License plate check digit

last SG · 2003-10-05T16:37:32+00:00

Can't be because Volvo had already absorbed Leyland at that time. If SBS was comprehensive with the LO 3x, then why would they order over 400 VO 3x? The VO i...

service_238 SG

In addition to that. YJ also have the same check digits. Smile

SBS9818A SG

SH-series registrations share the same check digits as TIBS registrations.

^tamago^ SG

i've done a mIRC script that will return the suffix of SBS and TIB licence plate numbers.. download the file by right-clicking then "Save Target As" and then selecting ur mirc directory (usually c:\mirc), load the script with the command "/load -rs liccheck.txt" on any mIRC window and press F11 to start.. u can also view the contents of the script alogrithm with Notepad or any text editor.. will be trying to do a JavaScript version so Pocket PC users can verify the suffixes on Pocket IE on the go when bus-spotting..

some math:
TIB(4XX) = SBS(25XX/91XX)
TIB(5XX) = SBS(26XX/92XX)
TIB(6XX/10XX) = SBS(XX/27XX/93XX)
TIB(7XX/11XX) = SBS(1XX/28XX/94XX)
TIB(8XX/12XX) = SBS(2XX/29XX/95XX)
TIB(9XX) = SBS(3XX/96XX)
SBS(4XX) = SBS(97XX)
SBS(5XX) = SBS(98XX)
SBS(X - 1000) = TIB(X) = SH(X)

reference suffixes:
SBS 0B, 100X, 200R, 300K, 400E, 500A, 600U, 700P, 800J, 900D, 1000U, 1700H, 1800C, 1900Y, 2000M, 2100H, 2500L, 2600G, 2700B, 2800X, 2900R, 3000G, 3700U, 3800P, 3900J, 7000E, 7100A, 7200U, 9000S, 9100L, 9200G, 9300B, 9400X, 9500R, 9600K, 9700E, 9800A
TIB 0H, 100C, 200Y, 300S, 400L, 500G, 600B, 700X, 800R, 900K, 1000B, 1100X, 1200R

SBS9818A SG

Wah tamago you very free hor? Thanks. Will run it and check its accuracy.

^tamago^ SG

duplicate post

^tamago^ SG

dun worry, it took only an hour to finish it=) the JavaScript is the more important one as i can double check my regos list with suffixes on the go.. as u can see, the file is veri small and the operative alogrithm (liccheck command) is very easy to do..

for those who know IRC scripting, this is how it goes:

alias liccheck {

/.set %licchk_inputPrefix $1
/.set %licchk_inputNumber $2

/.set %licchk_SBS_CycledOrder ZXTRMKHECAYUSPLJGDB
/.set %licchk_TIB_CycledOrder ECAYUSPLJGDBZXTRMKH

if %licchk_inputNumber isnum 1-9999 {
if ( ( %licchk_inputPrefix == SBS ) || ( %licchk_inputPrefix == TIB ) ) {
if %licchk_inputNumber isnum 1000-9999 {
/.set %licchk_inputNumber1 $mid(%licchk_inputNumber,1,1)
/.set %licchk_inputNumber2 $mid(%licchk_inputNumber,2,1)
/.set %licchk_inputNumber3 $mid(%licchk_inputNumber,3,1)
/.set %licchk_inputNumber4 $mid(%licchk_inputNumber,4,1)
}
if %licchk_inputNumber isnum 100-999 {
/.set %licchk_inputNumber1 0
/.set %licchk_inputNumber2 $mid(%licchk_inputNumber,1,1)
/.set %licchk_inputNumber3 $mid(%licchk_inputNumber,2,1)
/.set %licchk_inputNumber4 $mid(%licchk_inputNumber,3,1)
}
if %licchk_inputNumber isnum 10-99 {
/.set %licchk_inputNumber1 0
/.set %licchk_inputNumber2 0
/.set %licchk_inputNumber3 $mid(%licchk_inputNumber,1,1)
/.set %licchk_inputNumber4 $mid(%licchk_inputNumber,2,1)
}
if %licchk_inputNumber isnum 1-9 {
/.set %licchk_inputNumber1 0
/.set %licchk_inputNumber2 0
/.set %licchk_inputNumber3 0
/.set %licchk_inputNumber4 $mid(%licchk_inputNumber,1,1)
}

