Finding a number with conditions while using for loop

Write a program in a script file that finds the smallest odd integer that is divisible by 13 and whose square root is greater than 120. Use a for-loop in the program. The loop should start from 1 and stop when the number is found. The program prints the message “The required number is:” and then prints the number. Hint: Use rem(n,2)~=0 for finding odd integers, and rem(n,13)==0 for finding the number to be divisible by 13.

I know this is easily solved through other methods but I don't know what to do.

I've rewritten my code many times and don't know what I am doing.

[David Hill]

Show us what you have done.

[Bryan Cordeiro]

clc

clear

n=120^2;

for i=1:n -------- IM NOT SURE WHAT NUMBER SHOULD BE IN PLACE OF THE N HERE

if rem(n,2)~=0 && rem(n,13)==0;

fprintf('The required number is: %d\n', n);

else

n=n+1;

end

end

but it doesn't end the loop... so I'm not sure if its the right way to do it

Keep in my I have very limited knowledge of matlab

[Bryan Cordeiro]

it prints The required number is: 14417 which seems to be the right answer however im not sure if the code is right due to it repeating the the line a bunch of times. How do I end the loop when it finds the number?

Answer 1

[Image Analyst]

You need a continue. And you should print i, NOT n. Then, after the fprintf() put a break

if sqrt(i) < 120
    continue; % Skip to end of loop and continue iterating when sqrt(i) < 120
end
fprintf('The required number is: %d.\n', i);
break; % Break out of loop.

And you could just have n be some huge number, like 100 billion or whatever.

[Bryan Cordeiro]

I tried implementing that but its not working. Could you show me the full code, if you were solving the problem?

[Bryan Cordeiro]

clear
clc
n=120^2;
for i=1:n
    if rem(n,2)~=0 && rem(n,13)==0 && sqrt(n)>120
        fprintf('The required number is: %d\n', n);
        break
    else
        n=n+1;
    end
end  

I did this and it worked, I got the right number and it displays the text once, so it works. But I just want to know if its the right way to do it

[David Hill]

The MATLAB way is with no loop as John discussed below, but your homework requires a for-loop. I would get rid of the else statement and just process the loop from 1:15000.

[Image Analyst]

No, it's not. Never change the loop iterator or it's starting and ending values in the for loop body. You won't get the expected results. For example

n = 9;
for k = 1 : n
	fprintf('Iteration %d, n = %d.\n', k, n);
	n=n+1;
end

gives

Iteration 1, n = 9.

Iteration 2, n = 10.

Iteration 3, n = 11.

Iteration 4, n = 12.

Iteration 5, n = 13.

Iteration 6, n = 14.

Iteration 7, n = 15.

Iteration 8, n = 16.

Iteration 9, n = 17.

You can see that you still do iterations 1 through 9 despite the fact that you unwisely changed the ending value inside the loop from 9 to (eventually) 17. You got 9 iterations, not 17.

Here is how I'd do it:

fprintf('Beginning to run %s.m ...\n', mfilename);
for k = 1 : 100000
	if sqrt(k) < 120 || rem(k, 13) ~= 0
		continue; % Skip to end of loop and continue iterating when sqrt(i) < 120
	end
        % If you get to here, you've found it!
	fprintf('The required number is: %d.\n', k);
	break; % Break out of loop.
end
fprintf('Done running %s.m.\n', mfilename);

[Bryan Cordeiro]

thanks alot for the comments explaining. I appreciate it!

Answer 2

[John D'Errico]

Why use a loop? ;-)

Lets see, the smallest INTEGER with the desired property is 14404. We can get that as:

n = 120;
>> m = n^2 + mod(-n^2,13)
m =
       14404
       
>> sqrt(m)
ans =
           120.01666550942
           
>> rem(m,13)
ans =
     0

But, then I see that you needed to find the smallest ODD integer, and since the smallest such integer is even, we need to find a solution yielding the smallest odd integer. This will work, but it is sort of a kludge:

n = 120;
m = n^2 + mod(-n^2,13);
if rem(m,2) == 0
  m = m + 13;
end
m
m =
       14417

Well, yes. This is a homework problem.

Hmm. How would I solve it using a loop? After all, you are making a credible effort.

The smallest number that satisfies the listed conditions MUST be one of the integers in the set: [120^2 + (0:(2*13-1))]. So we never need to loop over more than 26 elements beyond 120^2. THINK ABOUT IT! As such, I could set this up as a for loop, over 26 numbers, then breaking out of the loop when we find success.

Or, you could just use a while loop, which requires far less thought.

n = 120;
m = n^2;
while ~isequal(mod(m,[13,2]),[0 1])
  m = m + 1;
end
m =
       14417

You should see the isequal test reduces two tests into one vectorized test. As well, since we started out at 120^2, we absolutely know that sqrt(m) must be greater than 120.

Answer 3

[baiel kurstanbek]

Write a program in a script file that finds the smallest even integer that is divisible by a and by b whose square root is greater than c. Use a loop in the program. The loop should start from 1 and stop when the number is found.

Sample Input: 3

5

7

Sample Output: 60

[Image Analyst]

@baiel kurstanbek Read this link:

How do I get help on homework questions on MATLAB Answers? - MATLAB Answers - MATLAB Central

Hint: Look up the functions sqrt(), rem(), mod(), gcd(), break, for, and while. Start a new question (not part of @Bryan Cordeiro's question) with your attempt at code if you still have questions.

Usually divisible means that the ratio of the two numbers is an integer like

output = 60/3

output = 20

however it looks like you've either made a typo (like 60 should have been 70), or you're allowing absolutely anything since 60 is not an integer multiple of 7

output = 60/7

output = 8.5714

So in that case (allowing results with fractional parts), any number is divisible by any other number and you don't need to check for divisibility.