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For Loop that counts up and down from a start index inside range

I'm having trouble developing a for-loop that starts at a given integer (start) and then counts up and down from this starting point inside a given range (min to max). I want the result to be in order by distance to the given start. If a for-loop is not the right tool here, i am open for other suggestions!
Since this is hard to explain i have a example down below:
var start = 4;
var max = 10;
var min = 0;
var stepSize = 1;
for(var i=startIndex; i>=min && i<=max; ???) {
Console.WriteLine(i)
}
The result should look like this:
4 5 3 6 2 7 1 8 0 9 10
The really tricky part for me is the end where the alternation has to stop in order to reach the max value of 10 and not go below 0 in the meantime.
I started experimenting and had a solution that would work for the first bit:
const int start = 4;
const int stepSize = 1;
for (var (i,j,k) = (start,1,5); Math.Abs(j)<=11; k=(k+j),j=-j-Math.Sign(j)*stepSize,i=(k+(start-5))%11) {
Console.WriteLine(i);
}
This works fine for the first part where alternation is needed, but it breaks once the alternation needs to stop. The result from this experiment is:
4 5 3 6 2 7 1 8 0 9 -1

Here it is with a for loop, but like others have already alluded to, I think other structures are better suited for your task:
public static void Main() {
int start = 4;
int max = 10;
int min = 0;
int stepSize = 1;
for(int i=start, j=start; i>=min || j<=max; i-=stepSize, j+=stepSize) {
if (i==start) {
Console.Write(i + " ");
}
else {
if (j <= max) {
Console.Write(j + " ");
}
if (i >= min) {
Console.Write(i + " ");
}
}
}
}

The sequence you are after:
4 5 3 6 2 7 1 8 0 9 10
Could be specified as the following:
4 - 3 - 2 - 1 - 0
- 5 - 6 - 7 - 8 - 9 10
- - - - - - - - - - --
4 5 3 6 2 7 1 8 0 9 10
Which using something like MoreLinq, could be expressed like this:
Sequence(4, 0).Interleave(Sequence(5, 10))
Obviously, you control the start, max and min by controlling the values of both sequences. So, say you want to start from 7, going up and down towards 1 as the minimum value, and 14 as the maximum. The call would be like:
Sequence(7, 1).Interleave(Sequence(8, 14))
Lastly, this also allows you to easily change from a "up-down" to a "down-up" by swapping the intervals. Using the example above:
Sequence(8, 14).Interleave(Sequence(7, 1))
Note that Sequence also has an overload that takes a step argument, so you can additionally expand the examples below with arbitrary step increments and not only 1.

Since you explicitly stated that you want a for loop that produces the pattern, how about this approach:
Find the larger "distance" first and let the loop run until this point, not starting with 0. In the loop check in both directions and do the calculations inline and call the method once beforehand with the original value.
int x = 4;
int min = 0;
int max = 10;
int largest = Math.Max(Math.Abs(min - x), Math.Abs(max - x));
YourMethod(x);
for (int i = 1; i <= largest; i++)
{
if (x + i <= max)
YourMethod(x + i);
if (x - i >= min)
YourMethod(x - i);
}

A simple but elegant solution could generate the number sequence you want like so:
static IEnumerable<int>
UpDownCount(int start, int lower, int upper)
{
int i = start, j = start;
yield return start;
while (true)
{
bool ilimit = ++i <= upper,
jlimit = --j >= lower;
if (ilimit)
yield return i;
if (jlimit)
yield return j;
if (!(ilimit || jlimit))
yield break;
}
}
Then you can use that sequence with LINQ or a simple foreach loop:
foreach (int i in UpDownCount(4, 0, 10))
Console.WriteLine(i);

C# Multiplication Table logic

I'm making a c# program that shows the multiplication table, by using my code I can print only complete table, how can I print random table as below.
my code bit:
using System;
public class Exer
{
public static void Main()
{
int j,n;
Console.Write("\n\n");
Console.Write("Display the multiplication table:\n");
Console.Write("-----------------------------------");
Console.Write("\n\n");
Console.Write("Input the number (Table to be calculated) : ");
n= Convert.ToInt32(Console.ReadLine());
Console.Write("\n");
for(j=1;j<=10;j++)
{
Console.Write("{0} X {1} = {2} \n",n,j,n*j);
}
}
}
Expected Output:
1 x 2 = 2
2 x 2 = 4
3 x 2 = 6
1 x 4 = 4
2 x 4 = 8
3 x 4 = 12
1 x 6 = 6
2 x 6 = 12
3 x 6 = 18
1 x 8 = 8
2 x 8 = 16
3 x 8 = 24
1 x 10 = 10
2 x 10 = 20
3 x 10 = 30