/.set %licchk_inputNumberChecksum $calc( %licchk_inputNumber1 * 12 + %licchk_inputNumber2 * 2 + %licchk_inputNumber3 * 11 + %licchk_inputNumber4 * 1 )

/.set %licchk_inputNumberChecksum $calc( %licchk_inputNumberChecksum % 19 )
if ( %licchk_inputNumberChecksum == 0 ) {
/.set %licchk_inputNumberChecksum 19
}

if ( %licchk_inputPrefix == SBS ) {
/.set %licchk_outputSuffix $mid(%licchk_SBS_CycledOrder, %licchk_inputNumberChecksum, 1)
}
if ( %licchk_inputPrefix == TIB ) {
/.set %licchk_outputSuffix $mid(%licchk_TIB_CycledOrder, %licchk_inputNumberChecksum, 1)
}

/.return %licchk_outputSuffix

}
else {
/.return $null
}
}
else {
/.return $null
}

}

EXPLANATION:
for e.g. SBS9846, take thousands place times 12 (cos from SBS0 shift by 12 to the right to get suffix of SBS1000, and so on), take hundreds place times 2, take tens place times 11, take ones place times 1, add them up and find the remainder when divided by 19 (number of suffixes in cycled order).. find the corresponding position in the SBS's cycled ordering given above..

1. mutliply each digit by their 'reference' number and add up:
9 * 12 + 8 * 2 + 4 * 11 + 6 * 1 = 174

2. divide by 19 and get remainder.
if it is divisible by 19 exactly, take it as 19.
174 divided by 19, the remainder is 3 ( 9 * 19 + 3)

3. now, SBS's cycled order is "ZXTRMKHECAYUSPLJGDB" (SBS1 is Z, 2 is X etc.) and the 3th alphabet is T.. thus T is the suffix for SBS9846.

REVISION 1:

A bug has been found, related to the fact that 0 remainder does not mean taking 0th alphabet but rather the last (19th) alphabet and an additional 3 lines of code are added.. the old bug meant numbers with suffix "B" had "Z" returned instead for SBS and suffix "H" had "E" returned instead for TIB.. users can download the new text file and overwrite the existing one with mIRC closed (important!)! then you can launch mIRC and use the new revised version.. download it at the same location here..

SBS9818A SG

Now running it for some checks, every result it comes up with is correct. Thanks.

^tamago^ SG

first suffixes of major prefix series.. if u need to get the other numbers' suffixes in these series, pls derive on ur own first using the same old method before i compile them into my script.. what is shown here is indicative that it can be or will be registered but need not have been registered by someone already, especially the S?, SF? and SZ? series.. it follows that those series whose first suffix is 'Z' is the same series as SBS and 'E' is same as TIB..

from one series to another corrsponding series, the jump is 2 places right from the same cycled ordering (e.g. "ECAYUSPLJGDBZXTRMKH" )..

SBA1T
SBB1M
SBC1H
SBD1C
SBE1Y
SBF1S
SBG1L
SBH1G
SBJ1X
SBK1R
SBL1K
SBM1E
SBN1A
SBP1P
SBQ1J
SBR1D
SBS1Z
SBT1T
SBU1M
SBW1C
SBX1Y
SBY1S
SBZ1L

SCA1G
SCC1X
SCD1R
SCE1K
SCF1E
SCG1A
SCH1U
SCJ1J
SCK1D
SCL1Z
SCM1T
SCN1M
SCP1C
SCQ1Y
SCR1S
SCU1B
SCV1X
SCW1R
SCX1K
SCY1E
SCZ1A

SDA1U
SDB1P
SDD1D
SDE1Z
SDF1T
SDG1M
SDH1H
SDJ1Y
SDK1S
SDL1L
SDM1G
SDN1B
SDP1R
SDQ1K
SDR1E
SDS1A
SDT1U
SDU1P
SDV1J
SDW1D
SDX1Z
SDY1T
SDZ1M

