Sum of digits in recursion

Question

Super-sum S of an integer x is defined as x if x is single digit number otherwise Super-sum of x is defined as Super-sum of digit-sum of x. Given two numbers n,k find the Super-sum of the number formed when n is concatenated k times.Note that k is only multiplied with the number when it is at least a 2 digit number

Input: 1987 4

Output: 1

Is there a faster method than this?

s,k=input().split()
summ=0
for ele in s:
    summ+=int(ele)
s=summ*int(k)

while s>=10:
    s=str(s)
    summ=0 
    for ele in s:
        summ+=int(ele)
    s=summ
print(s)

Demonking28 · Accepted Answer · 2017-11-12T13:16:16.287

n,k=map(int,input().split())
if n<10:
    print(n)
else:
    if ((n*k)%9==0):
        print(9)
    else:
        res=(n*k)%9

Any number greater than 9 will have digits repeated that's why you need to take mod of 9 for example 13 will have sum of 1+3 =4 and 13 %9=4 .There will be a special case when mod of 9 will be zero and this will be at number 9,18,27,36 etc which are divisible by 9 and their sum will always be 9 hence return 9 .

score 0 · Answer 2 · answered Sep 18 '21 at 20:47

The formula used in the accepted answer should actually be proved. So, let's do this.

First, let's prove that for any positive integer number N, N modulo 9 is the same sum of its digits modulo 9.

Having proved the statement above, we easily conclude that N % 9 provides the needed digit as per challenge description. The only exception is if N % 9 == 0, then a digit is 9.

Sum of digits in recursion

2 Answers2