In how many ways can he form a rectangle using all the tiles each time?

A contractor has exactly $1088$ square tiles. In how many ways can he form a rectangle using all the tiles each time?

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Basically there asking how many ways can he multiply two numbers together to get 1088. Because the multiplication of two numbers 'a' and 'b' forms a rectangle with length a, and side b, and area ab. – Ethan Jan 13 '13 at 1:16

Hint: How many $m,n\in \mathbb{N}$ such that $m\times n = n\times m= 1088$ can you find? Note that $1088=2^6\cdot17$

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$$1088 = 2\times2\times2\times2\times2\times2\times17$$ So \begin{align} 1088 & = 1\times1088 \\ & = 2\times544 \\ & = 4 \times272 \\ & = 8 \times136 \\ & = 16\times68 \\ & = 17\times64 \\ & = 32\times 34. \end{align} There's your list of rectangles.

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Basically there asking how many ways can he multiply two numbers together to get 1088. Because the multiplication of two numbers 'a' and 'b' forms a rectangle with length a, and side b, and area ab.

In general the solution to this problem given n tiles will be how many divisors does n have less then or equal to $n^{1/2}$.

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is the rectangle $(a,b)$ equal to the rectangle $(b,a)$ if not then here is a way to solve it: $1088=2^6*17$. since you need to keep the tiles whole and you use all of them the height needs to be a divisor of $1088$. Therefore the number of rectangles $(a,b)$ is the number of divisors of $1088$.(if the number where a perfect square it would be the number of divisors minus 1)

How do we count them? from the fundamental theorem of arithmetic we know every integer number greater than 1 has a single representation as a product of primes.

Therefore how many combinations can we make with six 2's and one 17?? we can chose to add $0,1,2,3...$or $6$ 2's and we can chose to use $0$ or $1$ 17's. Using the fundamental theorem of combinatorics we know the number of combinations is $7*2=14$. Now: if you don't want to count $(a,b)$ and $(b,a)$ the result is $7$ because we are counting all of them twice.

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