Maxematics

$144_b = b^2 + 4b + 4 = (b + 2)^2$

Since a perfect square can never be prime, thre are no bases that are prime

Try to do long division with $1 \div 13$. If you do that you should get that $\frac{1}{13} = 0.\overline{076923}$

Therefore, $\mathbf{(A, B, C) = (9, 2, 3)}$

6 can be written with 6 dice in 2 ways:

6 * 1 * 1 * 1 * 1 * 1

2 * 3 * 1 * 1 * 1* 1

The first one has 6 ways to write it (6 spots for the 6 to go into)

The second one has 6 * 5 = 30 ways to write it (6 spots for the 3, 5 spots for the 2)

Therefore, there are 6 + 30 = 36 sequences of rolls

= 7! / (3! * 2! * 2!) = 5040/(6 * 2 * 2) = 5040/24 = 210 ways

Technically, the answer is 0 since sqrt[4](0) = 0, but if that's not it, then the second smallest integer is:

$\sqrt[4]{1323*n*84*779*5*3*441}$

$1323 = 3^3 7^2$

$84 = 2^23^17^1$

$779 = 19^141^1$

$441 = 3^27^2$

$\sqrt[4]{1323*n*84*779*5*3*441} = 2^23^85^17^519^141^1$

To make all exponents a multiple of four, n must be $2^25^37^319^341^3 = \mathbf{81073047338500}$

The first two sentences is just a complicated way of descibing a hexagon with a side length of 3. The area of a hexagon with a side length n is$\frac{3\sqrt{3}}{2}n^2$

Therefore,

$A = \frac{3\sqrt{3}}{2}6^2$

$=\frac{3\sqrt{3}}{2}36$

$3\sqrt{3}(18)$

$=\mathbf{54\sqrt{3}}$

The formula for # of diagonals for a n-sided poligon is $\frac{n (n - 3)}{2}$. Therefore,

$\frac{n(n-3)}{2} = 2n$

$n^2 - 3n = 4n$

$n^2-7n=0$

$n(n-7)=0$

$n = 0, 7$

And since there are no 0-sided polygons, the polygon has 7 sides

Case 1: x = 1

Then u + v + w + y + z = 1, and since none of them can be negative, exactly one of (u, v, w, y, z) is one, and the rest are zero. There are 5 ways to do this.

Case 2: x = 2

Then u + v + w + y + z, and using the same logic, the only case is u = v = w = y = z = 0.

Therefore there are 5 + 1 = 6 solutions

The Euclidea Algorith says that gcd(a, b) = gcd(a, b - a). Therefore, gcd(972, 1220) = gcd(972, 248) = gcd(228, 248) = gcd(228, 20) = gcd(8, 20) = gcd(8, 4) = gcd(0, 4) = 4