The modulus (or modulo) operator returns the remainder of a division statement. It only works with integer operands. Thus, it will only return an integer result. Six is able to divide forty-eight a maximum of eight times. 

And then 

Thus,

When using the plus-equals (+=) operator, the answer is cast to the type of the left-hand argument. Since a was an int, the value returned by the expression is converted to int before it is stored in a. Thus, the decimal point and any information behind the decimal is truncated.

It is probably easiest to give a running commentary on the code in order to understand it. See the comments in bold below:

int i = 85, j = 2, k = 4;

i--; // After this, i = 84

j++; // After this, j = 3

k -= 6; // After this, k = k - 6 or k = -2

/*

Consider this in several steps:

i / j = 84 / 3 = 28

i / k = 84 / -2 = -42

Thus, the final expression is:

i = 28 - 42 = -14

*/

i = i / j + i / k;

The easiest way to trace code like this is to provide running commentary in bold comments. See these below:

int a = 24, b = 17, c = 5, d = 7; 

/*

The modulus (%) is the remainder of a division. So . . .

b % d = 17 % 7 = 3 (17 divided by 7 is 2 remainder 3.)

a / b = 1 (Remember, integer division truncates the decimal portion.)

Thus, we have:

b = 3 + 1 = 4

*/

b = b % d + a / b;

a %= b; // This is the same as a = a % d, which is a = 24 % 4 =  0 (No remainder)

a -= 12; // Remember that this is the same as a = a - 12, which is a = 0-12 or -12.

The Unicode Standard is Java's attempt to simplify the translation of characters and symbols to input the computer can understand, while being able to work on all Java machines. The first 128 numbers (0-127, 0x0000-0x007F) are for basic Latin. Under this character encoding scheme, the letter 'W' is denoted by U+0057. 

The standard for writing Unicode values is U+####, where the number is in hexadecimal format, not a base 10 number.

Taking a look at the first line in main, the variable bin is defined as true, which we know to be 1. We see in the next line that the variable var is 2 * bin +1. The trick here is that you can actually do math with boolean values, so the value of var is 3.

The value of triple is 5.

The the variable final asks us to AND bin and var first, then AND triple.

To do this, we can convert all three numbers into binary notation to make it easier.

1 in binary is 001.

3 in binary is 011.

5 in binary is 101.

1&3 =1.

1&5=1.

Final answer is 1.

1. 2893 / 10 = 289, since x is an int, it does not store the decimal value

2. 2893 % 10 = 3 because the remainder of 2893/10 is 3

The expression 1+1 * 2 + 4 / 5 can be rewritten as

1 + (1*2) + (4/5) = 1 + 2 + 0

4/5 = 0 because a is an int and when dividing ints, there are no decimals and since 4/5=0.8 we keep the ones digit which is zero.

You have to remember PEMDAS.

Multiplication and Division take precedence first

 (double multiplied by an int equal a double)

Addition and Subtraction after

The line, char newvar = str[4] is declaring a new variable which is of the type character. Further more, when it declares it, it is saying that it is the same as the character at the 4th location in the string above.

The value 4 actually represents the 5th value, since indices start at 0.

Therefore, at str[4] we have the character 'o'.