Random 8 digit number?

I need to know the formula to create a column of random 8 character ID
numbers, preferably with a set prefix of 3 numbers. I'd be REALLY happy if
I could create it with random text and numerals, but it doesn't have to be.

(where 123 is the constant)
i.e.:
12386756
12337288
12378645
12321356

or

123g568e
1237t7er
12397u23

Thanks in advance for your help.

Kenny

Use:

=RANDBETWEEN(12300000,12399999)

--
Gary's Student


"KennyS" wrote:

> I need to know the formula to create a column of random 8 character ID
> numbers, preferably with a set prefix of 3 numbers. I'd be REALLY happy if
> I could create it with random text and numerals, but it doesn't have to be.
>
> (where 123 is the constant)
> i.e.:
> 12386756
> 12337288
> 12378645
> 12321356
>
> or
>
> 123g568e
> 1237t7er
> 12397u23
>
> Thanks in advance for your help.
>
> Kenny
>
>
>

Kenny,

Are you looking for a list of numbers that have no repeats?

It can start with ="123" & (something) but I think it'll have to be vba to
enure no repeats.

Beege

"KennyS" <[email protected]> wrote in message
news:[email protected]...
>I need to know the formula to create a column of random 8 character ID
> numbers, preferably with a set prefix of 3 numbers. I'd be REALLY happy
> if
> I could create it with random text and numerals, but it doesn't have to
> be.
>
> (where 123 is the constant)
> i.e.:
> 12386756
> 12337288
> 12378645
> 12321356
>
> or
>
> 123g568e
> 1237t7er
> 12397u23
>
> Thanks in advance for your help.
>
> Kenny
>
>

You have already post this question two days ago.

English version

Kenny,


Put in A1: 1234567890abcdefghijklmnopqrstuvwxyz
and in A2:
=3D"123"&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,IN=
T(=ADRAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)



Best regards,


Michel BRUYERE
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
Version fran=E7aise


Kenny,


Placer en A1 : 1234567890abcdefghijklmnopqrstuvwxyz
et en A2 :
=3D"123"&STXT(A$1;TRONQUE(ALEA()*36)+1;1)&STXT(A$1;TRONQUE(ALEA()*36)+1;1)&=
ST=ADXT(A$1;TRONQUE(ALEA()*36)+1;1)&STXT(A$1;TRONQUE(ALEA()*36)+1;1)&STXT(A=
$1;TR=ADONQUE(ALEA()*36)+1;1)



Salutations.=20


Michel BRUYERE

Replys to the other post:

English version

Kenny,

Put in A1: 1234567890abcdefghijklmnopqrstuvwxyz
and in A2:
="123"&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)

Best regards,

Michel BRUYERE


Hi Kenny,

Adding to Michel's solution, you can name the formula
MID(A$1,INT(RAND()*36)+1,1) and thereby shorten the overall formula.

Go to Insert > Name > Define > enter name > Refers to >
=MID(A$1,INT(RAND()*36)+1,1) > OK.

So if you named the formula X, then your worksheet formula would be:

="123"&X&X&X&X&X

HTH
Regards,
Howard

Maybe those did not get thru to KennyS

"Gary''s Student" <[email protected]> wrote in message
news:[email protected]...
> Use:
>
> =RANDBETWEEN(12300000,12399999)
>
> --
> Gary's Student
>
>
> "KennyS" wrote:
>
>> I need to know the formula to create a column of random 8 character ID
>> numbers, preferably with a set prefix of 3 numbers. I'd be REALLY happy
>> if
>> I could create it with random text and numerals, but it doesn't have to
>> be.
>>
>> (where 123 is the constant)
>> i.e.:
>> 12386756
>> 12337288
>> 12378645
>> 12321356
>>
>> or
>>
>> 123g568e
>> 1237t7er
>> 12397u23
>>
>> Thanks in advance for your help.
>>
>> Kenny
>>
>>
>>

I did not see my previous post listed, hence I saw none of the replies.

Thanks to all who replied!
KS


"Michel Bru" <[email protected]> wrote in message
news:[email protected]...
You have already post this question two days ago.

English version

Kenny,


Put in A1: 1234567890abcdefghijklmnopqrstuvwxyz
and in A2:
="123"&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(*RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)



Best regards,


Michel BRUYERE
==============================================================
Version française


Kenny,


Placer en A1 : 1234567890abcdefghijklmnopqrstuvwxyz
et en A2 :
="123"&STXT(A$1;TRONQUE(ALEA()*36)+1;1)&STXT(A$1;TRONQUE(ALEA()*36)+1;1)&ST*XT(A$1;TRONQUE(ALEA()*36)+1;1)&STXT(A$1;TRONQUE(ALEA()*36)+1;1)&STXT(A$1;TR*ONQUE(ALEA()*36)+1;1)



Salutations.


Michel BRUYERE

Thanks L. This worked great!
KS
"L. Howard Kittle" <[email protected]> wrote in message
news:[email protected]...
> Replys to the other post:
>
> English version
>
> Kenny,
>
> Put in A1: 1234567890abcdefghijklmnopqrstuvwxyz
> and in A2:
> ="123"&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)&MID(A$1,INT(RAND()*36)+1,1)
>
> Best regards,
>
> Michel BRUYERE
>
>
> Hi Kenny,
>
> Adding to Michel's solution, you can name the formula
> MID(A$1,INT(RAND()*36)+1,1) and thereby shorten the overall formula.
>
> Go to Insert > Name > Define > enter name > Refers to >
> =MID(A$1,INT(RAND()*36)+1,1) > OK.
>
> So if you named the formula X, then your worksheet formula would be:
>
> ="123"&X&X&X&X&X
>
> HTH
> Regards,
> Howard
>
> Maybe those did not get thru to KennyS
>
> "Gary''s Student" <[email protected]> wrote in
> message news:[email protected]...
>> Use:
>>
>> =RANDBETWEEN(12300000,12399999)
>>
>> --
>> Gary's Student
>>
>>
>> "KennyS" wrote:
>>
>>> I need to know the formula to create a column of random 8 character ID
>>> numbers, preferably with a set prefix of 3 numbers. I'd be REALLY happy
>>> if
>>> I could create it with random text and numerals, but it doesn't have to
>>> be.
>>>
>>> (where 123 is the constant)
>>> i.e.:
>>> 12386756
>>> 12337288
>>> 12378645
>>> 12321356
>>>
>>> or
>>>
>>> 123g568e
>>> 1237t7er
>>> 12397u23
>>>
>>> Thanks in advance for your help.
>>>
>>> Kenny
>>>
>>>
>>>
>
>