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   Divide by 3 - itoa()

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Topic: Divide by 3 - itoa() (Read 5535 times)

kens
Junior Member

Posts: 59

Divide by 3 - itoa()
« on: Jul 23^rd, 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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Posts: 1261

Re: Divide by 3 - itoa()
« Reply #1 on: Jul 23^rd, 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
Newbie

Posts: 1

Re: Divide by 3 - itoa()

div3.c
« Reply #2 on: Aug 18^th, 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 18^th, 2012, 8:12pm by vinay_sahu »

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