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I saw this Khan Academy video on the visualization of Euclid's algorithm. The problem was to find the HCF of $32$ and $12$.

At $6:53$, the instructor reduced the original problem to $12$ and $8$ as: $$ \left\{ \begin{array}{c} 32 = 12 + 12 + 8 \\ 12 \end{array} \right. $$

Why did he leave the numbers $9, 10$, and $11$ and directly jumped to $8$? How can I be sure that the numbers won't be a factor of $32$ and $12$?

Alessio K
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Niraj Raut
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  • Please consider reading this https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question – HelloWorld Sep 03 '20 at 06:12
  • @Dimitris Was this question on the wrong site? Should this be on metastack? – Niraj Raut Sep 03 '20 at 06:15
  • I think that you should expand you answer more here, instead of linking to an external source. You could write down the Khan Academy's example using mathjax (https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) so that it is easier for the readers to understand your problem and thus answer your question. More than that, this is the correct subdomain for this question – HelloWorld Sep 03 '20 at 06:20

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He tried to express $32$ as a sum of $12$'s. This is why he "skipped" $9,10,11$. In addition to that note than none of $9,10,11$ divide 32. Hence, he reduced the problem to finding th HCF of 8 and 12, because it would work for 12 and 32 as well, since 12 is a divisor of 32. So In fact, he didnt "jump" to 8, he just divided 32 with 12, and 8 was the remainder.

Indeed he writes $32 = 12 + 12 + 8$.

He says "if you found a break that can cover 12 and the you found another break for 8 with the condition it covers 12 too (5:45)". And that is of course, 4.

HelloWorld
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His argument is that if a whole number of bricks fits inside of $12$, then a whole number of bricks also fits inside of $12+12$. Is it clear to you why this is true?

Now if a whole number of the same kind of brick fits inside $32$, then we can remove the whole number of bricks that fit inside $12+12$ from the bricks that fit inside $32$, and the number of bricks that fit inside of the remainder, $8$, will also be a whole number.

So whatever the size of the brick is, if a whole number of that brick fits inside both $12$ and $32$, then a whole number of that brick will fit inside $8$.

If you follow this argument, there is no need to rule out $11$, $10$, and $9$. You ask how you can be sure none of these numbers will be a factor of both $32$ and $12$, but that is the wrong question. After all, $8$ is not a factor of both $32$ and $12$ either. So being a factor of both $32$ and $12$ is not the reason for considering $8$. To repeat, the reason for considering $8$ is that any brick that fits inside of both $12$ and $32$ also fits inside of $8$.

Will Orrick
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  • How do we know that $8$ is the longest brick? – Niraj Raut Sep 03 '20 at 07:18
  • We don't. In fact, bricks of length $8$ do not work. Later in the argument, the longest brick that works ends up being $4$, of course. At this stage of the argument one thing we can say is that, whatever length ends up working, it can't be bigger than $8$. The reason for this is that the brick fits a whole number of times into $24$ and a whole number of times into $32$. So it fits a whole number of times into $32-24=8$. Obviously a brick of length greater than $8$ is not going to fit a whole number of times into $8$. – Will Orrick Sep 03 '20 at 07:34
  • Here's a different visualization that might help you. Draw a $12\times32$ rectangle on graph paper. Remove a $12\times 12$ square from the rectangle, leaving a $12\times20$ rectangle. Cut off another $12\times12$ square to get a $12\times 8$ rectangle. Now cut off an $8\times8$ square, leaving a $4\times8$ rectangle. Finally, cut off a $4\times4$ square, leaving a $4\times4$ square. The visuals at my answer here may help. – Will Orrick Sep 03 '20 at 08:09
  • I got a video. It's the same as your answer. My only concern is the reasoning which you gave that it cannot be bigger than $8$. I didn't understand that reasoning. I feel that I am just overlooking the obvious. – Niraj Raut Sep 03 '20 at 08:26
  • After thinking relentlessly for 2 hours, I finally got it. Thanks :-) – Niraj Raut Sep 03 '20 at 09:52
  • Hey, just one question: In this video, why does the maximum square tile give us the HCF? – Niraj Raut Sep 03 '20 at 12:21
  • Take the example of $32$ and $12$. and form a rectangle with sides of these lengths. Starting at a corner, mark off points spaced $4$ units apart all along the perimeter of this rectangle. Form a square grid by joining corresponding points on opposite sides by line segments. The first image in the post I linked to in an earlier comment should help visualize this. The conclusion is that the rectangle can be tiled with squares whose side is the HCF of the sides of the rectangle. – Will Orrick Sep 03 '20 at 15:09
  • "The conclusion is that the rectangle can be tiled with squares whose side is the HCF of the sides of the rectangle." — Why does HCF give the largest square tile that can fit the rectangle? I didn't understand it. – Niraj Raut Sep 03 '20 at 15:11
  • Let us continue this discussion in chat. – Will Orrick Sep 03 '20 at 15:14
  • I wasn't sure whether you would see this in chat, so I'm adding it here. In chat I said that there is a one-to-one correspondence between common divisors and tiling squares, but I neglected to emphasize that the term "divisors" here includes numbers that are not integers. For example, $3.5$ is a common divisor of $14$ and $21$. The one-to-one correspondence only holds if we allow non-integer divisors. When the rectangle sides are integers, however, you can see that all of the smaller rectangles the procedure produces also have integer sides, and the HCF will be an integer, as expected. – Will Orrick Sep 03 '20 at 17:57
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If you do some divisibility checks, you can see that because $3+2 = 5$ and $5 \nmid 9$. Additionally, neither of the numbers end in zero, or consist of the same digit (note this check only works for testing divisibility by eleven for numbers less than 99). I assume that's what's happened here.

simeondermaats
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