If you have 4444 to the power of 4444 you get a large number if you add all the digts in that number(and keep adding) you end up with one number.

eg number 12345 = 15 = 6

so what number do you end up with?

I believe it to be 7, can work it out based on a pattern but that is not a satisfactory answer. if think you can prove it iteratively.....how.

any ideas?

any suggestions much appreciated

Steven


I do know that the answer is not dr[n^{a mod 3} ] where a,n are integers and dr[ ] is the digital root.
odd ones are 11¹¹ , 2¹⁰ and 2¹¹

Dear Steve,


I agree with you that the digital sum of 4444⁴⁴⁴⁴ is 7 and this is why ...

The digital sum of any number K is the remainder when K is divided by 9, (or to say the same thing another way) it is the congruence class of K mod 9.

You probably know this already but if not you might be interested in the proof which is given on the NRICH web site in an article on Tests for Divisibility by Tim Rowland.

(1) Proof of your result using the binomial theorem (BT)

4444⁴⁴⁴⁴ = [9(493) + 7]⁴⁴⁴⁴ = multiple of 9 + 7⁴⁴⁴⁴ using the BT.

This means that the digital sum of our number is the same as the digital sum of 7⁴⁴⁴⁴ .

Now we use the fact that 7³ = 1 + multiple of 9.

7⁴⁴⁴⁴ = 7(7³ )¹⁴⁸¹ = 7(1 + multiple of 9)¹⁴⁸¹ = 7 + multiple of 9. (using the BT again)

So now we know that the digital sum of 7⁴⁴⁴⁴ is 7 and hence the digital sum of 4444⁴⁴⁴⁴ is 7.



(2) Another proof, essentially the same but using the language of modulus arithmetic.

Let ds(K) denote the digital sum of the number K.

ds(4444) = ds(16) = ds(7)

Alternatively we can write:

4444 º 16 (mod 9) º 7 (mod 9).

Hence 4444⁴⁴⁴⁴ º 7⁴⁴⁴⁴ (mod 9)

7³=343 º 1 (mod 9)

7⁴⁴⁴³=(7³)¹⁴⁸¹ º 1 (mod 9)

Hence 7⁴⁴⁴⁴=7(7⁴⁴⁴³) º 7×1 º 7 (mod 9)

and so 4444⁴⁴⁴⁴ º 7⁴⁴⁴⁴ º 7 (mod 9)
Please let me know if this makes sense to you. I hope it does but if not I'll have another go at explaining it more simply.

Cheers

Anon.