Call the first k digits 'a' and the last 3 digits 'b'.

(Solution for N below in bold is a 999 digit number with k=996 and b=200)

n = 1000a + b     100 <= b <= 999

2n = 2000a + 2b

2n = (10^k)*b + a

1999 a = (10^k - 2)*b

a = b * (10^k - 2) / 1999

1999 / (10^k - 2) ~~ 1.999 · 10^(3-k)

For k > 3, the value of (10^k - 2) / 1999 is between

 5.002401 * 10^(k-4)  and 5.002502 * 10^(k-4)

a/b = (10^k - 2)/1999

Note that 1999 is prime

Series:  8, 98, 998, 9998, 99998 etc mod 1999 is

    8, 98, 998, 3, 48, 498, 1000, 23, ....

And this finally reaches 0 if k = 996, or 1995, or 2994, or 3993

So for k = 996

(10^996 - 2)/1999 equals  (thank you to Full Precision Calculator)

50025012506253126563281640820410205102551275637818909454

72736368184092046023011505752876438219109554777388694347

17358679339669834917458729364682341170585292646323161580

79039519759879939969984992496248124062031015507753876938

46923461730865432716358179089544772386193096548274137068

53426713356678339169584792396198099049524762381190595297

64882441220610305152576288144072036018009004502251125562

78139069534767383691845922961480740370185092546273136568

28414207103551775887943971985992996498249124562281140570

28514257128564282141070535267633816908454227113556778389

19459729864932466233116558279139569784892446223111555777

88894447223611805902951475737868934467233616808404202101

05052526263131565782891445722861430715357678839419709854

92746373186593296648324162081040520260130065032516258129

06453226613306653326663331665832916458229114557278639319

65982991495747873936968484242121060530265132566283141570

78539269634817408704352176088044022011005502751375687843

92196098049024512256128064032016008004002

This is 993 digits.  We must multiply it by a 3 digit number so that the number of digits becomes 996.  Times 100 for example is too small, this would only be 995 digits.  199 gets close but the smallest 3 digit number that will work is 200.

a = 200 times the 993 digit number above; and then b is 200

The final number is   N =

Multiplying this by 2 yields:

200100050025012506253126563281640820410205102551275637818

909454727363681840920460230115057528764382191095547773886

943471735867933966983491745872936468234117058529264632316

158079039519759879939969984992496248124062031015507753876

938469234617308654327163581790895447723861930965482741370

685342671335667833916958479239619809904952476238119059529

764882441220610305152576288144072036018009004502251125562

781390695347673836918459229614807403701850925462731365682

841420710355177588794397198599299649824912456228114057028

514257128564282141070535267633816908454227113556778389194

597298649324662331165582791395697848924462231115557778889

444722361180590295147573786893446723361680840420210105052

526263131565782891445722861430715357678839419709854927463

731865932966483241620810405202601300650325162581290645322

661330665332666333166583291645822911455727863931965982991

495747873936968484242121060530265132566283141570785392696

348174087043521760880440220110055027513756878439219609804

902451225612806403201600800400

Input:  f(N)

Output: 2.0

-----------------------------

def f(n):

    s = str(n)

    new = int(  s[-3:] + s[:-3]  )

    return new/n

for n in range(1000,10000):

    s = str(n)

    new = int(  s[-3:] + s[:-3]  )

    if new/n == 2:

        print(n, new, new/n)

for k in range(1,11):

    print(k, 10**k - 2, (10**k - 2)%1999 )

mymodvals = [8]

modval = 8

for k in range(2,10000):

    modval = (9*10**(k-1) + modval)%1999

    if modval == 0:

        print(k, modval)

    mymodvals.append(modval)

print(mymodvals[:10])

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