A large brick is built out of small bricks with size $\displaystyle 5\times 10\times 20$. Show that the same large brick can also be built using the same small bricks with all edges of equal lengths being parallel.
just a simple Q n
I am going to give you my answer as I imagined ..okay
= $\displaystyle 5\times 10\times 20$
= $\displaystyle 5\times (5\times2)\times (5\times2\times2)$
=$\displaystyle (5\times5\times5)\times(2\times2\times2)$
= $\displaystyle 5^3\times2^3$
=$\displaystyle (5\times2)^3$
= $\displaystyle 10^3$
so the length of a side of the new small brick is 10 .
Trying to explain: We have lots of $\displaystyle 5\times 10\times 20$ ("small") bricks. We have built a large brick using them. We have to show that it's possible to build the same large brick using the same smaller ones in a way that the edges of equal lengths are parallel. So if we show that the sides of the large brick are multiple of 5, 10 and 20 respectively then we're done.
I think you still don't get it. We don't know how "big" the large brick is and we don't have to know that and also it's impossible to find it as it can be as big as we want. We have infinitely many small bricks to build it. But the thing is that we get that large brick (I mean built using lots of small, $\displaystyle 5\times 10\times 20$ bricks.) And we have to prove something. We have to prove that the same large brick can also be built using the same small bricks with all edges of equal lengths being parallel. Is there somebody who understood it?
I think we do need to know how big the larger brick is .
I agree with you but there is no such information given .
We can assume the larger brick should be made using smaller bricks only without concerning its size .
but we can see the answer varies with the size of larger brick ,see
we can understand the two sides of the large brick is a multiple of 10 .(since smaller brick has sides of 10*20)
remaining side of the larger brick is at least a multiple of 5 .
since there is no information given about the larger brick we can select a smaller brick which has a side of 5 i.e $\displaystyle 5\times5\times5$ brick
Hello james_bondYes! Since you replied to my initial question, I understand perfectly what the question means - it's just that I can't solve it for you as yet.
It's all to do with the orientation of the small bricks -which way they are pointing. If we imagine fixed x, y and z axes and a small brick is placed so that its plane faces are parallel to the y-z, z-x and x-y planes then there are six possible orientations of the brick. A large brick is then made up of an unknown number of small bricks placed in various orientations, but (presumably) without leaving any gaps. The problem is then to prove that we can dismantle the large brick and re-arrange the small ones to re-create the large one, in such a way that all the small bricks are 'pointing the same way'. This is what you meant by
As you said, this amounts to proving that the sides of the large brick are multiples of 5, 10 and 20. But I can't see how to do that at present.all edges of equal lengths being parallel.
Grandad