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calculate Check Number in Java

Not sure if anyone can explain this to me or help me.
I have a 15 Digit Number of which I want to multiply each even number by 2 unless the even number is greater than 9. If it is this needs to be subtracted by 9 to give me an integer that again I can multiply by 2. Once I have all the even numbers multiplied by 2 i need to add them all together with the odd numbers.
Does that make sense.
UPDATE ***
so i have a number say 49209999856459. for that number I am looking to get the even integer so for example the first even one would be 4 then the second would be 2 and so on.
If one of those even numbers are multiplied by 2 then it might be above 9 so I want to subtract 9 to then use the remainder as the even number in its place.
SO !!!
Multiply by 2 the value of each even digit starting from index 0 and then each even index. In each case, if the resulting value is greater than 9, subtract 9 from it (which reduces larger values to a single digit). Leave the values of the digits at the odd indexes unchanged.
public String calculateCheckNumber()
String firstFifteen = longNumber.substring(0,15) ;
int i, checkSum = 0, totalSum = 0;
for (i = 0; i<firstFifteen.length(); i += 2) {
while (i < 9)
i *= 2;
if (i > 9)
i -= 9 ;
}
Was one option I was trying but it honestly I cant seem to get my head around it.
Any Help would be greatly appreciated.

Well, here is one approach. This uses the ternary (?:) operator to condense the operations. Edited base on clarification from the OP. The example you gave is actually a 14 digit string. But the following will work with any number of digits if they start out in a string. If you have a long value, then you can create the character array using:
long v = 49209999856459L;
char[] d = Long.toString(v).toCharArray();
Here is the main algorithm.
String s = "49209999856459";
int sum = 0;
char[] d = s.toCharArray();
for (int i = 0; i < d.length; i++) {
int v = d[i] - '0';
// The even digit will only be greater than 9 after
// doubling if it is >= 5 before.
sum += ((i % 2) == 1) ? v : (v >= 5) ? v+v-9 : v+v;
}
System.out.println(sum);
Prints
86

Creating a sliding number puzzle board using arrays in Java

So I'm kind of new to Java and decided to create a sliding number puzzle of some sort. Here's what I have :
int[] puz = {1,2,3,
4,5,6,
7,8,9}
for(int i=0; i<puz.length; i++){
System.out.println(puz[i]);
}
The 1 is supposed to be the blank spot but I'll figure that out later. My problem is that the code prints:
1
2
3
4
5
6
7
8
9
when I want it to print:
1 2 3
4 5 6
7 8 9
I've also tried doing a nested loop that I'm too embarrassed to show on here due to how hideous it was.
Would I try using a 2d array instead?

I guess you could try...
int puz = {1,2,3,4,5,6,7,8,9};
int n = Math.ceil(Math.sqrt(puz.length));
for (int i = 0; i < puz.length; i++) {
System.out.print(puz[i] + ((i + 1) % n == 0 ? "\r\n" : " ");
}

Try creating a variable counter and increment it every time you iterate through the loop. Using a modulus operator, divide it by 3 and when remainder is 0, create a new line.
int puz = {1,2,3,4,5,6,7,8,9};
int counter = 1;
for(int i=0; i<puz.length; i++){
System.out.print(puz[i]);
if (counter % 3 == 0){
System.out.println("");
}
counter++;
}

The trick here is to use the modulus operator. This operator divides one number by another, and returns the remainder. In java (and everywhere else as far as I know), % is the modulus operator. If you want every third number to have a line break after it, simply divide by three using modulus division, like so:
int[] puz = {1,2,3,4,5,6,7,8,9};
//For what it's worth, you don't have this semicolon in your question, so I added it in.
for(int i=0; i<puz.length; i++){
System.out.print(puz[i] + " ");
if(i % 3 == 2){//It's equal to 2 because you start at 0 and not 1.
System.out.println("");
}
}
This code, when executed, prints the following, which is what you wanted:
1 2 3
4 5 6
7 8 9

Write a program in java to print the sum of all numbers from 50 to 250(inclusive of 50 and 250) that are multiples of 3 and not divisible by 9

/* when I run this code there is no error in fact output generated is also correct but I want to know what is the logical error in this code? please can any one explain what is the logical error. */
class abc
{
public static void main(String arg[]){
int sum=0;
//for-loop for numbers 50-250
for(int i=50;i<251;i++){
// condition to check if number should be divided by 3 and not divided by 9
if(i%3==0 & i%9!=0){
//individual number which are selected in loop
System.out.println(i);
//adding values of array so that total sum can be calculated
sum=sum+i;
}
}
//final display output for the code
System.out.println("the sum of intergers from 50 to 250 that are multiples of 3 and not divisible by 9 \n"+sum);
}
}

My philosophy is "less code == less bugs":
int sum = IntStream.rangeClosed(50, 250)
.filter(i -> i % 3 == 0)
.filter(i -> i % 9 != 0)
.sum();
One line. Easy to read and understand. No bugs.