SFA1X
SFB1R
SFC1K
SFD1E
SFE1A
SFF1U
SFG1P
SFH1J
SFJ1Z
SFK1T
SFL1M
SFM1H
SFN1C
SFP1S
SFQ1L
SFR1G
SFS1B
SFT1X
SFU1R
SFV1K
SFW1E
SFX1A
SFY1U
SFZ1P

SH1E
SHA1Y
SHB1S

SZA1J
SZB1D
SZC1Z
SZD1T
SZE1M
SZF1H
SZG1C
SZH1Y
SZJ1L
SZL1G
SZM1B
SZN1X
SZP1K
SZQ1E
SZR1A
SZS1U
SZT1P
SZU1J
SZV1D
SZW1Z
SZX1T
SZY1M
SZZ1H

EA1H
EB1C
EC1Y
ED1S
EE1L
EF1G
EG1B
EH1X
EJ1K
EK1E
EL1A
EM1U
EN1P
EP1D
EQ1Z
ER1T
ES1M
ET1H
EU1C
EV1Y
EW1S
EX1L
EY1G
EZ1B

S1Y
SB1L
SC1G
SD1B
SE1X
SF1R
SG1K (no one holds this..?)
SJ1U
SK1P
SL1J
SM1D
SN1Z
SP1M
SQ1H
SR1C
SS1Y
ST1S
SU1L
SV1G
SW1B
SX1X
SY1R
SZ1K

CB1B
CSS1L

F1R (someone owns this..!)
FA1X
FB1R
FC1K
FD1E
FE1A
FF1U
FG1P
FH1J
FJ1Z
FK1T
FL1M
FM1H
FN1C
FP1S
FQ1L
FR1G
FS1B
FT1X
FU1R
FV1K
FW1E
FX1A
FY1U
FZ1P

G1K
GA1J
GB1D
GC1Z
GD1T
GE1M
GF1H
GG1C (someone holds this!)
GH1Y
GJ1L
GK1G
GL1B
GM1X
GN1R
GP1E
GQ1A
GR1U
GS1P
GT1J
GU1D
GV1Z
GW1T
GX1M
GY1H
GZ1C

PA1C
PB1Y
PC1S
PZ1X

Q1H
QA1R
QB1K
QC1E

QE1X (odd one out)

QS1Y (odd one out)

QX1U
QY1P

QBA1T
QBB1M
QBC1H
QBD1C
QBE1Y
QBF1S
QBG1L
QBH1G
QBJ1X
QBK1R
QBL1K
QBM1E
QBN1A
QBP1P
QBQ1J
QBR1D
QBS1Z
QBT1T
QBU1M
QBV1H
QBW1C
QBX1Y
QBY1S
QBZ1L

QCA1G
QCE1K
QCF1E
QCG1A
QCH1Y
QCJ1J
QCK1D
QCL1Z
QCM1T
QCN1M
QCP1C
QCQ1Y
QCR1S
QCU1B
QCV1X
QCW1R
QCX1K
QCY1E
QCZ1A

QDA1U
QDB1P
QDC1J
QDD1D
QDE1Z
QDF1T
QDG1M
QDH1H
QDJ1Y
QDK1S
QDL1L
QDM1D
QDN1B
QDP1R
QDQ1K
QDR1E
QDS1A
QDT1U
QDU1P
QDV1J
QDW1D
QDX1Z
QDY1T
QDZ1M

QEA1H
QEB1C
QEC1Y
QED1S
QEE1L
QEF1G
QEG1B
QEH1X
QEJ1K
QEK1E
QEL1A
QEM1U
QEN1P
QEP1D
QEQ1Z
QER1T
QES1M
QET1H
QEU1C
QEV1Y
QEW1S
QEX1L
QEY1G
QEZ1B

QFA1X
QFB1R
QFC1K
QFD1E
QFE1A
QFF1U
QFG1P
QFH1J
QFJ1Z
QFK1T
QFL1M
QFM1H
QFN1C
QFP1S
QFQ1L
QFR1G
QFS1B
QFT1X
QFU1R
QFV1K
QFW1E
QFX1A
QFY1U
QFZ1P

RU1Z

TIB1E

W1B
WA1U
WB1P
WC1J

Y1R
YA1X
YB1R
YC1K
YD1E
YE1A
YF1U
YG1P
YH1J
YJ1Z
YK1T
YL1M
YM1H
YN1C
YP1S
YQ1L
YR1G
YS1B
YT1X
YU1R
YV1K
YW1E
YX1A
YY1U
YZ1P

TRA and TRB cannot be found

S?? and Q?? series are actually identical when it comes to suffix cycle, but if there is a certain S?? series doesn't mean there will be a Q? series and vice versa..

