The Question
Find the sum of the digits of 50!
For example, 5! = 5x4x3x2x1 = 120, so the sum of the digits is 1+2+0=3, and the digital sum is ds(5!)=3.
This is at the level of a middle secondary national mathematics competition.
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ha ha... This is tough for me! :D
Very tempting to get a brute forced answer by matlab....
But Ill do it by hand :)
Trying to cheat with Google and it sends me to: The 5 Steps I Followed To Find Love After 50 . Must find better ways to cheat. ;)
erm... google is not the right place to cheat from.
If you scan my articles, you will find a clue as to the best site to help with such questions. ;-)
Okay, cheating with a factorial table. The ds are: 1,2,6,6,3,9,9,9,27,27, 36, 27, 27, 45, 45. I don't see the patter, but it will be a multiple of 3.
Actually, it will have to be a multiple of 9, because 50! is a multiple of 9, and the sum of the digits of any multiple of 9 is itself a multiple of 9. This is more or less easy to show. As for finding the actual sum of the digits of 50! by analytic means, I'm still not sure how I'd do it.
So 50! = 30414093201713378043612608166064768844377641568960512000000000000 and the sum is....216. But I can't get there other than by brute force.
Praise the speed of modern computers! Yeah I was stumped too. I know there is probably a clever way, but I was stuck just with "brute force" as well
Yes, it's very simple to compute that in many programming languages (I use Python) or tools (like Mathematica). I'm not even sure if there is an efficient analytical way to do it. I have play around a bit with it, but so far to no avail.
Is there a prize for this?
Does it say so anywhere?
No, some questions are for the joy of learning.
My apology. I was just clarifying. Sorry sir.
That's OK. I have 1000s of questions I can put on - just doing it slowly as the number of interested people grows. The prize contests are fun and I will carry on with them, but I don't want to do that for every question.
To be honest, some people can earn more than the prizes by giving good answers and getting upvotes, so every question has the potential to earn something by participating - especially if commenting from within chainBB forum.
In the future, I personally would like to see this Mathematics forum work somewhat like Stackexchange, with others posting questions and answers - but that will take time. :-)
I think the sum of digits of every factorial after 5! is 9. Is this correct? I think this works as long as you continue the process. I remember from Number Theory that mod 9 and factorials are friends. I forget and I want to know why! Thanks :)
It's not clear to me that you are supposed to keep summing. So, the sum of the digits in 15! is 45, and I don't think you are to add those together until you get to a one digit number. If you do, you are correct - they all sum to 9.
yeah, a bit of clarification on definitions.
The digital sum is the sum of all the digits.
The digital root is the continued digital sum until you get to a single digit. So, yes, the digital root of n! (n>5) will always be 9.