Change this:
if(i%3==0 & i%9!=0){
to this:
if(i%3==0 && i%9!=0){
& = bitwise and operator
&& = logical operator
Difference between & and && in Java?

The only problems I saw were:
The variable sum was undeclared
Use && in place of &
int sum = 0;
for (int i = 50; i <= 250; i++) {
if (i % 3 == 0 && i % 9 != 0) {
System.out.println(i);
sum = sum + i;
}
}
System.out.println("the sum of intergers from 50 to 250 that are multiples of 3 and not divisible by 9 \n" + sum);

Well, instead of touching every single value from 50 to 250 like you would do here for(int i=50;i<251;i++), you can consider something like this...
int i = 48;
int sum = 0;
while(i < 250) {
i += 3;
if(i%9 != 0)
sum += i;
}
This is somewhat optimized in the sense that I am skipping over values that I know are not possible candidates.
But, there is a much bigger issue in your code. The following code block prints true, sure. But, it is a bad idea to depend on the & since that is not its job. The & is for bitwise AND whereas the && is for logical AND, which is what you are trying to do.
boolean t = true;
boolean f = false;
System.out.println(f&t);
Why?
In Java, if it is a && operation, as soon as you find the first false, you are sure that the expression will evaluate to false. Meanwhile, in your implementation, it would need to evaluate both sides. f&t will evaluate to false, but the JVM would need to look at both the f and t variables. Meanwhile, on using &&, it wouldn't even need to look at the t.

How to get count value assign the table

I get some number from URL, if the number is b/w 1 to 4 then I want's the result to be 4, .. and number b/w 4 to 8 then I want result 8 ......And so on.
This is my code. and this is get from url in count value.
adltavailable = Integer.parseInt(count.get(i).toString());
for(int x =0; x<adltavailable; x++)
{
c = "Adultavailable";
category.add(c);
}
//Here is assign the table
int k = 0;
int size = category.size();
while(k < size)
{
for(int z=0; z<size; z++)
{
if(category.get(z).equals("Adultavailable"))
{
mycirimgs[k].setBackgroundResource(R.drawable.adultadd);
}
k++;
}
}
I get total seats value from url .And the value is assigned in table.If suppose i got 3 seats means I assign the table in 3 seats is not look like good.But this three seats assign 4 instead of 3.like wise.So I want the result If I get total seats 1 or 2 or 3 or 4 means I assign the table in 4 seats and 5 or 6 or 7 or 8 means I assign the table in 8 seats like wise. Thanks giving ur support.

I get some number from URL, if the number is b/w 1 to 4 then I want's the result to be 4, .. and number b/w 4 to 8 then I want result 8 ......And so on.
To round x to the next multiple of 4, write
(x + 3) & ~3
where + 3 rounds up and & ~3 clears the bottom two bits making it a multiple of 4.

To round up your input to the next multiple of 4, you can try the following:
Integer result;
if (input % 4 == 0) //Input is already a multiple of 4.
{
result = input
}
else // round up
{
result = ((input / 4) + 1)*4
}
Hope this is what you were looking for.

How to "invert" numbers in Java

Lets say I have a number 1-5, now if I have 2, I want 4 as an output, if I had 3 then have 3 as the output, if I have 1 then 4 as the output. Here is a chart of what I want:
1-10 Chart:
Give 1 return 9
Give 2 return 8
Give 3 return 7
Give 4 return 6
Give 5 return 5
What algorithm do I use for such a thing?

I don't see that you need an algorithm as much. What you have is:
InverseNumber = (myCollection.Length - MySelection);
Thats all you need for even numbers.
With a collection of 1 - 6 for example:
Give 2; 6 - 2 = 4. Also if given 4, 6 - 4 = 2.
You will need a slightly different problem for odds:
1 - 5; with 1 given 1 is at index 0, the opposite is 5, 2 given and the inverse ( 5 - 2) is 3. But if 3 is given, there is no inverse. So you might want to also add a catch for:
if (((myCollection.Length *.5).Round) == mySelection) { //Inverse does not exist!!!}
If you are using just integers, and not arrays of numbers then just replace the myCollection.Length with the upperbound integer.

I think the following code will work for what you need:
int a[] = new a[length_needed];
int counter = length_needed;
for(int c = 0; c < length_needed; c++) {
a[c] = counter;
counter--;
}
int number_inputed;
for(int c = 0; c < length needed; c++) {
if(c == number_inputed) System.out.println(a[c]);
}

Let's say you are giving max number as input. Then you are going to have 0-n numbers. For ex., if 9 is the max number you will have 0-9.
Then you can do something like this:
public static void main(String[] a) {
int max = a[0]; // read values from cmd line args
int forWhichNum = a[1]; //for which number we need its inverse
Sop(max- forWhichNum);
}

Integer value = 2;
Integer maxValue = 6;
Integer reverseCounter = 0;
for (int i = maxValue; i > 0; i--) {
reverseCounter++;
if (i == value) {
return reverseCounter;
}
}