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I have this java project(using eclipse) that calculates check digit numbers. I'm stuck on how to code this one.
The check has a check digit so that it makes the sum of the sights including the check digit divisible by 7. Assume there are always 4 digits plus the check digit. The sample is 3875 with the number being 5 Second trial problem is 5862 and check number needs to be found. How do I go about doing this? I got to entering each digit and adding them but how can i do the rest?
This is for an into to computer science class so please no super complex stuff as if we didn't learn it I cant use it.
My teacher sucks by the way we learn none of this. I already did part a I need part b. Thanks. Here's an image to hell clarify.

First of all, you need to develop some "programmer logic", these problems help to develop it.
Airline tickets divide the actual identification number by 7 and assign the remainder to the check digit. Number can be of any length
Example:
12358 #3
Let's break this example:
12358 / 7 = 1765
and the reminder is 3
Let's do the same with the 2nd number on the example:
45349 / 7 = 45346
and the reminder is 3
So, your logic is correct.
An American Express traveler's check has a digit so that it maskes the sum of the digits, including the check digit, evenly divisible by 7.
Example:
3875 #5
In this problem the thing is a little different, you need to sum the digits:
3875 -> 3 + 8 + 7 + 5 = 23
Now you need to get the reminder of 23 / 7
23 / 7 = 3
And a reminder of 2
7 - 2 = 5
That's your checkDigit
5862
5862 -> 5 + 8 + 6 + 2 = 21
21 / 7 = 3
Reminder = 0
checkDigit = 7 - 0 = 7
So the formula is:
Split the number into digits
Sum the digits
Get the mod 7 of the sum
Make a rest of 7 - reminder

HashMaps - Unclear on the hash function and double hashing

I have been asked to implement a hash map using an array. I need to insert the following keys:
15, 7, 26, 39, 11, 9, 27, 5, 18, 2, 54, 22, 4
into an array of size 19 using the hash function:
(3x + 7) % 19
Using linear probing, I would expect to get the following array (correct me if I'm wrong):
Index: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Key: 4 11 5 18 7 26 39 27 2 15 9 22 54
where 26 had a collision with 7 at index 9, and so was inserted at index 10, and 39 then had a collision with 26 at index 10 and so was inserted at index 11.
I am now attempting to insert the same elements in an array implementation of a HashMap, using double hashing instead of linear probing. The 2nd hash function I am given is:
11 - (x % 11)
I have two questions:
Does this mean that my array will be of size 11 or still 19?
Do I initially use the original hash function and if the given index is free, insert the element there, otherwise if there is a collision, use the 2nd hash function and insert the element there?

According to Wikipedia the secondary hash function is used indirectly in the probing function:
h(0, x) = (3x + 7) % 19
h(j, x) = ((3x + 7) + j(11 - (x % 11))) % 19
Where j is the collision counter.

Regex that blocks single digit numbers 2,3,4,5

I am pretty new to regular expressions and for some particular case need a regular expression which will block numbers:
2,3,4 and 5
... out of:
0 to 21
Specifically, this should block only single digit 2,3,4 and 5 and not 12,13,14,15 or 21 and 22 for that matter.
I tried [^\d2-5] but then its also blocking 12,13,14,15,21,20,22 which is not desired since only 4 numbers specifically 2,3,4 and 5 are to be blocked.
Any help on this will be really helpful.

For the range [0;21] excluding [2;5] range you can use the following:
^(?:[016789]|1\d|2[01])$
Demo
If you just need to exclude [2;5] range, then the following might suit you:
^(?:[016789]|[1-9]\d+)$
Demo

If I understand your question correctly, you can try finding those specific digits by delimiting them with word boundaries.
For instance:
String singleDigitsToBeBlocked = "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21";
Pattern p = Pattern.compile("\\b[2-5]\\b");
Matcher m = p.matcher(singleDigitsToBeBlocked);
while (m.find()) {
System.out.printf("Blocked: %s%n", m.group());
}
Output
Blocked: 2
Blocked: 3
Blocked: 4
Blocked: 5

what about just checking for patterns that include everything EXCEPT those single digits?
(\b[0-9]{2,})|([01])|([6-9])

