SOLVED

FILL SERIES

%3CLINGO-SUB%20id%3D%22lingo-sub-414011%22%20slang%3D%22en-US%22%3EFILL%20SERIES%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-414011%22%20slang%3D%22en-US%22%3E%3CP%3EHI%2C%3C%2FP%3E%3CP%3E%26nbsp%3B%3C%2FP%3E%3CP%3EI%20WANT%20TO%20FILL%20SERIES%20NUMBERS%20AS%20ATTACHED.%20PLEASE%20CAN%20YOU%20HELP%20ME%20OUT%20ON%20THIS.%3C%2FP%3E%3CP%3E%26nbsp%3B%3C%2FP%3E%3CP%3E04760000000000001%3C%2FP%3E%3CP%3E04760000000000001%3C%2FP%3E%3CP%3E04760000000000002%3C%2FP%3E%3CP%3E04760000000000002%3C%2FP%3E%3CP%3E04760000000000003%3C%2FP%3E%3CP%3E04760000000000003%3C%2FP%3E%3CP%3E%26nbsp%3B%3C%2FP%3E%3CP%3ETHANKS%3C%2FP%3E%3CP%3EKRISH%3C%2FP%3E%3C%2FLINGO-BODY%3E%3CLINGO-LABS%20id%3D%22lingo-labs-414011%22%20slang%3D%22en-US%22%3E%3CLINGO-LABEL%3EExcel%3C%2FLINGO-LABEL%3E%3CLINGO-LABEL%3EFormulas%20and%20Functions%3C%2FLINGO-LABEL%3E%3C%2FLINGO-LABS%3E%3CLINGO-SUB%20id%3D%22lingo-sub-414427%22%20slang%3D%22en-US%22%3ERe%3A%20FILL%20SERIES%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-414427%22%20slang%3D%22en-US%22%3E%3CP%3E%3CA%20href%3D%22https%3A%2F%2Ftechcommunity.microsoft.com%2Ft5%2Fuser%2Fviewprofilepage%2Fuser-id%2F317114%22%20target%3D%22_blank%22%3E%40krish86%3C%2FA%3E%26nbsp%3B%3CA%20href%3D%22https%3A%2F%2Ftechcommunity.microsoft.com%2Ft5%2Fuser%2Fviewprofilepage%2Fuser-id%2F100377%22%20target%3D%22_blank%22%3E%40Vijay%3C%2FA%3E%3C%2FP%3E%3CP%3E%26nbsp%3B%3C%2FP%3E%3CP%3EHii%20Krish%2C%3C%2FP%3E%3CP%3E%26nbsp%3B%20%26nbsp%3BI%20have%20solve%20your%20problem%2C%20the%20method%20which%20i%20have%20followed%3A-%3C%2FP%3E%3CP%3EFill%20a%20normal%20serie%20in%20A%20Column%20like%20as%201%2C1%2C2%2C2%2C3%2C3%2C4%2C4%2C5%2C5%2C6%2C6%2C7%2C7%3C%2FP%3E%3CP%3Ethen%20put%20this%20formula%20%3DREPT(0%2C13-LEN(A1))%20in%20B1%20Because%20series%20is%20start%20A1%2C%3C%2FP%3E%3CP%3Eand%20again%20put%20this%20formula%20%3D%220476%22%26amp%3BB1%26amp%3BA1%20and%20Drug%20drop%20the%20formula%20and%20your%20series%20is%20done%3C%2FP%3E%3CP%3E%26nbsp%3B%3C%2FP%3E%3CP%3E04760000000000001%3CBR%20%2F%3E04760000000000001%3CBR%20%2F%3E04760000000000002%3CBR%20%2F%3E04760000000000002%3CBR%20%2F%3E04760000000000003%3CBR%20%2F%3E04760000000000003%3CBR%20%2F%3E04760000000000004%3CBR%20%2F%3E04760000000000004%3CBR%20%2F%3E04760000000000005%3CBR%20%2F%3E04760000000000005%3CBR%20%2F%3E04760000000000006%3CBR%20%2F%3E04760000000000006%3CBR%20%2F%3E04760000000000007%3CBR%20%2F%3E04760000000000007%3CBR%20%2F%3E04760000000000008%3CBR%20%2F%3E04760000000000008%3CBR%20%2F%3E04760000000000009%3CBR%20%2F%3E04760000000000009%3CBR%20%2F%3E04760000000000010%3CBR%20%2F%3E04760000000000010%3CBR%20%2F%3E04760000000000011%3CBR%20%2F%3E04760000000000011%3CBR%20%2F%3E04760000000000012%3CBR%20%2F%3E04760000000000012%3CBR%20%2F%3E04760000000000013%3CBR%20%2F%3E04760000000000013%3C%2FP%3E%3C%2FLINGO-BODY%3E%3CLINGO-SUB%20id%3D%22lingo-sub-414526%22%20slang%3D%22en-US%22%3ERe%3A%20FILL%20SERIES%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-414526%22%20slang%3D%22en-US%22%3E%3CP%3ETHANKS.%20IT%20WORKS%3C%2FP%3E%3C%2FLINGO-BODY%3E%3CLINGO-SUB%20id%3D%22lingo-sub-414660%22%20slang%3D%22en-US%22%3ERe%3A%20FILL%20SERIES%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-414660%22%20slang%3D%22en-US%22%3E%3CP%3E%3CA%20href%3D%22https%3A%2F%2Ftechcommunity.microsoft.com%2Ft5%2Fuser%2Fviewprofilepage%2Fuser-id%2F317114%22%20target%3D%22_blank%22%3E%40krish86%3C%2FA%3E%26nbsp%3B%2C%20another%20variant%20without%20helper%20columns%20is%20to%20type%20first%20two%20numbers%20(as%20text%2C%20all%20cells%20are%20formatted%20as%20General)%2C%20if%20they%20start%20from%20L4%20in%20third%20cell%20(L6)%20is%20the%20formula%3C%2FP%3E%0A%3CPRE%3E%3DLEFT(L5%2CLEN(L5)-8)%20%26amp%3B%20TEXT(RIGHT(L5%2C8)%2BMOD(ROW()-ROW(%24L%245)%2C2)%2CREPT(%220%22%2C8))%3C%2FPRE%3E%0A%3CP%3Eselect%20the%20column%20starting%20from%20L6%20till%20end%20of%20your%20range%20and%20Ctrl%2BD%20(or%20drag%20this%20cell%20down).%3C%2FP%3E%0A%3CP%3E%3CSPAN%20class%3D%22lia-inline-image-display-wrapper%20lia-image-align-inline%22%20style%3D%22width%3A%20683px%3B%22%3E%3CIMG%20src%3D%22https%3A%2F%2Fgxcuf89792.i.lithium.com%2Ft5%2Fimage%2Fserverpage%2Fimage-id%2F106322i0991D4B45C1ACB71%2Fimage-size%2Flarge%3Fv%3D1.0%26amp%3Bpx%3D999%22%20alt%3D%22image.png%22%20title%3D%22image.png%22%20%2F%3E%3C%2FSPAN%3E%3C%2FP%3E%0A%3CP%3EInstead%20of%20eight%20that%20could%20be%20any%20number%20which%20is%20more%20than%20max%20number%20of%20digits%20at%20the%20end%20and%20less%20than%2012.%3C%2FP%3E%0A%3CP%3E%26nbsp%3B%3C%2FP%3E%0A%3CP%3E%26nbsp%3B%3C%2FP%3E%3C%2FLINGO-BODY%3E%3CLINGO-SUB%20id%3D%22lingo-sub-415043%22%20slang%3D%22en-US%22%3ERe%3A%20FILL%20SERIES%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-415043%22%20slang%3D%22en-US%22%3E%3CP%3E%3CA%20href%3D%22https%3A%2F%2Ftechcommunity.microsoft.com%2Ft5%2Fuser%2Fviewprofilepage%2Fuser-id%2F521%22%20target%3D%22_blank%22%3E%40Sergei%20Baklan%3C%2FA%3E%26nbsp%3BGOOD%20ONE.%20THANKS%3C%2FP%3E%3C%2FLINGO-BODY%3E%3CLINGO-SUB%20id%3D%22lingo-sub-415231%22%20slang%3D%22en-US%22%3ERe%3A%20FILL%20SERIES%3C%2FLINGO-SUB%3E%3CLINGO-BODY%20id%3D%22lingo-body-415231%22%20slang%3D%22en-US%22%3E%3CP%3E%3CA%20href%3D%22https%3A%2F%2Ftechcommunity.microsoft.com%2Ft5%2Fuser%2Fviewprofilepage%2Fuser-id%2F317114%22%20target%3D%22_blank%22%3E%40krish86%3C%2FA%3E%26nbsp%3B%2C%20you%20are%20welcome%3C%2FP%3E%3C%2FLINGO-BODY%3E
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New Contributor

