You are doing 157/32 which is dividing two integers with each other, which always result in a rounded down integer. Therefore the (int) Math.ceil(...) isn't doing anything. There are three possible solutions to achieve what you want. I recommend using either option 1 or option 2. Please do NOT use option 0.
Option 0
Convert a and b to a double, and you can use the division and Math.ceil as you wanted it to work. However I strongly discourage the use of this approach, because double division can be imprecise. To read more about imprecision of doubles see this question.
int n = (int) Math.ceil((double) a / b));
Option 1
int n = a / b + ((a % b == 0) ? 0 : 1);
You do a / b with always floor if a and b are both integers. Then you have an inline if-statement which checks whether or not you should ceil instead of floor. So +1 or +0, if there is a remainder with the division you need +1. a % b == 0 checks for the remainder.
Option 2
This option is very short, but maybe for some less intuitive. I think this less intuitive approach would be faster than the double division and comparison approach:
Please note that this doesn't work for b < 0.
int n = (a + b - 1) / b;
To reduce the chance of overflow you could use the following. However please note that it doesn't work for a = 0 and b < 1.
int n = (a - 1) / b + 1;
Explanation behind the "less intuitive approach"
Since dividing two integers in Java (and most other programming languages) will always floor the result. So:
int a, b;
int result = a/b (is the same as floor(a/b) )
But we don't want floor(a/b), but ceil(a/b), and using the definitions and plots from Wikipedia: 
With these plots of the floor and ceil functions, you can see the relationship.
You can see that floor(x) <= ceil(x). We need floor(x + s) = ceil(x). So we need to find s. If we take 1/2 <= s < 1 it will be just right (try some numbers and you will see it does, I find it hard myself to prove this). And 1/2 <= (b-1) / b < 1, so
ceil(a/b) = floor(a/b + s)
= floor(a/b + (b-1)/b)
= floor( (a+b-1)/b) )
This is not a real proof, but I hope you're satisfied with it. If someone can explain it better I would appreciate it too. Maybe ask it on MathOverflow.
Answer from martijnn2008 on Stack OverflowWhy floor, round and ceil return double?
Why Do Both Math.Floor(double) and Math.Ceiling(double) return a double and not an int?
Because a Double can contain significantly larger values than an integer, you can't cast a Double to an Int without risking overflow. There's no reason for such simple functions to add such risk.
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You are doing 157/32 which is dividing two integers with each other, which always result in a rounded down integer. Therefore the (int) Math.ceil(...) isn't doing anything. There are three possible solutions to achieve what you want. I recommend using either option 1 or option 2. Please do NOT use option 0.
Option 0
Convert a and b to a double, and you can use the division and Math.ceil as you wanted it to work. However I strongly discourage the use of this approach, because double division can be imprecise. To read more about imprecision of doubles see this question.
int n = (int) Math.ceil((double) a / b));
Option 1
int n = a / b + ((a % b == 0) ? 0 : 1);
You do a / b with always floor if a and b are both integers. Then you have an inline if-statement which checks whether or not you should ceil instead of floor. So +1 or +0, if there is a remainder with the division you need +1. a % b == 0 checks for the remainder.
Option 2
This option is very short, but maybe for some less intuitive. I think this less intuitive approach would be faster than the double division and comparison approach:
Please note that this doesn't work for b < 0.
int n = (a + b - 1) / b;
To reduce the chance of overflow you could use the following. However please note that it doesn't work for a = 0 and b < 1.
int n = (a - 1) / b + 1;
Explanation behind the "less intuitive approach"
Since dividing two integers in Java (and most other programming languages) will always floor the result. So:
int a, b;
int result = a/b (is the same as floor(a/b) )
But we don't want floor(a/b), but ceil(a/b), and using the definitions and plots from Wikipedia:
With these plots of the floor and ceil functions, you can see the relationship.
You can see that floor(x) <= ceil(x). We need floor(x + s) = ceil(x). So we need to find s. If we take 1/2 <= s < 1 it will be just right (try some numbers and you will see it does, I find it hard myself to prove this). And 1/2 <= (b-1) / b < 1, so
ceil(a/b) = floor(a/b + s)
= floor(a/b + (b-1)/b)
= floor( (a+b-1)/b) )
This is not a real proof, but I hope you're satisfied with it. If someone can explain it better I would appreciate it too. Maybe ask it on MathOverflow.
157/32 is int/int, which results in an int.
Try using the double literal - 157/32d, which is int/double, which results in a double.
Might be stupid question, but this has always puzzled me about Java.
Why functions Math.floor(), Math.round() and Math.ceil() return double value?
I thought main point of those is to convert floating point number into whole number. So why not returning long or int? That way we don't have to manually cast everytime.