(2 + 87) N = X(X+1)00 N.
X6XX N = (X+1)0XX N but not equals to (X+1)(0+4)XX N.
I'll give you now the pattern of checksum letters.
Applying the above 2 forumlas,
Let's say that XX01 N, whereby N = A.
XX00 C
XX01 A
XX02 Y
XX03 U
XX04 S
XX05 P
XX06 L
XX07 J
XX08 G
XX09 D
XX10 Z
XX11 X
XX12 T
XX13 R
XX14 M
XX15 K
XX16 H
XX17 E
XX18 C
XX19 A
XX20 U
XX21 S
XX22 P
XX23 L
XX24 J
XX25 G
XX26 D
XX27 B
XX28 Z
XX29 X
XX30 R
XX31 M
XX32 K
XX33 H
XX34 E
XX35 C
XX36 A
XX37 Y
XX38 U
XX39 S
XX40 L
.
.
.
.
.
XX71 Y
XX72 U
XX73 S
XX74 P
XX75 L
XX76 J
XX77 G
XX78 D
XX79 B
XX80 X
XX81 T
XX82 R
XX83 M
XX84 K
XX85 H
XX86 E
XX87 C
XX88 A
XX89 Y
XX90 S
XX91 P
XX92 L
XX93 J
XX94 G
XX95 D
XX96 B
XX97 Z
XX98 X
XX99 T
X(X+1)00 Y
