[bonanova]

Johnny received a Jumbo Set of 1000 building blocks for his birthday.

He begins to assemble them into "bricks" of dimensions a x b x c

where of course a, b and c are integers, and a <= b<= c.

He notices that for each brick he builds, some of the blocks are

"outside" blocks [visible] and some are "inside" blocks [hidden

by the outside blocks.] He imagines that for some bricks the

numbers of outside and inside blocks must be equal; call such

a brick an Equal Brick.

With his set of blocks, how many different Equal Bricks can Johnny make?

Now for small bricks [a=1 or 2] there are no Inside Blocks.

And for larger bricks, the Inside Blocks will eventually dominate.

So there must be a finite number of Equal Bricks.

If Johnny wants to make all possible Equal Bricks, how many

Jumbo Sets of blocks will he need?

[Guest]

one Equal Brick: 8-10-12

[bonanova]

one Equal Brick: 8-10-12

Right on.

[Yoruichi-san]

Since each side of the 'inside block' is the respective side of the entire block minus 2, and the volume of the 'inside block' should be half the volume of the entire block, then using the formula (a-2)(b-2)(c-2)=abc/2 along with the inequality, I get all the possible Equal blocks as (you may recognize the output format...*cough*):

(a == 5 && ((b == 13 && c == 132) || (b == 14 && c == 72) || (b == 15 &&

c == 52) || (b == 16 && c == 42) || (b == 17 &&

c == 36) || (b == 18 && c == 32) || (b == 20 &&

c == 27) || (b == 22 && c == 24))) || (a ==

6 && ((b == 9 && c == 56) || (b == 10 && c ==

32) || (b == 11 && c == 24) || (b == 12 && c ==

20) || (b == 14 && c == 16))) || (a == 7 && ((b == 7 &&

c == 100) || (b == 8 && c == 30) || (b == 9 && c ==

20) || (b == 10 &&

c == 16))) || (a == 8 && ((b == 8 && c == 18) || (

b == 9 && c == 14) || (b == 10 && c == 12))))

Summing the (a*b*c)s of these, I get 53552, so it would require 54 sets of 1000 blocks.

I feel like there should be a better way of doing this though...I'm not really seeing any obvious patterns in the numbers...

If someone has a more elegant way of solving this, I'd love to see it...

[bonanova]

If someone has a more elegant way of solving this, I'd love to see it...

Since each side of the 'inside block' is the respective side of the entire block minus 2, and the volume of the 'inside block' should be half the volume of the entire block, then using the formula (a-2)(b-2)(c-2)=abc/2 along with the inequality, I get all the possible Equal blocks as (you may recognize the output format...*cough*):

(a == 5 &&

((b == 13 && c == 132) ||

(b == 14 && c == 72) ||

(b == 15 && c == 52) ||

(b == 16 && c == 42) ||

(b == 17 && c == 36) ||

(b == 18 && c == 32) ||

(b == 20 && c == 27) ||

(b == 22 && c == 24))) ||

(a == 6 &&

((b == 9 && c == 56) ||

(b == 10 && c == 32) ||

(b == 11 && c == 24) ||

(b == 12 && c == 20) ||

(b == 14 && c == 16))) ||

(a == 7 &&

((b == 7 && c == 100) ||

(b == 8 && c == 30) ||

(b == 9 && c == 20) ||

(b == 10 && c == 16))) ||

(a == 8 &&

((b == 8 && c == 18) ||

(b == 9 && c == 14) ||

(b == 10 && c == 12))))

Summing the (a*b*c)s of these, I get 53552, so it would require 54 sets of 1000 blocks.

I feel like there should be a better way of doing this though...I'm not really seeing any obvious patterns in the numbers...

Yeah, I meant one at a time, so you only need enough for the biggest one [8580 blocks, 9 Sets].

Nice one.

Don't recognize the format, tho. LISP?

Only method I know is to set limits on a [4 and 9 are derivable limits] then solve for b and c for each a value.

a=4: 8[b + c] = 16. No solution with 4 <= b <= c

a=5: [b − 12][c − 12] = 120 - eight pairs of factors

a=6: [b − 8][c − 8] = 48 - five pairs of factors

a=7: [3b − 20][3c − 20] = 280 - four pairs of factors

a=8: [b − 6][c − 6] = 24 - three pairs of factors

a=9: [5b − 28][5c − 28] = 504. No factors with 9 <= b <= c

Not elegant.