



 
Shivdeep, If you were to repeat the process of adding the digits of a number to a new number, you'd get the remainder of division of 4444^{4444} by 9. For instance, 4444/9 = 493 remainder of 7. You can also compute the remainder by adding 4+4+4+4 = 16, 1+6 = 7. Now 4444^{4444} < 10000^{4444}, so 4444^{4444} has at most 4 × 4444 = 17776 digits, each of which is at most 9. So A is at most 9 × 17776 = 159984. The number smaller than 159984 with the largest digit sum is 99999, and 9+9+9+9+9 = 45. So B is at most 45, and the sum of digits in B is at most 3 + 9 = 12. You'll have to figure out what is the remainder of division of 4444^{4444} by 9. If you are lucky, it is not 1, 2 or 3, so there is only one number from 1 to 12 which has the same sum of digits. Claude
 


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