Is very simple
(?:1[0-9]|2[0-1]|[0-1]|[6-9])
You can test this here: http://www.regexr.com/39044

Print triangle of numbers in java

I've been having some trouble with a homework assignment for my Java class. In it, we're supposed to take in an integer between 1 and 13 and display three different triangles consisting of numbers. For example, if I were to enter 5, the result would be:
Triangle 1
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
Triangle 2
1
2 6
3 7 10
4 8 11 13
5 9 12 14 15
Triangle 3
5
4 9
3 8 12
2 7 11 14
1 6 10 13 15
I've already got the first Triangle going fine, but my big concern is the second triangle. I haven't attempted the third one yet. The other thing is that my Professor is picky about what method we use in creating the project. In other words, we can only use what he has taught us. He told us to use the System.out.printf("%3d", n) statement to space out the characters and we have to create them within a separate class.
The code for the first triangle is as follows:
void triangle1(int n)
{
int k = 1;
for (int i = 1; i <= n; i++)
{
for (int j = 0; j < 1; j++)
{
System.out.printf("%3d", n);
k += 1;
}
System.out.println();
}
}
So, pretty much, I need to follow that standard to create the other two triangles, but I'm really stuck on the second one and I don't know where to start. Any help will be much appreciated!

Here is the way I would approach it.
Programs print one line at a time, you cannot print half a line then start to print another line.
With that being said, you should recognize the pattern in the triangles.
1
2 6
3 7 10
4 8 11 13
5 9 12 14 15
You have the first number n, then you see the next row starts with n + 1. The next number starts in the row is (n + 1) + t where t = 4. There is a pattern there.
The third row follows the same pattern.
The first number is (n + 1) then you can calculate the others by + (t - 1)
This can be done with a for loop, like you did in the first time.
For the last triangle you can use the same process, just change the signs and t would equal something different.
Algorithm writing is all about identifying patterns.

If you look closely, you'll see there's a repeating pattern between each number and the one that follows on a given line.
3 7 10 => [3 & 7 differ by 4][7 & 10 differ by 3]
4 8 11 13 => [4 & 8 differ by 4][8 & 11 differ by 3][11 & 13 differ by 2]
5 9 12 14 15 => [... differ by 4][... by 3][... by 2][... by 1]
You can use that information to make the second triangle. I'll leave the rest to you. I hope that helps!

Seems you are a CS-Student, thus I will not present a finished solution. I'll give you some hints how I would solve this.
This is what the print statement has to do:
for i=1 j=0 print 1
for i=2 j=0 print 2
for i=2 j=1 print 6
for i=3 j=0 print 3
for i=3 j=1 print 7
for i=3 j=2 print 10
Find a formula that calculates the correct output from i and j, its a simple linear combination.

What was wrong with my logic on Java insertion sorts?

I had a question on a test - I got it wrong, but I don't know why?
Question: An array of integers is to be sorted from biggest to smallest using an insertion sort. Assume the array originally contains the following elements:
5 9 17 12 2 14
What will it look like after the third pass through the for loop?
17 9 5 12 2 14
17 12 9 5 2 14 - Correct answer
17 12 9 5 14 2
17 14 12 9 5 2
9 5 17 12 2 14
In an insertion sort, I thought the source array is untouched; I felt, at the third pass, the destination array would be incomplete.
How did the test arrive at this answer?

Basically, after n passes of insertion sort, the first n+1 elements are sorted in the correct order and the rest are untouched. As you can see in the alternatives, the correct answer is the only one that fulfills that. Every step is only inserting relative to the already sorted numbers, so each step one extra number is correctly sorted.
Step 0 (Original, 5 assumed sorted)
5 9 17 12 2 14
Step 1, takes the 9 and puts it in the correct place before 5 (result, 9 5 sorted)
9 5 17 12 2 14
Step 2, takes the 17 and puts it in the correct place before 9 (result 17 9 5 sorted)
17 9 5 12 2 14
Step 3, takes the 12 and puts it in the correct place after 17 and before 9 (result 17 12 9 5sorted)
17 12 9 5 2 14

Insertion sort is typically done in-place. After k iterations the first k + 1 elements are sorted, hence the answer you've shown above.