# divisors of natural numbers

I have a problem with an exercise. I have the two following sentences I have to determine if they are true or false.

If $a<b$ and the remainder of $b$ and $a$ divided by $c$ are equal then $c\mid(b-a)$

If $a<b$ and the remainder of $b$ and $a$ divided by $c$ are equal then $c\mid b-a$

I think the first is true. But I can't understand what's the different when removing the brackets?

Thanks.

EDITED! I was mistaken copying the questing. Sorry! Now its correct

• There isn’t any difference when you remove the brackets. It’s the same thing. – user507623 Jan 26 '18 at 9:12
• I think the difference is how $c\mid b - a$ and $(c\mid b)-a$ but... does the latter make any sense? There is no difference when moving the brackets btw. It is like saying $c=b-a$ compared to $c=(b-a)$. They mean the same thing. But with your example, we replace $=$ with $\mid$. Thus, it is a trick question.... I think? – Mr Pie Jan 26 '18 at 10:01
• Looks like it is... And both are true... – maryum375 Jan 26 '18 at 10:40

The first one is true even if you remove the hypothesis that $a<b$.
• I don't think this is true. The statements $b\equiv c \pmod a$ and $c\mid (b-a)$ are quite unrelated, aren't they? – Claudius Jan 26 '18 at 9:36
• @Claudius I don't understand your remark since the relation $b\equiv c\pmod a$ was nowhere mentioned in the original question. – José Carlos Santos Jan 26 '18 at 9:40
• But "the remainder of $b$ and $c$ divided by $a$ are equal" translates to $b\equiv c \pmod a$... – Claudius Jan 26 '18 at 9:41