human
Count of numbers satisfying m + sum ( m ) + sum ( sum ( m ) ) = N | function that returns sum of digits in a number.
initially sum of digits is 0.
loop runs till all digits have been extracted.
last digit from backside.
sums up the digits.
the number is reduced to the number removing the last digit.
returns the sum of digits in a number.
function to calculate the count of such occurrences.
counter to calculate the occurrences.
loop to traverse from n - 97 to n.
calls the function to calculate the sum of digits of i.
calls the function to calculate the sum of digits of a.
if the summation is equal to n then increase counter by 1.
returns the count.
driver program to test the above function.
calls the function to get the answer"
How can the above be solved in Python?
 Give me a code to count of numbers satisfying m + sum ( m ) + sum ( sum ( m ) ) = N and the other steps required?
assistant
#Here is code to count of numbers satisfying m + sum ( m ) + sum ( sum ( m ) ) = N and the other steps required.
 def sum ( n ) :
	rem = 0
	sum_of_digits = 0
	while ( n > 0 ) :
		rem = n % 10
		sum_of_digits += rem
		n = n // 10
	return sum_of_digits
def count ( n ) :
	c = 0
	for i in range ( n - 97 , n + 1 ) :
		a = sum ( i )
		b = sum ( a )
		if ( ( i + a + b ) == n ) :
			c += 1
	return c
n = 9939
print ( count ( n ) )