# Thread: Find sum of the digits of B?

1. ## Find 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.

2. 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.

3. ## A 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.

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

Please share the solution and answer. I'd love to know it too.

4. 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!

Originally Posted by Coffee Cat
5 + 16208 * 9 + 5 + 6 = 145888. It has 6 digits. The sum of the digits is 34.
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?

5. Lemma: The maximum possible sum of the digits of an -digit number is . In other words,

And so, the solution
Using this useful result,

Hence

Now, , but since that last digit sum cannot be or greater it must be .

6. ## :)

Awesome! Thanks, fardeen_gen!

7. ## Easier 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!

8. Originally Posted by manjil
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!
Yours has got some things incorrect o.O

The largest possible sum of numbers is the sum of 99,999, which is 45 (>44)

And why is the sum B at most the sum of digits of 39 ?

9. 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!

Thanks for pointing that out!

But don't you agree that my proof is better than the others posted?

10. Originally Posted by manjil
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!
It has a sloppy estimate of the number of digits.

RonL

11. Originally Posted by manjil
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!

Thanks for pointing that out!

But don't you agree that my proof is better than the others posted?
It doesn't prove anything. So basically, it's not better...