# plz send logic to write given program.

Discussion in 'C Programming' started by pitamber kumar, Mar 2, 2008.

1. ### pitamber kumarGuest

Write a program to find the number of and sum of all intergers greater
than 100 & less than 200 that are divisible by 7.

pitamber kumar, Mar 2, 2008

2. ### Sjouke BurryGuest

pitamber kumar wrote:
> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

We are not supposed to do your homework.
And dont think you will ever learn anything by cheating.

Sjouke Burry, Mar 2, 2008

3. ### Johannes BauerGuest

pitamber kumar schrieb:
> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

i=101; while [ \$i -lt 200 ]; do if [ \$((\$i%7)) -eq 0 ]; then echo
\$((\$i/7-14)): \$i; fi; i=\$((\$i+1)); done

You're welcome.

Kind regards,
Johannes

--
"PS: Ein Realname wÃ¤re nett. Ich selbst nutze nur keinen, weil mich die
meisten hier bereits mit Namen kennen." -- Markus Gronotte aka Makus /
Kosst Amojan / maqqusz / Mr. G / Ferdinand Simpson / Quartillia
Rosenberg in dse <45608268\$0\$5719\$-online.net>

Johannes Bauer, Mar 2, 2008
4. ### AmandilGuest

On Mar 2, 3:25 pm, pitamber kumar <> wrote:
> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

Ok, I wrote it. Now you write one yourself, post it, and I'll see how
they compare.

In general, do not post homework questions. If you want to hide the
fact that it's a homework problem, show us what you have done, and

Hint: Use a loop to run through the numbers from 100 to 200. Check
each number if it's divisible by 7: If it is, add it to a running
total.

-- Marty

Amandil, Mar 2, 2008
5. ### Keith ThompsonGuest

pitamber kumar <> writes:
> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

Done.

(No, we will not do your homework for you.)

--
Keith Thompson (The_Other_Keith) <>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

Keith Thompson, Mar 2, 2008
6. ### Kenny McCormackGuest

In article <>,
Keith Thompson <> wrote:
>pitamber kumar <> writes:
>> Write a program to find the number of and sum of all intergers greater
>> than 100 & less than 200 that are divisible by 7.

>
>Done.
>
>
>(No, we will not do your homework for you.)

Actually, you just did. You just haven't published your results.

Kenny McCormack, Mar 2, 2008
7. ### osmiumGuest

"pitamber kumar" writes:

> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

Look at the modulo operator. Perversely, this is represented by '%'.

osmium, Mar 2, 2008
8. ### Ioannis VranosGuest

pitamber kumar wrote:
> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

Why?

Ioannis Vranos, Mar 2, 2008
9. ### santoshGuest

pitamber kumar wrote:

> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

This is trivially easy. Show us your effort and you'll likely get good
help. But we don't do homework on demand.

santosh, Mar 2, 2008
10. ### Eric SosmanGuest

pitamber kumar wrote:
> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

Here you go. Invoke it exactly as described in the
"Usage" message, or it may misbehave.

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main(int l,char**s){static const short o[]={23,4,5,-2,-5,-11,14,
9,-5,1,-6,};int n,m;if(l!=2){fprintf(stderr,"Usage:\n\t%s \"Do your"
" own homework next time, vile cheater!\"\n",*s);return EXIT_FAILURE;
}for(m=n=puts(*++s)<0;m<(int)sizeof o;m+=sizeof(short)){n+=o[m/sizeof
(short)];putchar(toupper((unsigned char)(n[*s])));}puts("");return 0;}
--
Eric Sosman
lid

Eric Sosman, Mar 2, 2008
11. ### Johannes BauerGuest

Eric Sosman schrieb:

> Here you go. Invoke it exactly as described in the
> "Usage" message, or it may misbehave.

Humm, I did. It reports, after printing argv[Â¹]:

XIV, MMCVII

That's not quite the solution - am I misunderstanding something here?

