When 4444^4444 is written in decimal notations, the sum of its digits is A. Let B be the sum of the digits of A. Find the sum of the digits of B.

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- Aug 7th 2008, 04:47 AMfardeen_genFind sum of the digits of B?
When 4444^4444 is written in decimal notations, the sum of its digits is A. Let B be the sum of the digits of A. Find the sum of the digits of B.

- Aug 8th 2008, 02:37 AMKiwi_Dave
My solution is not very elegant. It may not even be right. But here goes.

First I seek an upper bound to the solution.

$\displaystyle log(4444^{4444}) = 16.2$

Furthermore $\displaystyle 10^{16.2}<1.6*10^{16}$

Now we conclude that $\displaystyle 4444^{4444}$ has 17 digits where the most significant digit is 1 and the other digits are unknown (but not more than 9).

Hence $\displaystyle A < 1 + 9*16 = 145$

Now for all numbers up to 145 the one with the largest digit sum is 139 which sums to 13. Hence B<13.

Now summing the digits of any number gives a result that is equal to the starting number (modulo 9).

Hence $\displaystyle 4444^{4444} = 7^7 = 823543$ (mod 9)

This is greater than 13 so I sum the digits repeatidly until I get an acceptable answer. The result is 823543, 25, 7. OK so I think the answer is 7.

When you receive the model answer can you please post it? - Aug 8th 2008, 03:50 AMCoffee CatA long train of thought
I think Kiwi_Dave is on the right track. We can use logarithms in this problem to determine the number of digits in 4444^4444.

log (4444^4444) = 4444 log 4444 = 16210.7079

which means 4444^4444 has 16211 digits. The sum of these 16211 digits is A. The sum of the digits of A is B.

Without restrictions save for the number of digits, the maximum possible value of A is 16211 * 9 = 145899. Of course, that has a digital root of 9, so it couldn't be the value of A.

We know that the digital root of 4444^4444 is 7 (because the digital root of 4444 is 7, and the powers of 7 have digital roots of 7, 4, 1 respectively for exponents of the form 3k-2, 3k-1 and 3k respectively). Its last digit is 6 (owing to the even exponent).

As for its first digit, we can calculate this using logarithms again.

Let's say n is the first digit. So n * 10^16210 < 4444^4444

So log n + 16210 < 16210.7079

We find that n = 5

So the 1st digit is 5, the last digit is 6. Together they have a digital root of 2. So the remaining (16211-2) = 16209 digits have a digital root of 5.

Let's say 16208 digits are 9's and another digit is 5. So the highest possible value of A is

5 + 16208 * 9 + 5 + 6 = 145888. It has 6 digits. The sum of the digits is 34.

My answer is 34.

Of course, the 16208 digits aren't necessarily 9 and the other digit 5... so I was just sharing this train of thought, lol (Hi) I still don't know how to peg a convincing answer for this caper, so I'll think about it for the night (Thinking)

Please share the solution and answer. I'd love to know it too. - Aug 8th 2008, 03:36 PMKiwi_Dave
I think one of the experts will show us an answer that does not involve logs. But between us we will get close! I mistook a comma for a period on my calculator!

By changing a few digits we could get A = 144999. The sum of the digits is 36 so the answer cannot be greater than 36.

However, $\displaystyle 4444^{4444}=7=16=25=34$ (mod 9). So the answer must be 7,16,25 or 34? - Aug 8th 2008, 09:20 PMfardeen_gen
**Lemma:**The maximum possible sum of the digits of an http://alt2.artofproblemsolving.com/...00c274bdaa.gif-digit number is http://alt1.artofproblemsolving.com/...5a41448dae.gif. In other words,

http://alt1.artofproblemsolving.com/...66eb769185.gif

And so, the solution

Using this useful result,

http://alt1.artofproblemsolving.com/...28a9c3ece3.gif

Hence

http://alt2.artofproblemsolving.com/...254aeb3561.gif

http://alt1.artofproblemsolving.com/...bc98d3073a.gif

Now, http://alt1.artofproblemsolving.com/...4ac69996ff.gif, but since that last digit sum cannot be http://alt1.artofproblemsolving.com/...5146201319.gif or greater it must be http://alt1.artofproblemsolving.com/...ffdf29b237.gif. - Aug 9th 2008, 05:19 AMCoffee Cat:)
Awesome! Thanks, fardeen_gen!

- Aug 14th 2008, 10:01 PMmanjilEasier and elegant!
See my reply to this at MathLinks :: View topic - A 103 this is a problem from the 1975 IMO. I believe my solution is both more elegant and easy!

- Aug 15th 2008, 05:02 AMMoo
- Aug 15th 2008, 07:20 AMmanjil
Anyway I was wrong on that count but the proof still holds if we replace 179,999 by 99,999 and 44 by 45.

Also the sum of the digits of B can't be more than 39 because 3+9=12 which is the largest possible sum!(Rofl)

Thanks for pointing that out!

But don't you agree that my proof is better than the others posted? - Aug 15th 2008, 08:28 AMCaptainBlack
- Aug 15th 2008, 08:04 PMMoo