human
3 | Function that returns true if n is an Osiris number.
3 rd digit.
2 nd digit.
1 st digit.
Check the required condition.
Driver code"
Solution in Python:
assistant
#Here is code to 3 and the other steps required.
 def isOsiris ( n ) :
	a = n % 10
	b = ( n // 10 ) % 10
	c = n // 100
	digit_sum = a + b + c
	if ( n == ( 2 * ( digit_sum ) * 11 ) ) :
		return True
	return False
if __name__ == ' _ _ main _ _ ' :
	n = 132
	if isOsiris ( n ) :
		print ( " Yes " )
	else :
		print ( " No " )