Maxematics

5a = 57

a = 11.4

4(11.4) + 5b = 63

45.6 + 5b = 63

5b = 17.4

b = 3.48

3(11.4) + 4(3.48) + 5c = 34

34.2 + 13.92 + 5c = 34

48.12 + 5c = 34

5c = -14.12

c = -2.824

2(11.4) + 3(3.48) + 4(-2.824) + 5d = 15

22.8 + 10.44 - 11.296 + 5d = 15

21.944 + 5d = 15

5d = -6.944

d = -1.3888

11.4 + 2(3.48) + 3(-2.824) + 4(-1.3888) + 5e = 47

11.4 + 6.96 - 8.472 - 5.5552 + 5e = 47

4.3328 + 5e = 47

5e = 42.6672

e = 8.53344

(a, b, c, d, e) = (11.4, 3.48, -2.824, -1.3888, 8.53344)

We an represent this problem as $\frac{1}{2}ab = 5(a+b)$, where a and b are the two legs. We can now simplify this equation further.

$\frac{1}{2}ab = 5a+5b$

$ab=10a+10b$

$ab-10a-10b=0$

We can now use Simon's Favourite Factoring Trick to simplify this.

$ab-10a-10b+100=100$

$(a-10)(b-10)=100$

The factor pairs of 100 are (1, 100), (2,50), (4, 25), (5, 20), and (10, 10). We now choose a and b to make thepart in the parentheses equal to one of these pairs. Therefore, all possibilities for the legs (a and b) are (11, 110), (12, 60), (14, 35), (15, 30), and (20, 20). We can calculate the area for each of these.

$\frac{1}{2}(11)(110)=605$

$\frac{1}{2}(12)(60)=360$

$\frac{1}{2}(14)(35)=245$

$\frac{1}{2}(15)(30)=225$

$\frac{1}{2}(20)(20)=200$

Therefore, the sum of all areas is 605 + 360 + 245 + 225 + 200 = 1635

Let's first find out how many cubes have how many colors. All cubes must have 0-3 black sides. 8 of the cubes have 3 black spots (all of the corners), 48 have 2 black spots (12 edges each with 4 cubes), and 96 have one black spot (4 by 4 center for each side). If we roll a cube with 3 black sides, it will have a 1/2 chance of landing on a black side, if we roll a cube with 2 black sides, it will have a 1/3 chance of landing on a black side, and if we roll a cube with 1 black side, it will have a 1/6 chance of landing on a black side. Therefore, the total probability is:

$\frac{8 \times \frac{1}{2} + 48 \times \frac{1}{3} + 96 \times \frac{1}{6}}{216}$

$\frac{4+16+16}{216}$

$\frac{36}{216}$

$\mathbf{\frac{1}{6}}$

Therefore, you will have a 1/6 chance of rolling a black face.

If $x \mathbin{\spadesuit} y = \frac{x^2}{y}$, then $a \mathbin{\spadesuit} (a + 1) = \frac{a^2}{a+1}$. Therfore, $\frac{a^2}{a+1}=9$. We can now use completing the square to solve for a.

$\frac{a^2}{a+1}=9$

$a^2 = 9a+9$

$a^2-9a-9=0$

$a^2-9a+20.25 = 29.25$

$(a-4.5)^2 = 29 \frac{1}{4}$

$(a-\frac{9}{2})^2 = \frac{117}{4}$

$a-\frac{9}{2}=\frac{\pm\sqrt{117}}{2}$

$\mathbf{a = \frac{9\pm3\sqrt{13}}{2}}$

So the two solutions are (9 + 3sqrt(13))/2, (9 - 3sqrt(13))/2

We can rewrite this as $\frac{1}{1\times2} + \frac{1}{2\times3} + \frac{1}{3\times4} + ... + \frac{1}{(p-2)\times (p-1)}$. All of these terms are in the form of $\frac{1}{n(n+1)}$, so let's first try to simplify that.

$\frac{1}{n(n+1)}$

$\frac{1+n-n}{n(n+1)}$

$\frac{1+n}{n(n+1)} - \frac{n}{n(n+1)}$

$\frac{1}{n} - \frac{1}{n+1}$

Now, we can rewrite all of out fractions as $(\frac{1}{1} - \frac{1}{2})+(\frac{1}{2} - \frac{1}{3})+(\frac{1}{3} - \frac{1}{4}) + ... +(\frac{1}{p-2} - \frac{1}{p-1})$

We can finally simplify this to $1 - \frac{1}{p-1}$ or $\mathbf{\frac{p-2}{p-1}}$, which is the final answer.

You could also solve this by induction, because you could notice that $\frac{1}{2}+\frac{1}{2\times3}=2/3$ and that  $\frac{1}{1\times2}+\frac{1}{2\times3} + \frac{1}{3\times4}=3/4$.

$x = 0.031\overline{4}$

$1000x = 31.\overline{4}$

$1000x = 31 \frac{4}{9}$

$1000x = \frac{283}{9}$

$\mathbf{x = \frac{283}{9000}}$

Therefore, $\mathbf{0.031\overline{4} = \frac{283}{9000}}$

$x = 0.29\overline{6}$

$100x = 29.\overline{6}$

$100x = 29 \frac{2}{3}$

$100x = \frac{89}{3}$

$\mathbf{x = \frac{89}{300}}$

Therefore, $\mathbf{0.29\overline{6} = \frac{89}{300}}$

For a number to be 2-digits in a certain base, the cub of the base must be more than the number. We can test out all numbeers until one works.

2^3 = 8 < 301

3^3 = 27 < 301

4^3 = 64 < 301

5^3 = 125 < 301

6^3 = 216 < 301

7^3 = 343 > 301

This means that 7 is the smallest value for B that works.