Judge whether a number n can be divisible by 3 and 5 at the same time. An algorithm for expressing problems with traditional processes

Numbers divisible by 3, the sum of each number is a multiple of 3.

A number divisible by 5. The last digit is 0 or 5.

= => A number divisible by 3 and 5 at the same time. The sum of each number is a multiple of 3, and the last digit is 0 or 5.

Or more simply, it can be divisible by 15.

Algorithm:

1, to determine whether it is a multiple of 3.

-take it apart

Add up all the numbers and divide by 3.

-Remainder ==0?

Yes, go to the next step; No, the exit number

2. Judge whether the last digit is 0 or 5.

—— Is the last digit ==0 or ==5?

——————— Yes, exit is; No, the exit number

or

1, and judge whether it is a multiple of 15.

———n N is the remainder of 15.

-Remainder ==0?

——————— Yes, exit is; No, the exit number

Procedure:

# Including

empty

Master ()

{

Internationalorganizations (same as international organizations)

I，m，k= 10，s = 0；

Internationalorganizations (same as international organizations)

a[ 10]；

Printf ("Enter a positive integer:");

scanf("%d "，& ampm)；

a[0]= 0；

for(I = 1； I< 10; i++)

{ a[I]=(m-s)% k； k * = 10； s+= a[I]； }

If (s%3==0)

{if (a [1] = || a [1] = = 5) printf ("\ nm can be divisible by both 3 and 5 \ n");

other

Printf("\nm cannot be divisible by 3 and 5 at the same time \ n ");

}

other

Printf("\nm cannot be divisible by 3 and 5 at the same time \ n ");

}