how to check whether the number ends with 9 or not in numbers 1to 100

Question

I want to print the numbers from 1 to100 in the format

1-------------9
10------------19
20-------------29
30--------------39
40-----------49
50-----------59
60-----------------69
70---------------79
80---------------89
90-----------------99

Answer 1

Just check whether the number modulus 10 equals 9:

if (number % 10 == 9) …

Answer 2

Check out the modulus operator.

Answer 3

if ( ( num % 10 ) == 9 )  
{  
  // I end in 9  
}

Answer 4

The modulus operator (%) will give you the remainder of a division. When you divide a number by 10, the "remainder" will give you the last digit.

int lastDigit = number % 10;           // this is the remainder of (number/10)
if (lastDigit == 9)
{
    // whatever you want to do here
}

Answer 5

if ( x == 0x09 || x == 0x13 || x == 0x1d || x == 0x27 || x == 0x31 || x == 0x3b || x == 0x45 || x == 0x4f || x == 0x59 || x == 0x63 )

or

if ( strchr( "\x09\x13\x1d\x27\x31\x3b\x45\x4f\x59\x63", x ) )

---

For those with no sense of the ridiculous, why are these more insane than x%10 == 9 ? in x%10==9 you have introduced two magic numbers rather than several, and turned a structured problem with no conditional behaviour (print rows, each row having columns) into a single loop with conditional behaviour.

Answer 6

int i = 0;
while (i < 100)
{
  i+=1; 
  while(i % 10 != 0)
  {
    Console.Write(i);
    i+=1;
  }
  Console.Write(Environment.NewLine);
}

Answer 7

for (i=1;i<=100;i++)
{
 if (i%10==0)
 {
  if (i==10)
  cout<<i-9<<"--------"<<i-1<<endl;
  else 
     cout<<i-10<<"--------"<<i-1<<endl;
    }
}

Answer 8

for (var i = 9; i < 100; i += 10)
    Console.WriteLine(((i==9)?1:i-9) + "----------" + i);

Should print pretty much exactly what you asked.

for (var i = 9; i < 100; i += 10)
    Console.WriteLine(i);

Will print just your result