Lemme have a crack at question 1. First, let
.
^{4444} > 4444^{4444} )
...meaning that X must have less than or equal to 17777 digits in it.
(TT Edit) A sum of the digits of any number
must be greater or equal to
. That is to say that, A (the sum of the digits of X) must have no more than 17776*9 = 159984 digits.
It is to note here that the number with the largest sum of digits below 159984 is 99999, which has a sum of 45. From this, we can see that B (the sum of the digits of A) must have no more than 5*9 = 45 digits.
Similarly, it is to note here that the number with the largest sum of digits below 45 is 39, which has a digit sum of 12. Hence, we can say that C (where C is the sum of the digits of B) must have less than or equal to 12 digits.
Remember now the divisibility test for 9. An integer is divisible by 9 if and only if the sum of its digits (in decimal notation), is divisible by 9. Quick proof for this is seen here,
, 10^2 \equiv 1 (\mod 9), ..., 10^n \equiv 1 (\mod 9) )
That is to say that:
 )
, for any natural number N.
Hence, we can see that:
 )
As 4444 = 9 x 493 + 7
 )
[1]
Now lets do some tests with the number 7.
 )
 )
 )
[2]
From [1],
 )
 )
 )
From [2],
 )
That means that,
 )
. Because C > 12, hence C must be 7.
Edit: I'm still learning how to use latex properly. Why is there a huge gap in the modulos? This is just a simple solution, there must be more elegant alternatives.
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