HI,

 

I WANT TO FILL SERIES NUMBERS AS ATTACHED. PLEASE CAN YOU HELP ME OUT ON THIS.

 

04760000000000001

04760000000000001

04760000000000002

04760000000000002

04760000000000003

04760000000000003

 

THANKS

KRISH

5 Replies
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Best Response confirmed by krish86 (New Contributor)
Solution

@krish86 @vijay

 

Hii Krish,

   I have solve your problem, the method which i have followed:-

Fill a normal serie in A Column like as 1,1,2,2,3,3,4,4,5,5,6,6,7,7

then put this formula =REPT(0,13-LEN(A1)) in B1 Because series is start A1,

and again put this formula ="0476"&B1&A1 and Drug drop the formula and your series is done

 

04760000000000001
04760000000000001
04760000000000002
04760000000000002
04760000000000003
04760000000000003
04760000000000004
04760000000000004
04760000000000005
04760000000000005
04760000000000006
04760000000000006
04760000000000007
04760000000000007
04760000000000008
04760000000000008
04760000000000009
04760000000000009
04760000000000010
04760000000000010
04760000000000011
04760000000000011
04760000000000012
04760000000000012
04760000000000013
04760000000000013

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THANKS. IT WORKS

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@krish86 , another variant without helper columns is to type first two numbers (as text, all cells are formatted as General), if they start from L4 in third cell (L6) is the formula

=LEFT(L5,LEN(L5)-8) & TEXT(RIGHT(L5,8)+MOD(ROW()-ROW($L$5),2),REPT("0",8))

select the column starting from L6 till end of your range and Ctrl+D (or drag this cell down).

image.png

Instead of eight that could be any number which is more than max number of digits at the end and less than 12.

 

 

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@Sergei Baklan GOOD ONE. THANKS

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@krish86 , you are welcome