A Very Big Block.

Number Theory Level 5

A bricklayer with too much time to spare has 200,000 bricks of size 1 $\times$ 2 $\times$ 3 units at his disposal.

He proposes to construct a solid cuboid block with these bricks (that is, not just the walls the entire interior of the cuboid is filled with bricks). Further, no brick is broken for the construction and no part of any brick is to project out of the cuboid.

In order to make the problem interesting, he plans to construct the block so that in addition to the length, breadth and height, all the face diagonals also have integer lengths.

Will he be able to finish the construction with the bricks available with him?

Image credit: Wikipedia Steve Evans .

Yes, and with more than 7% bricks to spare. No, but he has bricks to complete more than 95% of the construction. No, with the bricks available, less than 75% of the work is done. Yes, but less than 0.5% of the available bricks would be unused. Such a block can never be constructed, even with an infinite supply of bricks.

1 solution

Janardhanan Sivaramakrishnan
Jan 28, 2015

The smallest dimensions of a cuboid with face diagonals of integer lengths is $(240,117,44)$ . The face diagonals would have lengths $(267,244,125)$ .

We would need $120 \times 39 \times 44$ bricks arranged with the side with 2 units along x/first direction and side with length 3 along y/second direction.

Thus, a total of $120 \times 39 \times 44 = 205920$ bricks are needed.

Definitely, the bricklayer is short of bricks. However, with the available bricks he can complete $\frac{200000}{205920}=97.125\% \boxed{>95\%}$ of the construction.

Note that the said layout is just one of the options, there are many more ways to arrange the bricks. However, a smaller block cannot be constructed.

http://en.wikipedia.org/wiki/Euler_brick

Harrison Wang - 6 years, 4 months ago

how did u solve the problem.....u just wrote the answer directly

Akhil Reddy Bandi - 5 years, 8 months ago

Yeah, a simpler way to look at it without explicitly counting the number of bricks needed would be the examining the factors of the side lengths; since 240 is clearly divisible by 3, 44 by 2, and 117 by 1, there exists an arrangement of the bricks that can form an Euler brick given sufficient bricks.

Jake Lai - 6 years, 4 months ago

Yes. But, the question was would 200000 bricks be enough. In this case, you need to count the number of bricks.

Janardhanan Sivaramakrishnan - 6 years, 4 months ago

Is there a way to solve this without knowing about the "Euler brick"

Trevor Arashiro - 5 years, 11 months ago

@Trevor Arashiro – Are you into Python?

Peter Byers - 5 years, 11 months ago

@Trevor Arashiro – I think if you can solve the diophantine system of equations formed by the contraints, then you may solve this. However you will also have to prove that it is the smallest configuration.

Anupam Nayak - 5 years, 5 months ago

Such a block can never be constructed, even with an infinite supply of bricks.

why cant this be the answer then ?

A Former Brilliant Member - 5 years, 2 months ago

I made a mistake, typed 147 instead of 117 and keyed in the "haven't completed 75%" one. Gah...

Aloysius Ng - 5 years, 4 months ago

A Very Big Block.

Image credit: Wikipedia Steve Evans .

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