Thread: Find the remainder when 6341723110832864 is divided by 6 and 12?

$6341723110832864 \bmod 2 = 0$

$6341723110832864 \bmod 3 = (6341723110832860 + 4) \bmod 3 = ((6341723110832860 \bmod 3) + (4 \bmod 3)) \bmod 3 = (1+1)\bmod 3 = 2$

so

$6341723110832860 \bmod 6 = (3*0 - 2*2) \bmod 6 = -4 \bmod 6 = 2$

similarly

$6341723110832864 \bmod 4 = (6341723110832800 + 64) \bmod 4 =$

$((6341723110832800 \bmod 4) + (64 \bmod 4)) \bmod 4 = 0+0 \bmod 4 = 0$

so

$6341723110832864 \bmod 12 = (4*2 - 3*0) \bmod 12 = 8$

I'm not sure how romsek found the remainder (mod 3). Remainder mod 3 is the same as the remainder of the sum of the digits (mod 3). So, just add up the digits: . Add up the digits again: and again: , so the remainder is 2 (mod 3).

As romsek said, you wind up with 2 (mod 6)

And his method for finding the remainders mod 2 and mod 4 is the same as mine.

Thanks Guys but I'm still confused in how you get to the conclusion of

6341723110832864 mod2 = 0

A number is even if the last digit is even. It is odd if the last digit is odd. That is what the remainder (mod 2) means. If it is zero, that means the number was even. If it is one, that means the number was odd.

[QUOTE=SlipEternal;827324]I'm not sure how romsek found the remainder (mod 3). Remainder mod 3 is the same as the remainder of the sum of the digits (mod 3). So, just add up the digits: .

I just noted that

$m \times 10 \bmod 3=1, ~\forall m>0$

then split the original number into

(big buncha digits)*10 + (one digit)

above mod 3 is

(big buncha digits)*10 mod 3 + (one digit) mod 3 =

(1 + (one digit)) mod 3

10m mod 3 is m mod 3. So for example with m=2, 10m is 2 mod 3, not 1.

Originally Posted by johng

10m mod 3 is m mod 3. So for example with m=2, 10m is 2 mod 3, not 1.

yep, you're right.

Guys I'm really appreciate all the comments and I can follow the idea of what have been done but looking at my book they solved it in a different manner and that is the way that I need to present my answers here is the examples I got from my book:

Hmm... Clearly r(4)=0 since the ending 2 digits, 64, is divisible by 4. r(3)=2. Then r(12)=4r(3)-3r(4)=4*2-3*0=8.