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Let $A$ be the sum of the digits of $4444^{4444}$ and $B$ the sum of the digits of $A$ . Find the sum of the digits of $B$.

The solution given in the book is as follows:

Since $4444^{4444}<10^{4444}$ it is easy to see that $A<44440$ . Therefore, $B<5.9=45$ which implies that the sum of digits is a one digit number . We also know that $A$ and $B$ (and the sum of its digits as well) are congruent to $4444^{4444}$ modulo 9 . That is, they all have remainder $7$ when divided by $9$. Thus the sum of digits of $B$ must be $7$.

However, I am not getting the idea of the solution...i mean how do they conclude "Since $4444^{4444}<10^{4444}$ it is easy to see that $A<44440$ ." Is it at all true as $10<4444$ and $ 4444^{4444}<10^{4444}$...Also, how does they conclude that "Therefore, $B<5.9=45$ which implies that the sum of digits is a one digit number ." I mean how do they say that sum of digits of $B$ is a one digit number?...I am not quite getting the idea of the solution.....

  • That is a well-known problem from IMO 1975. This question was discussed many times here. – RFZ Apr 20 '22 at 11:42
  • @lulu Actually I am not quite getting the idea... –  Apr 20 '22 at 11:45
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    Try to ask a more precise question. You've been handed multiple solutions....what exactly don't you understand? – lulu Apr 20 '22 at 11:50
  • The $.$ in $B<5.9=45$ is a multiplication. You could also say that if $A< 44440$ then the sum of digits of $A$ is less than $9+9+9+9+9$ since each digit is nine or smaller and the first is four or smaller – Henry Apr 20 '22 at 11:51
  • @lulu Well, the link u posted has the sum of digits of $A$ as $B$ and sum of digits of $B$ as $C$ and sum of digits of $C$ as $D$ and it calculates $D$.... but here the question asks to find $C$...how to do that?... –  Apr 20 '22 at 11:53
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    Well, try. Note however that something seems off, as $4444^{4444}$ is obviously not less than $10^{4444}$. In any event, here is another solution. – lulu Apr 20 '22 at 11:55
  • @lulu the solution link u posted also has something bit off about it ...i mean they say that $4444^{4444}<10000^{4444}=10^{4.4444}$ ...So, $4444^{4444}$ has less than $17776$ digits so, it has maximum sum of digits of $9.17775=159975$ ...But then it says $A$ becomes maximum if in $4444$ it was $9999$...so it says $B\leq 45$...how does they draw the conclusion? –  Apr 20 '22 at 12:09
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    No, they are correct. Since $A<159975$ the maximum sum of digits occurs for $A=99999$. Putting the $1$ in on the left is no good since it forces the second digit to be $≤5$. Thus you gain $1$ in the sum but lose at least $4$. – lulu Apr 20 '22 at 12:16
  • @lulu Thanks a lot! i get it now.... –  Apr 20 '22 at 12:29

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