Congruence modulo m: Find Smallest Int Value

[pc2-brazil]

Homework Statement

Find a formula for the integer with smallest absolute value that is congruent to an integer a modulo m, where m is a positive integer.

Homework Equations

An integer x is congruent to an integer a modulo m if and only if:
x \equiv a \pmod m

The Attempt at a Solution

From the definition:
x \mod m = a \mod m
or:
x - a = km
where k is an integer.
From the division "algorithm":
x = mq + a\mod m
where q is the quotient.
But I'm not sure on how to proceed from here. The textbook gives a strange answer: x \mod m if x \mod m \leq \left \lceil m/2 \right \rceil and (x \mod m) - m if x \mod m > \left \lceil m/2 \right \rceil
I would say that the smallest absolute value of x is when the quotient (q above) is 0. Thus:
x=a\mod m
According to the answer, x=a\mod m is only true if x \mod m \leq \left \lceil m/2 \right \rceil, but I can't figure out why.

Thank you in advance.

 

[HallsofIvy]

For example, If a= 5 and m= 7, then x= 5 (mod 7) if and only if x= 5+ 7k for integer k. k= 0 gives x= 5, of course, and k= -1 gives x= -2. The "smallest absolute value" is 2, not 5, because ⌈m/2⌉= 3 and 3< 5. If a= 3 then x= 3 (mod n7) if and only if x= 3+ 7k. k= 0 gives x= 3 while k= -1 gives 3- 7= -4. Now the "smallest absolute value" is 3.

In general x= a (mod m) (for 0\le a&lt; m) if and only if x= a+ mk for integer k. k= 0 gives x= a&gt; 0 and k= -1 gives a- m&lt; 0. Which of those has smaller absolute value depends on how close a is to m.