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# Math issue

Greenhorn
Posts: 14
Hi,

Im trying to divide 2 numbers lets say 127 by 50. When I do I get a value of 3 returned. However I want 4 to be returned, asin I want the remainder to be recognised. How can I do this?

Thanks

author and iconoclast
Posts: 24203
43
Hmm. 127 / 50 is 2, not 3 or 4, with a remainder of 27, so I'm not exactly sure what you're after here.

But if you divide two integers, you get the quotient as an integer, ignoring the remainder, and you can use the "%" operator to get just the remainder -- i.e.,

127 / 50 --> 2
127 % 50 --> 27

Sarah Shay
Greenhorn
Posts: 14
Yes its 2.5 roughly. But I want to round this up to 3 so I dont lose the decimal at the end

Ranch Hand
Posts: 323
you could cast the values to doubles, so as to do a floating-point division, then round up, then after that convert back to integers if needed. the Math class has some rounding functions you might find useful.

or you could do an integer division the way you are, then do aseparate modulus operation to find the remainder, and add 1 to the result of the division if needed depending on what the modulus was. it's your choice which of the methods is more complicated and roundabout, i think.

Ranch Hand
Posts: 3061
If you need the decimal places, you should probably use floats or doubles to start with instead of just ints. Then if you need to, you can round these off from there.

Layne

(instanceof Sidekick)
Posts: 8791
To round up integer division add divisor-1 before dividing.

127 + 9 / 10 = 13

127 + 49 / 50 = 3

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