Multiples of three P61930


Statement
 

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thehtml

A well-kown mathematical property states that a natural number is a multiple of three if and only if the sum of its digits is also a multiple of three. For instance, the sum of the digits of 8472 is 8 + 4 + 7 + 2 = 21, which is a multiple of three. Therefore, 8472 is also a multiple of three.

Implement a recursive function that tells if a strictly positive natural number n is a multiple of three or not.

bool is_multiple_3(int n);

Interface

C++
bool is_multiple_3(int n);
C
int is_multiple_3(int n);
Java
public static boolean isMultiple3(int n);
Python
is_multiple_3(n) # returns bool
 
is_multiple_3(n: int) -> bool

Solve this problem using a recursive function to return the sum of the digits of a natural number n.

int sum_of_digits(int n);

Interface

C++
int sum_of_digits(int n);
C
int sum_of_digits(int n);
Java
public static int sumOfDigits(int n);
Python
sum_of_digits(n) # returns int
 
sum_of_digits(n: int) -> int

Observation

Here, you are allowed to use the operations of division and integer remainder only with the number 10. Otherwise, this exercise would be totally trivial!

Public test cases

  • Input/Output

    is_multiple_3(8472) → true
    is_multiple_3(8473) → false
    is_multiple_3(4) → false
    • Information
    Author
    Salvador Roura
    Language
    English
    Translator
    Carlos Molina
    Original language
    Catalan
    Other languages
    Catalan
    Official solutions
    C C++ Java Python
    User solutions
    C C++ Java Python