find a 13digit Number N that is an integer multiple of 2 to the power 13 and its whole digits consists of only 8's nd 9's

• The answer is 8898989989888.
  
  Given, the number is a multiple of 2^13. So, the number is a multiple of 2, 4, 8, 16, 2^5, 2^6, ..., 2^13.
  
  Since the number is a multiple of 2, the last digit has to be 8.
  Since the number is a multiple of 4, the last two digits must be a multiple of 4. Since the last digit is fixed at 8, the only option for the next one is 8. So, we have the last two digits as 88.
  Since the number is a multiple of 8, the last three digits must be a multiple of 8. Again, the only option turns out to be 888.
  Since the number is a multiple of 16, the last four digits must be a multiple of 16. Again, the only option is 9888. (8888 is not divisible by 16)
  Continuing in this fashion, we can find all the digits from right to left.
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