Regards,
Johannes

--
"PS: Ein Realname wÃ¤re nett. Ich selbst nutze nur keinen, weil mich die
meisten hier bereits mit Namen kennen." -- Markus Gronotte aka Makus /
Kosst Amojan / maqqusz / Mr. G / Ferdinand Simpson / Quartillia
Rosenberg in dse <45608268\$0\$5719\$-online.net>

Johannes Bauer, Mar 2, 2008
12. ### Richard HeathfieldGuest

pitamber kumar said:

> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

We can simplify the program by making some elementary observations up
front.

The number of integers in the range 101 to 199 that are divisible by 7 is
the same as the number in the range 1 to 99, although possibly there is an
extra integer. As it happens, 98 is evenly divisible by 7, giving a
quotient of 14, so we deduce that there are fourteen integers in the range
1-99. The first integer in the range 101 to 199 that is divisible by 7 is
98+7=105, and the last is 196, so we haven't managed to squeeze another
one in there. (That doesn't happen till 301-399, in fact.)

Now we know there are fourteen terms in this arithmetic series. Clearly
they can be divided into seven pairs, each totalling 301, and there are
seven such pairs, giving 2107.

So our final C program is:

#include <stdio.h>
int main(void)
{
puts("(a) 14 (b) 2107");
return 0;
}

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
"Usenet is a strange place" - dmr 29 July 1999

Richard Heathfield, Mar 2, 2008
13. ### Eric SosmanGuest

Johannes Bauer wrote:
> Eric Sosman schrieb:
>
>> Here you go. Invoke it exactly as described in the
>> "Usage" message, or it may misbehave.

>
> Humm, I did. It reports, after printing argv[Â¹]:
>
> XIV, MMCVII
>
> That's not quite the solution - am I misunderstanding something here?

Well, I may have made a misteak -- but I don't think
I did. The output looks correct to me, if we assume the
O.P.'s "intergers" are integers. If they're not, then I'm
completely lost; I don't know how to deal with intergers.

--
Eric Sosman
lid

Eric Sosman, Mar 2, 2008
14. ### Martin AmbuhlGuest

pitamber kumar wrote:
> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

This is not a question about C. It doesn't even have anything to do
with programming. That makes if off-topic. However, I will provide you
<ot>
The first integer greater than 100 divisible by 7 is 105.
Every one of the integers 105+7k < 200 is divisible by 7.
for 105+7k < 200 to be true, 7k < 95, and k <= 13.
So there are 14 such numbers for k in {0, 1, ... 13).
This should not surprise you, since (200-100) = 100 = 14*7+2.
The sum of these numbers is
sum(105+7k) for k = 0, ..., 13
which is 14*105 + 7*sum(k) for k = 0, ..., 13
or 1470 + 7*13*14/2 = 2107

So here is the program

#include <stdio.h>
int main(void)
{
printf("There are 14 integers greater than 100 and less than 200\n"
"which are divisible by 7. Their sum is 2107.\n");
return 0;
}

That which is easy to do analytically rarely calls for a programming
solution.

</ot>

Martin Ambuhl, Mar 2, 2008
15. ### Johannes BauerGuest

Johannes Bauer schrieb:

> XIV, MMCVII
>
> That's not quite the solution - am I misunderstanding something here?

Yes I am! My program is broken. I didn't read the instructions throughly
enough. Oh well. :-/

Here it goes again:

q=0; z=0; i=101; while [ \$i -lt 200 ]; do if [ \$((\$i%7)) -eq 0 ]; then
z=\$((\$z+\$i)); q=\$((\$q+1)); fi; i=\$((\$i+1)); done; echo \$q \$z

Regards,
Johannes

--
"PS: Ein Realname wÃ¤re nett. Ich selbst nutze nur keinen, weil mich die
meisten hier bereits mit Namen kennen." -- Markus Gronotte aka Makus /
Kosst Amojan / maqqusz / Mr. G / Ferdinand Simpson / Quartillia
Rosenberg in dse <45608268\$0\$5719\$-online.net>

Johannes Bauer, Mar 2, 2008
16. ### Randy HowardGuest

On Sun, 2 Mar 2008 14:25:34 -0600, pitamber kumar wrote
(in article
<>):

> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

I got an error when I tried to send it to you.

