If you import library datetime

import datetime
your_date = "01/05/2000"
(day, month, years) = your_date.split("/")
date = datetime.date(int(years), int(month), int(day))
date_10000 = date+datetime.timedelta(days=10000)
print(date_10000)

No library script

your_date = "20/05/2000"
(day, month, year) = your_date.split("/")
days = 10000
year = int(year)
month = int(month)
day = int(day)
end=False
#m1,m3,m5,m7,m8,m10,m12=31
#m2=28
#m4,m6,m9,m11=30
m=[31,28,31,30,31,30,31,31,30,31,30,31]
while end!=True:
    if(((year % 400 == 0) or  (year % 100 != 0) and  (year % 4 == 0)) and(days-366>=0)):   
        days-=366
        year+=1
    elif(((year % 400 != 0) or  (year % 100 != 0) and  (year % 4 != 0)) and(days-366>=0)):
        days-=365
        year+=1
    else:
        end=True
end=False
if(((year % 400 == 0) or  (year % 100 != 0) and  (year % 4 == 0))):   
    m[1]=29
else:
    m[1]=28
while end!=True:
    if(days-m[month]>=0):
        days-=m[month]
        if(month+1!=12):
            month+=1
        else:
            year+=1
            if(((year % 400 == 0) or  (year % 100 != 0) and  (year % 4 == 0))):   
                m[1]=29
            else:
                m[1]=28
            month=0
    else:
        end=True

if(day+days>m[month]):
    day=day+days-m[month]+1
    if(month+1!=12):
        month+=1
    else:
        year+=1
        if(((year % 400 == 0) or  (year % 100 != 0) and  (year % 4 == 0))):   
            m[1]=29
        else:
            m[1]=28
        month=0
else:
    day=day+days
print(day,"/",month,"/",year)

Here's a solution I came up with that involves no libraries or packages, just loops and conditionals (accounts for leap years):

def isLeapYear(years):
  if years % 4 == 0:
    if years % 100 == 0:
      if years % 400 == 0:
        return True
      else:
        return False
    else:
      return True
  else:
    return False

monthDays = [31,28,31,30,31,30,31,31,30,31,30,31]
sum = 0
sumDays = []
for i in monthDays:
  sumDays.append(365 - sum)
  sum += i

timeInp = input("Please enter your birthdate in the format dd/mm/yyyy\n")
timeInp = timeInp.split("/")
days = int(timeInp[0])
months = int(timeInp[1])
years = int(timeInp[2])
totDays = 10000

if totDays > 366:
  if isLeapYear(years):
    if months == 1 or months == 2:
      totDays -= (sumDays[months - 1] + 1 - days) + 1
    else:
      totDays -= (sumDays[months - 1] - days) + 1
  else:
    totDays -= (sumDays[months - 1] - days) + 1
  
  months = 1
  days = 1
  years += 1

while totDays > 366:
  if isLeapYear(years):
    totDays -= 366
  else:
    totDays -= 365
  years += 1

i = 0
while totDays != 0:
  if isLeapYear(years):
    monthDays[1] = 29
  else:
    monthDays[1] = 28
    
  if totDays >= monthDays[i]:
    months += 1
    totDays -= monthDays[i]
  elif totDays == monthDays[i]:
    months += 1
    totDays = 0
  else:
    days += totDays 
    if days % (monthDays[i] + 1)!= days:
      days %= monthDays[i] + 1
      months += 1
    totDays = 0

  if months == 13:
    months = 1
    years += 1

  i += 1
  if i == 12:
    i = 0

print(str(days) + "/" + str(months) + "/" + str(years))

As the name suggests, isLeapYear() takes in a parameter years, and returns a boolean value.

Our first step to this problem, to make it easier, is to just first "translate" our date to the next year. This makes our future calculations easier. To do this, we can define an array sumDays that stores the amount of days each month takes to finish the year (go to new years). Then, we subtract this amount from totDays, account for leap years, and update our variables.

Next, is the easy part, just skipping forward by the years while we have enough days for a complete year.

Once we can not add another full year, we just go month by month until we run out of days.

I hope this helped! Please let me know if you need any further details or clarification (or if I made a mistake) :)

Sample Test Cases:

Input #1:

19/05/2008

Output #1:

5/10/2035

Input #2:

05/05/2020

Output #2:

21/9/2047

Input #3:

29/02/2020

Output #3:

17/7/2047

I checked most of my solutions with this website: https://www.countcalculate.com/calendar/birthday-in-days/result