Denote "sod(x)" as sum of the digits of x. Examples: sod(49) = 4 + 9 = 13; sod(123) = 1 + 2 + 3 = 6.

Find sod(sod(sod(4444^4444))).

At the outset, let A = sod(4444^4444), B = sod(A), and C = sod(B). We are required to evaluate C= sod(B).

Now, we have: 4444 (mod 9) = 7

--> 4444^4444(mod 9) = 7^4444

Now, 7^4444 (Mod 9) = ((7^3)^1481)*7 (mod 9) = (1^1481)*7 (Mod 9) = 7

Thus, A leaves a remainder of 7, when divided by 9. Since any number and its sum of digits will leave the same remainder, when divided by 9, it follows that B will also leave a remainder of 7, when divided by 9.

Now, we observe that:

4444^4444 < (100000)^4444 = (10^4)^4444 = 10^(17,776)

-->4444^4444 has less than 17,776 digits.

Therefore, A < 9* 17777 < 199999

So, B < 1 + 5* 9 = 46

But, B must possess the form 7(mod 9). Since B (mod 9) = C, we must have C (mod 9) = 7. If possible, let C >= 16.

Now, the maximum value of C, with 1<= B <= 46 occurs whenever B = 39, giving C = 12 < 16. This is a contradiction.

Since C< 16, and C (mod 9) = 7, it follows that C must be equal to 7.

Consequently, sod ( sod ( sod (4444 ^ 4444 ) ) ) = 7.

*Edited on ***July 10, 2008, 6:20 am**