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An integer d is a divisor of a ⇔ ____ | ____.
Equivalently, d is a divisor of a ⇔ ____ mod ____ = _____.
Is it possible for a divisor of a to be bigger than a?

The first blank would be d|a, and I am not sure about the second.
And for the third question, I believe so. But I am not sure how to explain why.

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closed as off-topic by Daniel Robert-Nicoud, user1551, user147263, Shailesh, John B Mar 31 '16 at 0:05

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  • 1
    $\begingroup$ Sure, $2$ is a divisor of $-18$. Also, $17$ is a divisor of $0$. But if we restrict to positive integers, the answer is no. $\endgroup$ – André Nicolas Mar 30 '16 at 20:05
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$a$ mod $d$ = 0 <=> $d | a$.

Divisor of $a$ can not be bigger than $a$, because $d | a$ <=> $a = c * d$. If $d > a$, $c < 1$, and it is impossible.

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$d$ is a divisor of $a$ if $a/d$ is whole, in other words if $a \pmod d = 0$.

If $d$ is a divisor of $a$ and $d>a$, what does $a/d$ look like?

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think mod operator as the remainder. If d divides a, the remainder should be zero by definition. Also think what will be the maximum divisor of an integer number?

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  1. $d|a$
  2. $a=0 \mod d$
  3. It depends. If we are talking about positive integers then no, but for rationals or reals it is possible.
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