Solution to Sudoku no 733

Sudoku no 733 (Easy)

Our younger son created this puzzle using a Javascript program which he wrote himself:

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Sudoku grader and solver 

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Solution to Maths Problem no 5

To solve this problem, I looked up pi to several decimal places then worked it out by trial and error with the PL/SQL code below:

SQL> @pi

SQL> set serveroutput on

SQL> declare

 2  top number;

 3  bottom number;

 4  pi constant number := 3.141592654;

 5  ratio number;

 6  difference number;

 7  save_difference number := 1;

 8  save_top number;

 9  save_bottom number;

 10  save_ratio number;

 11 begin

 12  for top in 1..999 loop

 13   for bottom in 1..999 loop

 14    ratio := top/bottom;

 15    difference := abs(pi-ratio);

 16    if difference < save_difference then

 17     save_difference := difference;

 18     save_top := top;

 19     save_bottom := bottom;

 20     save_ratio := ratio;

 21    end if;

 22   end loop;

 23  end loop;

 24  dbms_output.put_line('Top: '||save_top);

 25  dbms_output.put_line('Bottom: '||save_bottom);

 26  dbms_output.put_line('Ratio: '||save_ratio);

 27 end;

 28 /

Top: 355

Bottom: 113

Ratio: 3.14159292035398230088495575221238938053 

PL/SQL procedure successfully completed. 

SQL>  

So there you have it. The best fractional approximation to pi with a 3 digit numerator and denominator is 355/113, which is accurate to 6 decimal places i.e. 3.141593.

Maths Problem no 5

3.14 and 22/7 are both approximations for pi (or π).

22/7 is a fraction with a 2 digit numerator and a 1 digit denominator.

Find the best fractional approximation for pi with a 3 digit numerator and denominator.

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Solution to Sudoku no 732

Sudoku no 732 (Extreme)

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Solution to Maths Problem no 4

The answer is 548834.

I used the program below to work it out:

andrew@ASUS-Laptop:~/Maths_Problems$ cat problem4e.c
#include <math.h>
#include <stdio.h>
int main()
{
double a;
double b;
double c;
double d;
double e;
double f;
int abcdef;
for(a=1;a<=9;a++)
  {
  for(b=0;b<=9;b++)
    {
    for(c=0;c<=9;c++)
      {
      for(d=0;d<=9;d++)
        {
        for(e=0;e<=9;e++)
          {
          for(f=0;f<=9;f++)
            {
            abcdef=100000*a+10000*b+1000*c+100*d+10*e+f;
            if (abcdef==pow(a,6)+pow(b,6)+pow(c,6)+pow(d,6)+pow(e,6)+pow(f,6))
              printf("%d\n",abcdef);
            }
          }
        }
      }
    }
  }

return 0;
}
andrew@ASUS-Laptop:~/Maths_Problems$

I ran it as follows:

andrew@ASUS-Laptop:~/Maths_Problems$ ./problem4e
548834
andrew@ASUS-Laptop:~/Maths_Problems$ 

...then I checked the result:

andrew@ASUS-Laptop:~/Maths_Problems$ echo "5^6+4^6+8^6+8^6+3^6+4^6"|bc
548834
andrew@ASUS-Laptop:~/Maths_Problems$

Maths Problem no 4

Find a six digit positive integer abcdef where:

abcdef = a^6 + b^6 + c^6 + d^6 + e^6 + f^6.

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Solution to Sudoku no 731

Sudoku no 731 (Diabolical)

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Solution to Maths Problem no 3

I worked this out using the program below. The solution is at the bottom of the page:

andrew@ASUS-Laptop:~/Maths_Problems$ cat problem1.c
#include <math.h>
#include <stdio.h>
int main ()
{
int abc;
int def;
double abcdef;
double square_root;
int solution;
for(abc=100; abc<=998; abc++)
  {
  def=abc+1;
  abcdef=(1000*abc)+def;
  square_root=trunc(pow(abcdef,0.5));
  if(pow(square_root, 2)==abcdef)
    {
    solution=abcdef;
    printf("%d\n",solution);
    }
  }
return 0;
}
andrew@ASUS-Laptop:~/Maths_Problems$ ./problem1
183184
328329
528529
715716
andrew@ASUS-Laptop:~/Maths_Problems$

Maths Problem no 3

Find all 6 digit integers ABCDEF such that:

(1) ABCDEF is a perfect square.

(2) DEF = ABC + 1.

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Solution to Sudoku no 730

Sudoku no 730 (Tough)

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