Mathematical Solutions: Multiplying by 9

We've always wanted a faster way to solve math problems. That's why we always have our handy dandy calculator. But what if you don't? You wouldn't be always carrying your calculating device everywhere, would you?

If you want to answer math problems in a snap, here's a useful math shortcut that you might wanna try before you become too dependent on calculators.. (To the point that you'd actually type 1 + 1 when the answer is already obvious).


Multiplying a number by 9.

169 X 9 is equal to? Something. You stare at the numbers with annoyance. But seriously, try and calculate it by yourself. You can write it down on a paper or mentally solve it. NO CALCULATORS!

Done? Okay. Here's the trick. You can actually solve this by using the number 10. How?
Watch:

169 X 10 = 1690
1690 - 169 = 1521

Try checking if it has the same answer as 169 x 9. It's the same, isn't it? So how did it happen?

Remember that 10 - 1 is 9.
So, basically: 169 x 9 == 169 x (10 - 1)
And we all know dividing any number to 10 is easy.

Well, it's faster if you like subtraction and subtraction is much easier than multiplication, right?


There'll be more Mathematical Solutions so stay tuned :3

A Dilemma of a Programming Student: Midterms Week part 1

Choosing a programming course because you like programming doesn't always mean fun.. Because most of the subjects you'll encounter will not even involve programming..

And so here I am, facing my PC trying to understand stuff that doesn't interest me..

But I guess they'll come handy sometime.. I mean.. I'm sure by solving the series of unfortunate events of my life using Taylor's series, I can be enlightened and be able to appreciate life..

I'm not kidding.. Life, like numbers and other mathematical entities, has a pattern :P

Well, I guess that's it for today. I have to go and become a nerd who studies math stuff D:

P.S.
I'm gonna fail.. I think D:

Programming Challenge no. 1: Sum of Multiples of 3 and 5

So I stumbled upon this site called Project Euler (It was named after Leonhard Euler who had a huge contribution in discovering infinitesimal calculus and graph theory.. Not sure if I'm thankful about his discovery.. But I guess I should since it helped improve a lot of things..)

What am I doing in Project Euler? Well, you see.. I promised to answer some programming challenges for you and as I googled the words "programming challenges", I saw this site. It's not bad actually.. I kinda like it :3

/* Warning: Must have at least basic programming experience */

So let's start with their very first challenge:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.

You can go ahead and get your calculator. I'm not stopping you. However, I must warn you that you're gonna waste at least 15 minutes of your life pressing the multiples of 3 and 5.. and with a huge possibility of committing a mistake..

If I were you, I'd find an algorithm that will give me the final result. One way is solving it the programming style :P


For this challenge, I'll use C as my programming language.
I'll use Turbo C as my compiler. (This doesn't matter actually)
And my left brain to solve the problem! ('Cause for some reason, my right brain hates numbers..)

Analyzing the problem

We are asked to compute the sum of the multiples of 3 and 5. 

In their example, the multiples of 3 and 5 below 10 are given: 3, 5, 6 and 9. Adding these multiples will give us 23. We need to do the same thing. Only this time, we need to find the multiples below 1000.

It's tempting to list down all the multiples then add them together.. but that would be an inefficient way to solve this challenge..

What we need to do is analyze and take note what we need to solve the problem:

1. Only multiples of 3 and 5 are allowed.
2. Numbers like 15, 30.. etc. are multiples of 3 and at the same time, of 5.
3. Multiples should be below 1000.

So basically we need to add numbers below 1000 that are either multiples of 3 or 5 or both..

Creating the Pseudo code

Pseudo codes are useful in creating programs. Since they use simple statements instead of a specific programming language's syntax, they're very easy to read.

So here's the pseudo code for our solution:

sum as unsigned long    // We create a variable sum that holds unsigned long integers 

num as integer             // We create a counter num that will go thru 1 to 999

While num is less than 1000     // Set a condition that will prevent the loop from going beyond 1000

          if num is divisible by 3 or 5     // Check if number is a multiple of 3 or 5

             add num to sum                  // If it is, add it to sum

          add 1 to num                          // Iterate num

print sum                                          // Show output on screen

What will happen is, the computer will go through all the numbers from 0 to 999. Each will be checked if it's divisible by 3 or 5. If the number is indeed a multiple of 3 or 5, it'll be added to the sum.

Now that wasn't so hard right? Now, let's go to the exciting part: coding..

Coding

I'll just show you directly what the code is and explain each part.

void main(){
     unsigned long sum;
     int num;

This part contains the declaration statements. Here, we declared two variables: sum and num. 

sum shall contain the total value of the multiples. It's unsigned long since we expect a large number. num will be iterated from 1 to 999 so it's basically like a counter.

for(sum = 0, num = 0; num < 1000; i++){
      if((num % 3 == 0) || (num % 5 == 0)){
             sum = sum + num;
     }
}

This loop searches thru the numbers below 1000 and finds the multiples of 3. The condition uses || known as the symbol for OR because we need all the multiples of 3 and 5. Using && (known as AND) will give us only the multiples of both 3 and 5.

printf("%lu", sum);

Note:  The sum should be long since we're dealing with a huge number.

Conclusion

So? Did you understand how to solve this programming problem? If you're still having trouble comprehending the method, don't hesitate to comment.

By the way, if you want a shorter code, here's another version of it..

void main(){
     unsigned long sum;
     int num;
     for(sum = 0, num = 0; num < 1000; sum = sum + ((num % 3 == 0 || num % 5 == 0) ?           num : 0), num++);
     printf("%lu", sum);
}

The italic part of the code is using ternary operation. I'll talk about that next time. I don't want to overwhelm you.

You might be asking: So what's the answer? That would be known only if you program the code yourself :P
I don't wanna spoon-feed you with answers to the Project Euler's challenges. My goal here is to relay what I can about programming. Not become a source of direct answers from a site that challenges you to solve programming problems.

If you want your answers checked or you're having troubles with the code, you can email me directly.

I hope you learned something :)