The algorithm check if the number is a palindrome number or not?

Answer 1

step 1 : input n

step 2 : s = 0, a=n

step 3 : while(n>0)

begin

rem=n%10

s=s*10+rem

n=n/10

end

step 4 : if(s==a)

print 'it is a palindrome'

else

print 'it is not a palindrome'

step 5 : stop

Answer 2

LET number be any positive integer.

SET temp to number.

SET reverse to zero.

WHILE temp is not zero...

{

SET digit to temp % 10 (mask LSD using modulo operator)

SET temp to integer (temp / 10) (remove LSD)

SET reverse to (reverse * 10) plus digit

}

END WHILE

IF reverse equals number

{

RETURN true (number is palindrome)

}

ELSE

{

RETURN false (number is not palindrome)

}

Answer 3

This is a very small way to do it in java by using strings instead of int.

You can always parse it back to int if you want.

String i = "121";

String j = new StringBuffer(i).reverse().toString();

System.out.println(i.equals(j));

Answer 4

Write it backwards, and compare with the original number. If they are equal, the number is a palindrome.

Write it backwards, and compare with the original number. If they are equal, the number is a palindrome.

Write it backwards, and compare with the original number. If they are equal, the number is a palindrome.

Write it backwards, and compare with the original number. If they are equal, the number is a palindrome.

Answer 5

int rem,sum=0,number;

sf("%d", &number);

while(number>0)

{

rem=number%10;

number=number/10;

sum=sum+rem;

}

if(sum==number)

pf("palindrom");

else

pf("not palindrome");

Answer 6

A number is a palindrome if the reverse of the number is the same as the original. The number must be a positive integer.

To reverse the digits in a number we start with an accumulator initialised to zero. The accumulator will eventually hold the reversed number.

While the number is greater than zero, we perform the following operations:

1. Multiply the accumulator by 10.
2. Add the modus of the number and 10 (the remainder after dividing by 10).
3. Divide the number by 10.

When the number is reduced to zero, the accumulator will hold the reverse of the original number.

To implement this algorithm we require two functions, one to reverse a number and the other to test if a number is a palindrome:

unsigned reverse (unsigned n) {

unsigned accumulator;

accumulator = 0;

while (n>0) {

accumulator = accumulator * 10;

accumulator = accumulator + (n%10);

n = n / 10;

}

return accumulator;

}

bool is_palindrome (unsigned n) {

return n==reverse(n);

}

To test the functions, let's loop through a sequence of numbers:

int main (void) {

unsigned n;

for (n=0; n<1000; ++n) {

if (is_palindrome (n))

printf ("%d is a palindrome\n", n);

else

printf ("%d is not a palindrome\n", n);

return 0;

}

Answer 7

This can be found in your textbook. You may also find it useful to do an image search to see what the flow chart looks like.

Answer 8

Write it backwards, and compare with the original number. If they are equal, the number is a palindrome.