The tables in your picture are NOT random, though. There is a definite pattern. The function, DisplayMath(), is receiving an integer parameter; 3, 4, and 5 in the picture.
The output in the pictures is most likely produced by two NESTED for loops.
The integer parameter is controlling the INNER for loop, which goes from 1 to the parameter, incrementing by 1.
The OUTER loop goes from 2 to 10, incrementing by 2.
Possible code:
public static void Main() {
DisplayMath(4);
}
public static void DisplayMath(int x) {
for(int y=2; y<=10; y=y+2) {
for(int z=1; z<=x; z++) {
Console.WriteLine("{0} X {1} = {2}",z,y,z*y);
}
}
}
Output:
1 X 2 = 2
2 X 2 = 4
3 X 2 = 6
4 X 2 = 8
1 X 4 = 4
2 X 4 = 8
3 X 4 = 12
4 X 4 = 16
1 X 6 = 6
2 X 6 = 12
3 X 6 = 18
4 X 6 = 24
1 X 8 = 8
2 X 8 = 16
3 X 8 = 24
4 X 8 = 32
1 X 10 = 10
2 X 10 = 20
3 X 10 = 30
4 X 10 = 40

As per the required output, you need to change the for loop logic a little.
Because there is a logic in the expected result that you need to get before start coding, that isn't a 'random' table
for(j=2; j<=10; j=j+2 )
{
for (mathNumber = 1; mathNumber <= n; mathNumber++ )
{
Console.Write("{0} X {1} = {2} \n",mathNumber,j,mathNumber*j);
}
}
As you can see the result must be the result of the n first numbers (being n the parameter you get from Console, but perhaps is better that you get it as argument) multiplyed by the numbers 2,4,6,8,10

Get min and max integer value containing X number of digits

What would be the best way to generate the lowest and highest integer values which contain x number of digits?
For example:
x = 1: Min = 0, Max = 9
x = 2: Min = 10, Max = 99
x = 3: Min = 100, Max = 999
x = 4: Min = 1000, Max = 9999
I feel like this should be a really easy thing to accomplish but I'm struggling to get my head around the math behind it.
Here's what I've got so far for generating the max value (based on this answer):
public static int MaxIntWithXDigits(this int x)
{
if (x == 0) throw new ArgumentException(nameof(x), "An integer cannot contain zero digits.");
try
{
return Convert.ToInt32(Math.Pow(10, x) -1);
}
catch
{
throw new InvalidOperationException($"A number with {x} digits cannot be represented as an integer.");
}
}
If anyone can help with generating the min value or suggest improvements on the code above I'd appreciate it.
Edit
I'm almost there - this seems to work for everything except when x = 1 (and min value expected as 0)
public static int MinIntWithXDigits(this int x)
{
if (x == 0) throw new ArgumentException(nameof(x), "An integer cannot contain zero digits.");
x -= 1;
try
{
return Convert.ToInt32(Math.Pow(10, x));
}
catch
{
throw new InvalidOperationException($"A number with {x} digits cannot be represented as an integer.");
}
}

Fox any x>1, the min value would be 10x-1. E.g.:
If x = 2, min=102-1 = 101 = 10
If x = 3, min=103-1 = 102 = 100
If x = 4, min=104-1 = 103 = 1000
For x=1, since it's the only value that is not a one followed by a series of zeroes, you'll need special treatment.
public static int MinIntWithXDigits(this int x)
{
if (x == 0)
{
throw new ArgumentException(nameof(x), "An integer cannot contain zero digits.");
}
if (x == 1) {
return 0;
}
try
{
return Convert.ToInt32(Math.Pow(10, x - 1));
}
catch
{
throw new InvalidOperationException($"A number with {x} digits cannot be represented as an integer.");
}
}
Side note:
If the result were restricted to positive integers instead of non-negative integers, the smallest integer with one digit would be 1, and the general formula would have applied for this case too.

