Friday, 17 June 2016

divisibility - What is the remainder when 318 is divided by 100?



I know people here might find it silly but I want to clear my understanding about remainder.



I had a fight with my brother over this question. I suggested that remainder is something that is left when two numbers can't further be divided.



This is what I think remainder should be:-
$$\dfrac{318}{100}=3+\dfrac{18}{100}$$
$$\dfrac{318}{100}=3+\dfrac{9}{50}$$So according to me 9 should be the remainder of $\dfrac{318}{100}$ as 9 cannot further be divided by $50$.




This is what my brother does enter image description here



So according to him remainder should be $18$. But this is not the remainder actually this something we call mod of two number (which is usually used by computers to calculate remainder of two numbers ). My brother says that 9 is not the remainder of $\dfrac{318}{100}$ but the remainder of $\dfrac{159}{50}$. Aren't they the same thing.



Can anyone tell me, who is correct.


Answer



The remainder of $n$ divided by $d$ is the unique integer $r$ satisfying $0\le r and $n=qd+r$ for some integer $q$. When $n= 318$ and $d=100$, $r=18$ satisfies the criterion (with $q=3$).


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...