SBS9818A SG

Ok, here’s a brainteaser, which SBS and TIB regoes have the same suffix, if any?

^tamago^ SG

TIB(4XX) = SBS(25XX/ 91XX)
TIB(5XX) = SBS(26XX/ 92XX)
TIB(6XX/ 10XX) = SBS(XX/ 27XX/ 93XX)
TIB(7XX/ 11XX) = SBS(1XX/ 28XX/ 94XX)
TIB(8XX/ 12XX) = SBS(2XX/ 29XX/ 95XX)
TIB(9XX) = SBS(3XX/ 96XX)
SBS(4XX) = SBS(97XX)
SBS(5XX) = SBS(98XX)
SBS(X - 1000) = TIB(X) = SH(X)

^tamago^ SG

If you find it confusing jus read the results from this derivation

Let the number be pqrs. For any such series of motor vehicles in Singapore,

12p + 2q + 11r + 1s = N

To repeat the suffix without repeating the number, we choose to do something to p and q (thousands and hundreds) such that the resulting number has the same remainder when divided by 19, i.e.
12( p + a ) + 2( q + b ) + 11r + 1s = N + ( 12a + 2b ) = N or N + 38 or N + 76 or N + 114 so

12a + 2b = 0 or 38 or 76 or 114 ...

Note that we did not choose 19 and 57 since the multipliers (12 and 2) are both even and it is impossible to obtain an odd number with any combination of a and b.

then, we find out:
1. 12 ¡Á 3 + 2 ¡Á 1 = 38; i.e. a = 3, b = 1, so you add 3100
2. 12 ¡Á 2 + 2 ¡Á 7 = 38; i.e. a = 2, b = 7, so you add 2700 (q has to be 0, 1 or 2!)
3. 12 ¡Á 4 + 2 ¡Á (-5) = 38; i.e. a = 4, b = -5, so you add 3500
4. 12 ¡Á 6 + 2 ¡Á 2 = 76; i.e. a = 6, b = 2, so you add 6200
5. 12 ¡Á 5 + 2 ¡Á 8 = 76; i.e. a = 5, b = 8, so you add 5800 (q has to be 0 or 1!)
6. 12 ¡Á 7 + 2 ¡Á (-4) = 76; i.e. a = 7, b = -4, so you add 6600
7. 12 ¡Á 9 + 2 ¡Á 3 = 114; i.e. a = 9, b = 6, so you add 9300 (p has to 0 and q has to be be between 0 and 6)
8. 12 ¡Á 1 + 2 ¡Á (-6) = 0; i.e. a = 1, b = -6, so you add 400 (p cannot be 9 and q has to be be between 6 and 9)
9. 12 ¡Á 10 + 2 ¡Á (-6) = 114; i.e. a = 10, b = -4 (upper limit)

Note than for any values of a and b, p + a must be less than or equals to 9 and also q + b must be less than or equals to 9. If it's q - b then it must be more than or equals to 0.

The significance of Equation 9 tells us that we have reached the upper limit since a alone is already 10 but p + a must be less than or equals to 9. The only possibility is that p is negative, which is impossible.

Equation 7 provides the little but useful possiblity that numbers from the range 1 to 699 have the possibility of finding the same suffixes with the range 9301 to 9999.

Likewise the other equations indicate that for any number from 1 to 9999, if you add a to the thousands place and add/subtract b to the hundreds place such that the above condition is met, then voila! You have found a recurring suffix series!

