Converting SELECT Statement into UPDATE

Hi All,
Running SQL Server 2008 R2.  I have the following SELECT query, which is returning the desired results.
SELECT DISTINCT
[x].[AccountNo],
[x].[AvgAccountLen],
CASE
WHEN LEN([AccountNo]) > 6 THEN LEFT([AccountNo], 6)
ELSE [AccountNo] + REPLICATE('0', [AvgAccountLen] - LEN([AccountNo]))
END AS [NewAccountNo]
FROM
SELECT DISTINCT
[AccountNo],
SELECT TOP 1
LEN([AccountNo])
FROM
[dbo].[Table]
WHERE
[AccountNo] > 0
GROUP BY
[AccountNo]
ORDER BY
COUNT(*) DESC
) AS [AvgAccountLen]
FROM
[dbo].[Table]
) AS [x]
WHERE
LEN([AccountNo]) <> [AvgAccountLen]
Below are results, which again are what I'm looking for.
AccountNo AvgAccountLen NewAccountNo
4200 6 420000
4250 6 425000
42000 6 420000
4030 6 403000
4460 6 446000
4250000 6 425000
4520000 6 452000
Long story short is that I've been left to clean up a partially-completed task.  I need to conduct an update on Table that pads (or trims) the account numbers accordingly.  Further, this process affects multiple entities which is why I can't simply
use a static pad/trim value of 6 (this particular entity returns 6, there could be other entities with 4, 8, etc.).  AvgAccountLen may not be the most appropriate column name either - it's a representation of the most frequently-occurring value length
(I have already confirmed that the result returned for this value is correct in each entity).  How would I go about writing a UPDATE statement to accomplish this?
Any help is greatly appreciated!
Best Regards
Brad

Can you provide your example data as a table to compliment your expected result?
I'm thinking something like this may help:
DECLARE @accounts TABLE (accountNo INT, avgAccountLen INT, newAccountNumber INT)
INSERT INTO @accounts (accountNo, avgAccountLen) VALUES
(4200 , 6),
(4250 , 6),
(42000 , 6),
(4030 , 6),
(4460 , 6),
(4250000, 6),
(4520000, 6)
UPDATE @accounts
SET newAccountNumber = LEFT(CAST(accountNo AS VARCHAR)+REPLICATE('0',avgAccountLen),avgAccountLen)
FROM @accounts
SELECT *
FROM @accounts
Don't forget to mark helpful posts, and answers. It helps others to find relevant posts to the same question.

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