Thread: flipping number question..

what is the math formula(algorithm) for flipping any number like:
47->74
1655 -> 5561

12345->54321

1190 -> 911

??

Originally Posted by itCbitC

Use a for loop and divide the number by 10 in each iteration till it's zero while accruing the remainder, which is printed out after the loop is finished. Just my 2c and others may have better ways.

Wow itCbitC

If you want an actual number that you can use, you could use something like this:

Code:

newnum = 0
for each digit in number:
    digit = number % 10
    newnum = newnum * 10 + digit

If all you want to do is reverse the data, you could do as itCbitC suggested, or, if you have a string representation of the number, you could just print the string in reverse.

solved it thanks