Hey!!
Was just wondering if its possible to do a simple C program displays all the numbers divisible by 6 in the range of 1 to 40 in rows of 3 columns...its kinda complicated to me..can anybody help me with a hint?? or explian??
thankies
Hey!!
Was just wondering if its possible to do a simple C program displays all the numbers divisible by 6 in the range of 1 to 40 in rows of 3 columns...its kinda complicated to me..can anybody help me with a hint?? or explian??
thankies
First, write a program that finds and "displays all the numbers divisible by 6 in the range of 1 to 40". Then think about the displaying in rows of 3 columns problem.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
The % operator is useful for both parts of the problem as laserlight has outlined it . . . .
dwk
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Thanks for who tried to help..
I did this so far...
Code:#include <stdio.h> void main() { int x=0; for(x=3;x<10; x+=3) printf("%i\t" ,x); printf("\n") ; for(x=12;x<19; x+=3) printf("%i\t" ,x); printf("\n") ; for(x=21;x<28; x+=3) printf("%i\t" ,x); printf("\n") ; for(x=30;x<37; x+=3) printf("%i\t" ,x); printf("\n") ; for(x=39;x<40; x+=3) printf("%i", x); }
But,I need the program to calculate it..can any body figure my mistake which I on't get....
thankies again
Well, that's one way to overly complicate something simple.
Like I said, the % operator is useful in two cases here. You can use it to check if a number is divisible by six.
You can also use it to space out your columns. Think of it this way: when the number of numbers you have printed is a multiple of 3, print a newline.Code:if(n % 6 == 0) { /* it's divisible by 6 */ }
dwk
Seek and ye shall find. quaere et invenies.
"Simplicity does not precede complexity, but follows it." -- Alan Perlis
"Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
"The only real mistake is the one from which we learn nothing." -- John Powell
Other boards: DaniWeb, TPS
Unofficial Wiki FAQ: cpwiki.sf.net
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Projects: codeform, xuni, atlantis, nort, etc.
Thanks! You helped me in the first part : D
Yeah, because I gave you the exact code required.
dwk
Seek and ye shall find. quaere et invenies.
"Simplicity does not precede complexity, but follows it." -- Alan Perlis
"Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
"The only real mistake is the one from which we learn nothing." -- John Powell
Other boards: DaniWeb, TPS
Unofficial Wiki FAQ: cpwiki.sf.net
My website: http://dwks.theprogrammingsite.com/
Projects: codeform, xuni, atlantis, nort, etc.
I think I have done the code...but,with some goofs lol :P
Code:#include <stdio.h> int main() { int x,count=0; for(count>=1 && count>=40;) { if(x %3==0) printf("%i/n",x); } }
If x is a number which is divisible by 6, then doing x = x + 6 also results in a number divisible by 6.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Code:#include <stdio.h> int main(void) { int i; int j; j = 0; for (i = 1; i <= 40; i++) { if (j < 3) { if (i % 6 == 0) { printf("\t%d\t", i); j++; } } else { if ( j == 3) { j = 0; printf("\n"); } } } return (0); }
broli86, read our homework policy.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Ok , I appologize.
Your solution skips over numbers where j happens to be 3.
And anyway, you're just overcomplicating as well. You can do it much more simply. Since we're giving away solutions . . . .
(Most of the time I wouldn't give away something like this, but the other program works too and has been on the internet for a long time already -- I'm sure the OP has seen it.)Code:#include <stdio.h> int main() { int num, count = 0; for(num = 1; num <= 40; num ++) { if(num % 6 == 0) { printf("%3d", num); if(++count % 3 == 0) putchar('\n'); } } return 0; }
dwk
Seek and ye shall find. quaere et invenies.
"Simplicity does not precede complexity, but follows it." -- Alan Perlis
"Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
"The only real mistake is the one from which we learn nothing." -- John Powell
Other boards: DaniWeb, TPS
Unofficial Wiki FAQ: cpwiki.sf.net
My website: http://dwks.theprogrammingsite.com/
Projects: codeform, xuni, atlantis, nort, etc.
I think we're skipping the most obvious solution. Salem hinted at it.