Results from this derivation:
EQUATION 1
1 to 899 = 3101 to 3999
1000 to 1899 = 4100 to 4999
2000 to 2899 = 5100 to 5999
3000 to 3899 = 6100 to 6999
4000 to 4899 = 7100 to 7999
5000 to 5899 = 8100 to 8999
6000 to 6899 = 9100 to 9999
EQUATION 2
1 to 299 = 2701 to 2999
1000 to 1299 = 3700 to 3999
2000 to 2299 = 4700 to 4999
3000 to 3299 = 5700 to 5999
4000 to 4299 = 6700 to 6999
5000 to 5299 = 7700 to 7999
6000 to 6299 = 8700 to 8999
7000 to 7299 = 9700 to 9999
EQUATION 3
500 to 999 = 4000 to 4499
1500 to 1999 = 5000 to 5499
2500 to 2999 = 6000 to 6499
3500 to 3999 = 7000 to 7499
4500 to 4999 = 8000 to 8499
5500 to 5999 = 9000 to 9499
EQUATION 4
1 to 799 = 6201 to 6999
1000 to 1799 = 7200 to 7999
2000 to 2799 = 8200 to 8999
3000 to 3799 = 9200 to 9999
EQUATION 5
1 to 199 = 5801 to 5999
1000 to 1199 = 6800 to 6999
2000 to 2199 = 7800 to 7999
3000 to 3199 = 8800 to 8999
4000 to 4199 = 9800 to 9999
EQUATION 6
400 to 999 = 7000 to 7599
1400 to 1999 = 8000 to 8599
2400 to 2999 = 9000 to 9599
EQUATION 7
1 to 699 = 9301 to 9999
EQUATION 8
600 to 999 = 1000 to 1399
1600 to 1999 = 2000 to 2399
2600 to 2999 = 3000 to 3399
3600 to 3999 = 4000 to 4399
4600 to 4999 = 5000 to 5399
5600 to 5999 = 6000 to 6399
6600 to 6999 = 7000 to 7399
7600 to 7999 = 8000 to 8399
8600 to 8999 = 9000 to 9399

^tamago^

SBS9818A SG

I know. I want sth where SBS xxx = TIB xxx. The numbers and suffix must be the same.

service_238 SG

My mind is in a mess after seeing the irc script. Confused

^tamago^ SG

sorry but this is impossible!

if there is a number N such that SBS N and TIB N have the same suffix, then every number in the SBS and TIB series will share the same suffix from 1 right up to 9999.. if not it will never ever happen.. for SBS, this is possible:

SBS xxx = SCL xxx = SDE xxx = SDX xxx = SFJ xxx = EQ xxx = GC xxx = GV xxx = YJ xxx

in these series, every corresponding number xxx has the same suffix throughout..

^tamago^ SG

unless u understanding the codes, or else jus read the explanation and example, and jus use the script can already.. i'm making one that will include every series in Singapore.. coming up next..

^tamago^ SG

the script incorporating all series in Singapore is ready for download here.. always remember to close mIRC before downloading..! it is now more user-friendly as in it does not pop up but displays the result in the same window and updates instantly as you type.. so it'll be better if you need to check a lot of regos at a time Smile

SBS9818A SG

Hmm. Thanks for the scripts!

^tamago^ SG

welcome Wink

^tamago^ SG

Version 0.1.1 is out.. the missing prefix series "PH", "QC?", "QD?", "QE?" and "QF?" are added to the script and the list above.. note that these are not popular series and it is not necessary to update yourself unless you need these series' suffixes..

these information is retrieved thanks to LTA Line at 1800-7320633.. follow this chart for the prefix:
21: A, 22: B, 23: C
31: D, 32: E, 33: F
41: G, 42: H, 43: I
51: J, 52: K, 53: L
61: M, 62: N, 63: O
71: P, 11: Q, 72: R, 73: S
81: T, 82: U, 83: V
91: W, 92: X, 93: Y, 12: Z

TIB822C SG

Hey there everyone Smile

TIB744U TIB1144U SBS144U SBS9444U

I know the suffix sequence is as follows:

K H E C A Y U S P L J G D B Z X T R M

Am I right? Smile

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