Use the modulo operator (%) to get the desired rightmost digits. Ex: The rightmost 2 digits of 572 is gotten by 572 % 100, which is 72. Hint: Use / to shift right by the desired amount. Ex: Shifting 572 right by 2 digits is done by 572 / 100, which yields 5. (Recall integer division discards the fraction). For simplicity, assume any part starts with a non-zero digit. So 0119998888 is not allowed.

Hi, you haven't provided the programing language in which you need the code, I'll just explain how to do it using Python, and you can apply a similar method for any programming language.

Answer:

def rightMost (num):

lenNum = len (str (num))

rightNum = num% (10* * (lenNum-1))

print (rightNum)

return (rightNum)

Explanation:

In this function we receive an integer number, then we find how many digits this number has, and finally, we find the rightmost digits; the main operation is modulo (takes the remainder after a division), what we want is to take all the digits except the first one, for that reason we find the modulo of the number when divided by ten to the power of the length of the number minus one, for example, if the number is 2734 we divided by 10^ (4-1), where four is the length of the number, this way we get 2734/1000 and the module of it is 734.