Дано 4 целых числа
найти произведение чисел, кратных 5. через программу паскаль ​

Const e=0.001;
var x,xn,h,an,s,f,y:real; n,k,i:integer;
begin
n:=10;
h:=(1-0.1)/(n-1);
x:=0.1-h;
for i:=1 to n do
 begin
 x:=x+h; an:=1;
 xn:=1; f:=1;
 k:=0;  s:=1;
 while an>e do
  begin
  xn:=xn*x*x;
  k:=k+1;
  f:=f*k;
  an:=(2*k+1)*xn/f;
  s:=s+an;
  end;
 y:=(1+2*x*x)*exp(x*x);
 writeln('x = ',x:4:1,'  s = ',s:6:4,'  y = ',y:6:4);
 end;
end.

Результат:
x = 0.1  s = 1.0303  y = 1.0303
x = 0.2  s = 1.1241  y = 1.1241
x = 0.3  s = 1.2911  y = 1.2911
x = 0.4  s = 1.5490  y = 1.5490
x = 0.5  s = 1.9260  y = 1.9260
x = 0.6  s = 2.4653  y = 2.4653
x = 0.7  s = 3.2320  y = 3.2320
x = 0.8  s = 4.3240  y = 4.3240
x = 0.9  s = 5.8894  y = 5.8895
x = 1.0  s = 8.1548  y = 8.1548

#include <stdio.h>#include <stdlib.h>#include <conio.h>
/* main program fucntion */void main(){  int     iMatrSize, // size of the matrix    iSum = 0,  // sum of the nessesary elements of the matrix    iCnt = 0,  // number of the nessesary elements of the matrix    **aMatr;   // the matrix  int i, j;
  scanf_s("%i", &iMatrSize);
  /* allocation memory for the array */  aMatr = (int**)malloc(sizeof(int) * iMatrSize);  for (i = 0; i < iMatrSize; i++)  {    aMatr[i] = (int*)malloc(sizeof(int) * iMatrSize);  }
  /* filling in the array */  for (i = 0; i < iMatrSize; i++)    for (j = 0; j < iMatrSize; j++)      aMatr[i][j] = rand() % 21 - 10;
  /* counting the sum of the elements */  for (i = 0; i < iMatrSize; i++)    for (j = 0; j < iMatrSize - i - 1; j++)      iSum += aMatr[i][j], iCnt++;
  /* outputing the array */  for (i = 0; i < iMatrSize; i++)  {    for (j = 0; j < iMatrSize; j++)      printf ("%3i ", aMatr[i][j]);    printf("\n");  }
  printf("Sum = %f\n", (float)iSum / iCnt);
  _getch();} /* End of 'main' function */