You can try something like this:
// let's return min and max in one go as a value tuple
public static (int min, int max) MaxIntWithXDigits(this int x) {
if (x <= 0 || x > 9)
throw new ArgumentOutOfRangeException(nameof(x));
int min = (int)Math.Pow(10, x - 1);
return (min == 1 ? 0 : min, min * 10 - 1);
}
Demo:
var result = Enumerable
.Range(1, 9)
.Select(x => $"{x} : {MaxIntWithXDigits(x).min} .. {MaxIntWithXDigits(x).max}");
Console.Write(string.Join(Environment.NewLine, result));
Outcome:
1 : 0 .. 9
2 : 10 .. 99
3 : 100 .. 999
4 : 1000 .. 9999
5 : 10000 .. 99999
6 : 100000 .. 999999
7 : 1000000 .. 9999999
8 : 10000000 .. 99999999
9 : 100000000 .. 999999999

c# rules about: a mod b, while a is bigger than b

I was examining a code about operator overloading, and I saw an interesting way of use of % to convert seconds to minutes if second variable is bigger than 59.
It seems while a<b (both int) a%b returns a, I havent read a rule about this before, I wanna learn why it returns a, is it about they are declared as integers, are there any more rules about %?
Thanks

% is the modulo operator. The modulo operation a % b returns the remainder z in the division
a / b = x, remainder z
In other words, it returns y in this equation:
x * b + y = a
*The second equation also shows why this works for fractional numbers too.**
Thus, the following examples (not complete) are true:
-5 % 4 = -1
-4 % 4 = 0
-3 % 4 = -3
...
1 % 4 = 1
2 % 4 = 2
3 % 4 = 3
4 % 4 = 0
5 % 4 = 1
...
101.4 % 100.3 = 1.1
In your example converting seconds to minutes and seconds, you probably read something like this:
int totalseconds = 72;
int minutes = totalseconds / 60; // == 1
int seconds = totalseconds % 60; // == 12
This is the same thing: the minutes are calculated as totalseconds / 60, which is an integral division that "rounds" to the lower integer for positive numbers, thus the result of 1. The seconds left are calculated as the remainder of 72 and 60.
You could also calculate the seconds like
int seconds = totalseconds - (minutes * 60)

In general, for any a and any b:
a / b = (SomeConstant * b) + RemainderOfb*
So if a = 2 and b = 5 means a < b then:
How many SomeConstant do you need to multiply b with, in order to get close to a? And once you get close, what is your remainder?
so if
a / b = (5 * 0) + 2
Means Remainder = 2 out of 5.
Then: a % b means the Remainder with respect to b, which is = 2.
And if a = 3 and b = 2 means a > b then:
If you remove all the "b"s from a, what is the Remainder? You can do
it the easy way: Subtract "a" from "b" till you can't subtract
anymore. Or:
a / b = SomeConstant * a (which is how many "a"s did you subtract) + Remainder
Substitute:
a / b = (1 * 2) + 1
Means Remainder = 1 out of the 2
Then a % b equals the Remainder with respect to a, which is = 1

a mod b is the remainder of a a / b, therefore if a < b AND a > -b the remainder is always a.
The following rules are valid for C#, but might differ in other languages, refer to Modulo operation page on Wikipedia for more information.
0 mod 3 = 0
1 mod 3 = 1
2 mod 3 = 2
3 mod 3 = 0
4 mod 3 = 1
-1 mod 3 = -1 // 2 in some languages
-2 mod 3 = -2 // 1 in some languages
-3 mod 3 = 0
-4 mod 3 = -1 // 2 in some languages
// if your language supports mod of floats
1.5 mod 3 = 1.5
5.5 mod 3 = 2.5
-1.5 mod 3 = -1.5 // 1.5 in some languages
-5.5 mod 3 = -2.5 // 0.5 in some languages

Divide X by Y, return number of items in last, partial set

I'm probably missing something really simple, but I'm trying to figure out how to calculate what's left over after I divide X by Y. I don't mean the remainder, I mean, e.g. if I divide 100 by 7 => 6 groups of 15 + one group of 10, how do I get 10?
I don't have code to show because I have no idea where start. X and Y are both integers.

It's not as simple as just using modulus. The fiddly bit is calculating your initial group size from the number of groups.
Try this:
int population = 100;
int numberOfGroups = 7;
int groupSize = (population + numberOfGroups - 1)/numberOfGroups;
Console.WriteLine(groupSize);
int remainder = population%groupSize;
Console.WriteLine(remainder);

I don't mean the remainder
Yes you mean remainder.
If you divide 100 by 15, you get 6 as quotient and 10 as remainder.
use Modulus operator like
int remainder = 100 % 15; // This will return 6
int quotient = 100/15; // This will return 10

It's a modulo operator, isntead of:
var result = x / y;
Try this:
var result = x % y;
EDIT.
Ok You are very unclear but i think one of below solutions is what You are trying to avhieve.
S1. do this:
int x = 100/15;
int z = 15 * x;
int y = 100 - z; // and You got Your 10
S2. do this:
int x = 100/7;
if ( x * 7 != 100)
{
int GroupSize = x+1;
int rest = 100 - GroupSize;
}