39nishant sir is it correct ?
Let A be the sum of the digits of the number 4444^{4444}, and B the sum of the digits of A. Compute the sum of the digits of B.
-
UP 0 DOWN 0 0 12
12 Answers
39hey dude nice one.
and ya welcome to targetiit .
current cut ,, will solve it as soon as i will be back.
62Hint: This is a famous question
First show that The sum of digits of B is a one digit number :)
62bhargav the answer is 7..
your method is also correct..
but please write it in a way that ppl dont get scared..
Your solution makes a simple simple problem look so so so scary :P
but yeah good work
62tell that S(x) stands for in your proof!
I can guess it bcos i know the proof..
39ok reposting solution again (deleting due to misclick)
ok we all know that The maximum possible sum of the digits of an n-digit number is 9n.
S(n) denotes sum of digits of `n`
now S(44444444)<9[(log1044444444+1)]<9.[log10100004444+1)]
=9(4444.4+1) <9.20000=180000
so
S(S(44444444))<S(!7999)=44
S(S(S(44444444)))≤S(39)=13
NEXTLY
44444444≡7mod 9
hence the sum of digits of B is 7
341You have to point out that the crux of the solution is that
n≡ S(n) mod 9.
that's what allows you to finally conclude that the sum of the digits of B is 7.
1if any one of you wants to know the source , google interesting problems and run into 6th link...