Number Theory?

Question

Try this out =D!

Given that D_n + D_(n-1) + ... + D_2 + D_1 mod 3 = 0

Prove that:
D_n * 10^(n-1) + D_(n-1) * 10^(n-2) + ... + D_2 * 10 + D_1 is divisble by 3 also.

Yes, moshi. This is the divisbility for rule for 3. I was trying to prove it for my math project. It wasn't required, but I feel like I should be acknowledged for this one time.

It took 10 minutes to prove this darn thing lol.

I wasn't so many answers though. Too easy I guess?

*I wasn't expecting so many answers though*

ciotappa: learn this on wikipedia.org

I didn't have a professor to teach me number theory. Searching around Yahoo!Answers helps also.

Try some books. You just reminded me of buying some books on number theory lol.

Answer 1

10 = 1(mod 3)
So 10 ^n = 1(mod 3) for all nonegative integers n.
Consequently D_n * 10^(n-1) + D_(n-1) * 10^(n-2) + ... + D_2 * 10 + D_1 = D_n + D_(n-1) + ... + D_2 + D_1 mod 3 = 0
also.
The same proof works if we replace 3 by 9 here.

Answer 2

Note that 10^m=1 mod 3 for any positive integer m. Then,
D_n * 10^(n-1) + D_(n-1) * 10^(n-2) + ... + D_2 * 10 + D_1
just reduces to
D_n + D_(n-1) + ... + D_2 + D_1
mod 3. Thus, this theorem is rather obvious. This is usually used to prove that any integer whose sum of digits is divisible by 3 is also divisible by 3. In this case, D_n stands for the nth digit from the right.

Answer 3

D_n * 10^(n-1) + D_(n-1) * 10^(n-2) + ... + D_2 * 10 + D_1
= { D_n * [10^(n-1) - 1] + D_(n-1) * [10^(n-2) - 1] + ... + D_2 * [10 - 1] + D_1[1 - 1] } + { D_n + D_(n-1) + ... + D_2 + D_1 }
Now [10^k-1 - 1] is divisible by three as it will just be 9999.....999 (a lot of nines). So all the first half is divisible by three. Also the second half is divisible by three as thats given. So the entire thing is divisible by three.

Answer 4

Number Theory lol buddy you are CRAZY I will help you next year when i get to that class