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 OverflowVideos
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.
Suppose you use the argument 4.5, and you want a number that is "greater than or equal to the argument and is equal to a mathematical integer".
There are lots of numbers that match those criteria: 5, 6, 7, 100, 2000, ....
The smallest of those (closest to negative infinity) is 5; and that's what ceil returns. The description is perfectly reasonable.
First, the answer is correct, but the reason for that complicated wording is negative numbers as arguments to ceil(), as Arvind mentioned in his comment.
Typically, we think left and right directions from 0 on the number line.
ceil( 4.5 ) returns 5, as we all expect without reading the documentation. But what is the result for ceil( -4.5 )? When we move on the number line towards negative infinity, the smallest number which is bigger than -4.5 is -4.0 which is the result of ceil( -4.5 ).