# Need help on an easy proof

• Feb 23rd 2010, 05:10 PM
Mathwizard4
Need help on an easy proof
need to prove that:
if m, d, and k are nonnegative integers and d does not equal 0, then (m + dk) mod d = m mod d.

I'm pretty sure you need to use the quotient remainder theorem.

• Feb 23rd 2010, 05:18 PM
icemanfan
$m + dk \mod d \equiv m \mod d + dk \mod d \equiv m \mod d + 0 \mod d \equiv m \mod d$. Is there any reason you can't use this argument?
• Feb 23rd 2010, 05:39 PM
Mathwizard4
how do you know (m + dk) mod d = m mod d + dk mod d?

is that a rule that works for all integers?

also

wouldn't m mod + 0 mod d = m mod d + d and not m mod d
• Feb 23rd 2010, 05:48 PM
icemanfan
Quote:

Originally Posted by Mathwizard4
how do you know (m + dk) mod d = m mod d + dk mod d?

is that a rule that works for all integers?

Yes, it does. The remainder when a + b is divided by d is congruent to the sum of the remainder of a when divided by d and the remainder of b when divided by d.
• Feb 23rd 2010, 05:52 PM
Mathwizard4

wouldn't m mod + 0 mod d = m mod d + d and not m mod d http://www.mathhelpforum.com/math-he...c/progress.gif ?
• Feb 23rd 2010, 05:55 PM
icemanfan
Quote:

Originally Posted by Mathwizard4
wouldn't m mod + 0 mod d = m mod d + d and not m mod d

Not sure what you're asking here. Do you mean $m \mod (d + d)$ or $m \mod d + d \mod d$?
• Feb 23rd 2010, 06:08 PM
Mathwizard4
how do you know....

$m \mod d + dk \mod d \equiv m \mod d + 0 \mod d$
• Feb 23rd 2010, 08:37 PM
Mathwizard4
can anyone else help?
• Feb 23rd 2010, 09:22 PM
discreteDilema