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   Divide by 3 - itoa()
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   Author  Topic: Divide by 3 - itoa()  (Read 5525 times)
kens
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Divide by 3 - itoa()  
« on: Jul 23rd, 2007, 8:33pm »
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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.
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Sameer
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Pie = pi * e

   


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Re: Divide by 3 - itoa()  
« Reply #1 on: Jul 23rd, 2007, 8:52pm »
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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
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple

Proof is an idol before which the mathematician tortures himself.
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vinay_sahu
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Re: Divide by 3 - itoa()   div3.c
« Reply #2 on: Aug 18th, 2012, 8:07pm »
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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();
}
 
« Last Edit: Aug 18th, 2012, 8:12pm by vinay_sahu » IP Logged
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