It said:

Error: Recipient incompatible with logic. Send aborted.

--
"The power of accurate observation is called cynicism by those
who have not got it." - George Bernard Shaw

Randy Howard, Mar 2, 2008
17. ### BartcGuest

"Johannes Bauer" <> wrote in message
news:...
> Johannes Bauer schrieb:
>
>> XIV, MMCVII
>>
>> That's not quite the solution - am I misunderstanding something here?

>
> Yes I am! My program is broken. I didn't read the instructions throughly
> enough. Oh well. :-/
>
> Here it goes again:
>
> q=0; z=0; i=101; while [ \$i -lt 200 ]; do if [ \$((\$i%7)) -eq 0 ]; then
> z=\$((\$z+\$i)); q=\$((\$q+1)); fi; i=\$((\$i+1)); done; echo \$q \$z

What language is this in? It's obviously not C, but apparently not Perl
either.

I had to convert to the following (with the help of a 10-min crash course in
Perl) to get it to work:

\$q=0; \$z=0; \$i=101; while ( \$i < 200 ) { if ( (\$i%7) == 0 )
{\$z=(\$z+\$i); \$q=(\$q+1)}; \$i=(\$i+1); } print \$q," ",\$z

--
Bart

Bartc, Mar 2, 2008
18. ### Ian CollinsGuest

Bartc wrote:
> "Johannes Bauer" <> wrote in message
> news:...
>> Johannes Bauer schrieb:
>>
>>> XIV, MMCVII
>>>
>>> That's not quite the solution - am I misunderstanding something here?

>> Yes I am! My program is broken. I didn't read the instructions throughly
>> enough. Oh well. :-/
>>
>> Here it goes again:
>>
>> q=0; z=0; i=101; while [ \$i -lt 200 ]; do if [ \$((\$i%7)) -eq 0 ]; then
>> z=\$((\$z+\$i)); q=\$((\$q+1)); fi; i=\$((\$i+1)); done; echo \$q \$z

>
> What language is this in? It's obviously not C, but apparently not Perl
> either.
>

bash/ksh shell.

>
> --

Your sig is missing the trailing space after the "--"

--
Ian Collins.

Ian Collins, Mar 3, 2008
19. ### Kenny McCormackGuest

In article <j3Hyj.16868\$>,
Bartc <> wrote:
>
>"Johannes Bauer" <> wrote in message
>news:...
>> Johannes Bauer schrieb:
>>
>>> XIV, MMCVII
>>>
>>> That's not quite the solution - am I misunderstanding something here?

>>
>> Yes I am! My program is broken. I didn't read the instructions throughly
>> enough. Oh well. :-/
>>
>> Here it goes again:
>>
>> q=0; z=0; i=101; while [ \$i -lt 200 ]; do if [ \$((\$i%7)) -eq 0 ]; then
>> z=\$((\$z+\$i)); q=\$((\$q+1)); fi; i=\$((\$i+1)); done; echo \$q \$z

>
>What language is this in? It's obviously not C, but apparently not Perl
>either.

It is equally clearly not Fortran.

Can we draw up an exhaustive list of languages that it isn't?

Kenny McCormack, Mar 3, 2008
20. ### MisterEGuest

"pitamber kumar" <> wrote in message
news:...
> Write a program to find the number of and sum of all intergers greater
> than 100 & less than 200 that are divisible by 7.

int main(int argc, char **argv)
{
int i,j=0;
for (i=101;i<200;i++)
{
printf("Number %d\n",i);
if (i%7==0) j+=i;
}
printf("Sum %d\n",j);
}

MisterE, Mar 3, 2008