MaxWong

What is m and n? The problem is still not complete!

Please mind what you are posting, do not just copy from some website and throw it here.

At least check if there are any mistakes in the problem before posting... *facepalm*

(As you edited your question, I will edit my answer.)

Area 

= (24^2 + 12^2 + 6^2)/2 * (sin 60 cos 60)

= sqrt(107163)/2

= 189sqrt(3) / 2 unit^2

Theorem: If the denominator of a simplest fraction is 2^a 5^b, then the fraction is terminating.

2010 = 2 * 3 * 5 * 67

We need to eliminate 3 and 67 from the denominator.

Therefore, \(n = 3\cdot 67 = 201\).

The question is not complete. 

(1 + i)(1 + 2i)(1 + 3i)

= (1 + 2i - i) (1 + 2i + i) (1 + 2i)

= ((1 + 2i)^2 + 1)(1 + 2i)

= (-3 + 4i + 1)(1 + 2i)

= (-2 + 4i)(1 + 2i)

= -2(1 - 2i)(1 + 2i)

= -2 (5)

= -10

\(\text{DF} = \sqrt{4^2+4^2} = 4\sqrt2\)(Pythagorean theorem)

\(\text{Let }l \text{ cm be the side length of the equilateral triangle.}\\ \dfrac{\sqrt3}{4} l^2 = 64\sqrt3\\ l^2 = 256\\ l = 16\\ \text{New length} = 12\text{ cm}\\ \text{New area} = \dfrac{\sqrt3}{4} (12^2) = 36\sqrt3\text{ cm}^2\\ \text{Area decrease} = 64\sqrt3 - 36\sqrt3= 28\sqrt3\text{ cm}^2\)

\(V = IZ\\ 1-i = I(1+3i)\\ I = \dfrac{1-i}{1+3i}\\ I = \dfrac{(1-i)(1-3i)}{(1+3i)(1-3i)}\\ I = \dfrac{-2-4i}{10}\\ I = -\dfrac{1}{5} -\dfrac{2}{5}i\)

We can solve this problem using dynamic programming, a technique taught in computer science, probably in discrete math also.

We can construct a 2-dimensional array called dp. each entry of the array is called dp_i,j, where (i,j) is the coordinates of the bug, and dp_i,j is the number of ways that the bug can reach (i,j).

First, an important observation is that \(dp_{i,j} = dp_{i-1,j} + dp_{i,j-1}\), for \((i,j) \neq (2,3)\).

This is the state transition formula of the dynamic programming.

And, there is only one way to go to (0, n) or (n, 0) for any n, which means \(dp_{0,n} = dp_{n,0} = 1\).

This is the base case of the dynamic programming.

Also, dp_2,3 = 0, as any attempt to go to (2,3) will fail due to the existence of the spider... :(

Now we can construct the array by hand... :)

First, list the base cases.

(Each entry of the matrix aligns with a lattice grid. Numbers are in relative position of the coordinates, which means bottom left corner is dp_0,0 and top right corner is dp_5,7.)

\(dp = \begin{pmatrix} 1&?&?&?&?&?\\ 1&?&?&?&?&?\\ 1&?&?&?&?&?\\ 1&?&?&?&?&?\\ 1&?&0&?&?&?\\ 1&?&?&?&?&?\\ 1&?&?&?&?&?\\ 1&1&1&1&1&1\\ \end{pmatrix}\)

Then use the transition formula to work out the remaining numbers:

\(dp = \begin{pmatrix} 1&8&26&70&180&342\\ 1&7&18&44&110&262\\ 1&6&11&26&66&152\\ 1&5&5&15&40&86\\ 1&4&0&10&25&46\\ 1&3&6&10&15&21\\ 1&2&3&4&5&6\\ 1&1&1&1&1&1\\ \end{pmatrix}\)

Therefore, there are 342 ways to go to (5,7).

Quite a challenging problem.

By exhaustion of the denominator, it is found that \(\dfrac{5}{6}\) has the smallest denominator :)

Simplify, try all fractions with denominator 1, then try all fractions with denominator 2, then 3 then 4, so on.