How to decide whether the result of an integer expression is an integer

Question

I have an expression of n like (30 - n(n - 1)) / 2n. And I want search the possible n which will be my answer only when the result is an integer. Is there any way to decide whether the result of this expression is an integer or not.

The only way that I can come up with is(in pseudo code ) :

for float n <- 1 to 100
  do float result = expression(n);
     int part = (int) result;
     if ( result - part < EPS )
       then good to go

because if it is an int, the expression will be an ineger. Am I wrong? — Joey.Z
– Joey.Z, Commented Aug 21, 2013 at 8:23
Sure but then you can use integer remainder to see if the division in exact ! — hivert
– hivert, Commented Aug 21, 2013 at 8:24

jxh · Accepted Answer · 2013-08-21 08:35:33Z

7

You can use % to compute the remainder.

int denom = 2 * n;
int numer = 30 - n * (n - 1);

if (denom) {
    if (numer % denom == 0) {
        then good to go
    }
} else {
    /*...denominator is 0! */
}

edited Aug 21, 2013 at 8:35

answered Aug 21, 2013 at 8:21

jxh

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2 Comments

paxdiablo Over a year ago

That won't work so well for n == 0 :-) But since OP stated n ranges from 1 to 100, it doesn't really matter.

jxh Over a year ago

A divide by 0 exception is as good as an assertion, but you are right that it would be friendlier to test for a non-zero denominator.

Sakthi Kumar · Accepted Answer · 2013-08-21 08:21:59Z

2

if (30 - n(n - 1)) mod 2n equals to zero

answered Aug 21, 2013 at 8:21

Sakthi Kumar

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Comments

James Kanze · Accepted Answer · 2013-08-21 09:01:58Z

1

If modf returns 0.0, the floating point number it is called on is an integer. This is the standard way of testing whether a floating point number is an integer; it works with all floating point values, even those that would overflow an int.

answered Aug 21, 2013 at 9:01

James Kanze

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