Maxematics

edit: The answer can't be 0 since it must be a positive integer. The second bolded number is the correct answer

There are 2 cases:

\(n = p_1 ^ 8\) or \(n = p_1 ^ 2 p_2 ^ 2\)

But wait! \(8 = 2^3\), so the second possibility won't work since it must not be divisible by a power of 2 greater than 4. Therefore, the smallest (and only) ice number with exactly 9 divisors is \(2^8 = \mathbf{256}\)

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

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