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        riddles >> cs >> Divide by 3 - itoa()
        
(Message started by: kens on Jul 23^rd, 2007, 8:33pm)

Title: Divide by 3 - itoa()
Post by kens on Jul 23^rd, 2007, 8:33pm

Without using /,% and * operators. write a function to divide a number by 3.
char *itoa(int) function is available.

Please tell me how to do this problem.

Title: Re: Divide by 3 - itoa()
Post by Sameer on Jul 23^rd, 2007, 8:52pm

You should really check before you post and the solution was just few posts away!!

http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs;action=display;num=1185094398

Title: Re: Divide by 3 - itoa()
Post by vinay_sahu on Aug 18^th, 2012, 8:07pm

hey,,it's too easy man.... ;D
# include<stdio.h>
# include<conio.h>
# include<stdlib.h>
# include<string.h>
main()
{
int i,j=0,len;
char buffer[30],ch;
printf("enter the no that you want to chek ");
scanf("%d",&i);
itoa(i,buffer,3);
len=strlen(buffer);
if(buffer[len-1]=='0')
printf("number is divisible by 3");
else
printf("number is not divisible by 3 ");
getch();
}