Math Extra Credit Help
This question legit stumped me for 2hours, I DON'T GET IT .
I really need this extra credit for my grade, can someone help me out with it? (and also explain this problem/answers)
A solid rectangular block is formed by gluing together "n" congruent one-centimeter cubes face to face.
When the block is viewed so that three of the faces are visible, exactly 231 of the 1-cm cubes cannot be seen.
Find the smallest possible value of "n"
March 28, 2012
4 Comments • Newest first
@konfooshus: no you are not thinking
here
the 231 cubes which are not visible must lie below exactly one layer of cubes. thus, they form a rectangular solid which is one unit shorter in each dimension. if the original block has dimensions l x m x n , we must have (l-1) x (m-1) x (n-1) =231 . The prime factorization of 231 = 3 x 7 x 11 , so we have a variety of possibilities; for instance, l - 1= 11 and m - 1 = 11 and n - 1 = 3 x 7 , among others. it should be fairly clear that the way to minimize l x m x n is to make l and m and n as close together as possible, which occurs when the smaller block is 3 x 7 x 11 . then the extra layer makes the entire block 4 x 8 x 12 , and n = 384 .
[quote=Keey]darvik is correct, however:
an alternate way to visualize the problem is to count the blocks that can be seen and subtract the blocks that cannot be seen. in the given block with dimensions l x m x n , the three faces have lm , mn , and ln blocks each. however, l blocks along the first edge, m blocks along the second edge, and n blocks along the tird edge were counted twice, so they must be subtracted. after subtracting these three edges, 1 block has not been counted - it was added three times on each face, but subtracted three times on each side. thus, the total number of visible cubes is lm + mn + ln - 1 - m - n + 1 , and the total number of invisible cubes is lmn - lm - mn - ln + l + m + n - 1 , which can be factored into (l-1)(m-1)(n-1)[/quote]
so do i plug in 3x7x11 into each of the variables ?
darvik is correct, however:
an alternate way to visualize the problem is to count the blocks that can be seen and subtract the blocks that cannot be seen. in the given block with dimensions l x m x n , the three faces have lm , mn , and ln blocks each. however, l blocks along the first edge, m blocks along the second edge, and n blocks along the tird edge were counted twice, so they must be subtracted. after subtracting these three edges, 1 block has not been counted - it was added three times on each face, but subtracted three times on each side. thus, the total number of visible cubes is lm + mn + ln - 1 - m - n + 1 , and the total number of invisible cubes is lmn - lm - mn - ln + l + m + n - 1 , which can be factored into (l-1)(m-1)(n-1)
[quote=Darvik]Okay well there are three factors of 231. You know that you'll still have a rectangular solid if you ignore the sides you can't see which has a volume of 231 cm^3.
The dimensions of this "hidden" block are 3x7x11. Then you can find out how many 1 cm blocks to make the whole rectangular solid which would be...
I think 4*8*12=384 blocks so n=384. I don't know if this is correct but it would be my answer.[/quote]
i tried something similar like that, by dividing 231 by 3, which gave me 77, so i tried to figure out the dimensions of the top/bottom surface of the rectangle, and the sides also, but accounting for the blocks that are shared by two different sides kinda threw me off.
lol