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

1. ## Find the remainder when 6341723110832864 is divided by 6 and 12?

I need to use the following rules to find the remainder but after solve it my answer is different from my books, so I need help and explanation in how to solve it.

the rules I got is:

r6 = 3r2 - 2r3 (mod 6)

r12 = 4r3 - 3r4 (mod 12)

xxx

2. ## Re: 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

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

3. ## Re: Find the remainder when 6341723110832864 is divided by 6 and 12?

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: $6+3+4+1+7+2+3+1+1+0+8+3+2+8+6+4 = 59$. Add up the digits again: $5+9=14$ and again: $1+4=5 = 3+2$, 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.

4. ## Re: Find the remainder when 6341723110832864 is divided by 6 and 12?

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

6341723110832864 mod2 = 0

5. ## Re: Find the remainder when 6341723110832864 is divided by 6 and 12?

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.

6. ## Re: Find the remainder when 6341723110832864 is divided by 6 and 12?

[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: $6+3+4+1+7+2+3+1+1+0+8+3+2+8+6+4 = 59$.

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

7. ## Re: Find the remainder when 6341723110832864 is divided by 6 and 12?

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

8. ## Re: Find the remainder when 6341723110832864 is divided by 6 and 12?

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.

9. ## Re: Find the remainder when 6341723110832864 is divided by 6 and 12?

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:

10. ## Re: Find the remainder when 6341723110832864 is divided by 6 and 12?

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.