1. Number of briquettes

There is a pile of briquettes, piled into a triangular pyramid. specific:
Put 1 on the first layer,
3 in the second layer (arranged in a triangle),
6 in the third layer (arranged in a triangle),
10 on the fourth layer (arranged in a triangle),
....
If there are 100 layers, how many briquettes are there?
Please fill in the number representing the total number of briquettes.

Note: What you submit should be an integer, do not fill in any extra content or explanatory text.

public class Main {
	public static void main(String[] args){
		int a=0;
		int sum=0;
		for(int i=1;i<=100;i++){
			a=a+i;
			sum=sum+a;
		}
		System.out.println(sum);
	}
	
}

The result is: 171700

2. Birthday candles
A certain person has held a birthday party every year since a certain year, and every time he blows out the same number of candles as his age.
Counting now, he blew out a total of 236 candles.
Excuse me, at what age did he start his birthday party?
Please fill in the age when he started the birthday party.
Note: What you submit should be an integer, do not fill in any extra content or explanatory text.

public class Main {
	public static void main(String[] args){
		for(int a=1;a<=100;a++){
			int sum=0;
			for(int i=a;i<=100;i++){
				sum += i;
				if(sum==236){
					System.out.println(a);
				}	
			}
		}
	}
}

The result is: 26

3. Make up the formula
        B       DEF
A + --- + ------- = 10
        C       GHI
 
In this formula, A~I represent numbers from 1 to 9, and different letters represent different numbers.
such as:
6+8/3+952/714 is a solution,
5+3/1+972/486 is another solution.
How many solutions are there in this formula?
Note: Your submission should be a whole number, do not fill in any extra content or explanatory text.

Violent cycle (must pay attention to the problems caused by division, so change the condition to multiplication)

public class Main {
	public static void main(String[] args){
		int count = 0;
		int a,b,c,d,e,f,g,h,i;
		for(a=1;a<=9;a++){
			for(b=1;b<=9;b++){
				for(c=1;c<=9;c++){
					for(d=1;d<=9;d++){
						for(e=1;e<=9;e++){
							for(f=1;f<=9;f++){
								for(g=1;g<=9;g++){
									for(h=1;h<=9;h++){
										for(i=1;i<=9;i++){
											int GHI=g*100+h*10+i;
											int DEF=d*100+e*10+f;
											if(a!=b&&a!=c&&a!=d&&a!=e&&a!=f&&a!=g&&a!=h&&a!=i
												   &&b!=c&&b!=d&&b!=e&&b!=f&&b!=g&&b!=h&&b!=i
												         &&c!=d&&c!=e&&c!=f&&c!=g&&c!=h&&c!=i
												               &&d!=e&&d!=f&&d!=g&&d!=h&&d!=i
												                     &&e!=f&&e!=g&&e!=h&&e!=i
												                           &&f!=g&&f!=h&&f!=i
												                                 &&g!=h&&g!=i
												                                       &&h!=i
												                                       &&((a-10)*c*GHI+b*GHI+c*DEF==0))
											{
												count+=1;
											}
													
										}
									}
								}
							}
						}
					}
				}
			}
		}
		System.out.println(count);
	}


}

The result is: 29

4. Sub-group
Nine athletes participate in the competition and need to be divided into 3 groups for preliminaries.
What are the grouping schemes?
We mark athletes as A,B,C,...I
The following program lists all the grouping methods.
The normal output of the program is:
ABC DEF GHI
ABC DEG FHI
ABC DEH FGI
ABC DEI FGH
ABC DFG EHI
ABC DFH EGI
ABC DFI EGH
ABC DGH EFI
ABC DGI EFH
ABC DHI EFG
ABC EFG DHI
ABC EFH DGI
ABC EFI DGH
ABC EGH DFI
ABC EGI DFH
ABC EHI DFG
ABC FGH DEI
ABC FGI DEH
ABC FHI DEG
ABC GHI DEF
ABD CEF GHI
ABD CEG FHI
ABD CEH FGI
ABD CEI FGH
ABD CFG EHI
ABD CFH EGI
ABD CFI EGH
ABD CGH EFI
ABD CGI EFH
ABD CHI EFG
ABD EFG CHI
..... (Omitted below, 560 lines in total).

First select A and sort the combination of 3 numbers

public class Main {
	public static String remain(int[] a)
	{
		String s = "";
		for(int i=0; i<a.length; i++){
			if(a[i] == 0) s += (char)(i+'A');
		}	
		return s;
	}
	
	public static void f(String s, int[] a)
	{
		for(int i=0; i<a.length; i++){
			if(a[i]==1) continue;
			a[i] = 1;
			for(int j=i+1; j<a.length; j++){
				if(a[j]==1) continue;
				a[j]=1;
				for(int k=j+1; k<a.length; k++){
					if(a[k]==1) continue;
					a[k]=1;
					System.out.println(s+" "+(char)('A'+i)+(char)('A'+j)+(char)('A'+k)+" "+remain(a) ); //Fill in the blank position
					a[k]=0;
				}
				a[j]=0;
			}
			a[i] = 0;
		}
	}

	public static void main(String[] args)
	{
		int[] a = new int[9];		
		a[0] = 1;
	
		for(int b=1; b<a.length; b++){
			a[b] = 1;
			for(int c=b+1; c<a.length; c++){
				a[c] = 1;
				String s = "A" + (char)(b+'A') + (char)(c+'A');
				f(s,a);
				a[c] = 0;
			}
			a[b] = 0;
		}
		
	}

}

Read the code carefully and fill in the missing content in the underlined part.
Note: Do not fill in any existing content or explanatory text.

5. Draw lots
What about different combinations of fewer countries?
The following procedure solves this problem.
The array a[] is the largest number of places that each country can send.
The results of the program execution are:
DEFFF
CEFFF
CDFFF
CDEFF
CCFFF
CCEFF
CCDFF
CCDEF
BEFFF
BDFFF
BDEFF
BCFFF
BCEFF
BCDFF
BCDEF
....
(Omitted below, a total of 101 lines)

Nested loop

public class Main {
	public static void f(int[] a, int k, int n, String s)
	{
		if(k==a.length){ 
			if(n==0) System.out.println(s);
			return;
		}
		
		String s2 = s;
		for(int i=0; i<=a[k]; i++){
			 f(a,k+1,5-s2.length(),s2); //fill in the blank position
			s2 += (char)(k+'A');
		}
	}
	
	public static void main(String[] args)
	{
		int[] a = {4,2,2,1,1,3};
		
		f(a,0,5,"");
	}

}

Read the code carefully and fill in the missing content in the underlined part.

Note: Do not fill in any existing content or explanatory text.

6.