Block Partition

Number Theory Level 4

Three rectangles of dimensions $a\times a, a\times b,$ and $b\times b$ with $a$ and $b$ coprime integers are available as the bases on which to build cuboid towers of any heights. The tower on each base will be built with $1\times 1\times 1$ cubic blocks. For example, if we want to make cuboids of heights 1, 2, 3 for the first, second, and third bases, respectively, the total number of the unit blocks used will be $a^2+2ab+3b^2.$ But it is not required to use all three of the bases in building cuboid towers: you may use only one or two of the three bases.

You try all combinations of the bases and heights and find that, given 47 unit blocks, you can never build cuboid towers as described above, but that there is always a way to do the job if there are more than 47 blocks.

What is $a+b$ ?

The answer is 7.

1 solution

Worranat Pakornrat
Apr 11, 2017

The question is asking for the Frobenius Number , or the greatest integer that can not be put in terms of summation of $a^2$ , $ab$ , and $b^2$ .

Since $LCM(a^2, b^2) = (ab)^2$ , then $ab| LCM(a^2, b^2)$ .

Hence, we can apply the formula:

$47 = LCM(a^2, ab) + LCM(b^2, ab) - a^2 - ab - b^2 = a^2 b + b^2 a - a^2 - ab - b^2 = (a+b+1)(a-1)(b-1) -1$ .

Thus, $48 = (a+b+1)(a-1)(b-1)$ , and these are the ways to factorize $48$ into three factors:

$48 = 1\times 2\times 24 = 1\times 3\times 16 = 1\times 4\times 12 = 1\times 6\times 8 = 2\times 3\times 8 = 2\times 4\times 6 = 3\times 4\times 4$ .

By trial & error, we will find out that $a + b+ 1 = 8$ ; $a-1 = 3$ ; $b-1 = 2$ . Thus, $a = 4$ and $b=3$ .

Finally, $a+b = \boxed{7}$ .

I find it hard to understand what you're saying. Can you express it clearer?

I first thought we were using those blocks to form a cuboid, and it took me a while to realize you were saying $47 \neq Xa^2 + Yab + Z b^2$ .

Calvin Lin Staff - 4 years, 1 month ago

Oh, Ok. Is it better now?

Worranat Pakornrat - 4 years, 1 month ago

Block Partition

The answer is 7.

1 solution

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