Maxematics

We can rewrite this as\(1000 \leq 47x + 3 \leq 2000\), which we can simplify

\(997 \leq 47x \leq 1997\)

\(21\frac{10}{47}\leq x\leq 42\frac{23}{47}\)

The only natural numbers that fit in that range are 22 through 42, and there are 21 numbers in that range.

To account for the conditional at the beginning, let's just give the two youngest kids four candies off the bat. That leaves you with 5 candies to divide between 5 kids, which is a simple stars and bars problem (if you don't know what it is, look it up it's really cool :) )

This means the answer is 9 choose 4 which is (9!)/(4!)*(5!) = (9 * 8 * 7 * 6) / 24 = 9 * 7 * 2 = 112 ways

\(\frac{1}{x} - \frac{1}{xy} - \frac{1}{xyz} = \frac{1}{3}\)

Multiply by x on both sides

\(1 - \frac{1}{y} - \frac{1}{yz} = \frac{x}{3}\)

x < 3

if x = 2:

\(\frac{1}{y}+\frac{1}{yz} = \frac{1}{3}\)

\(1 + \frac{1}{z} = \frac{y}{3}\)

Meaning z = 3, which cannot work in x > y > z

if x = 1:

\(\frac{1}{y}+\frac{1}{yz} = \frac{2}{3}\)

\(1 + \frac{1}{z} = \frac{2y}{3}\)

Also meaning z = 3, which doesn't work

Therefore, there is no solution.

It could have no multiple of 7: (15, 16, 18, 20, 22, 24, 26)

It must have at least 1 multiple of 5: (8, 9, 11, 13, 15, 17, 19)

It must have at least 2 multiples of 3: (13, 14, 15, 17, 19, 21, 23, 25)

It must have at least 1 multiple of 2: (2, 3, 5, 7, 9, 11, 13, 15)

Therefore the GCF is 5 * 3 * 3 * 2 = 90

\(b + 7 = 2b^2+b+1\)

\(2b^2 = 6\)

\(b^2 = 3\)

\(\mathbf{b = \sqrt{3}}\)

But since sqrt(3) < 7, the base wouldn't work for the left side of the equation. Therefore, there is no solution.

\(4 + \frac{1}{4 + \frac{1}{4 + \frac{1}{4}}}\)

\(4+\frac{1}{4+\frac{1}{\frac{5}{4}}}\)

\(4 + \frac{1}{4+\frac{4}{5}}\)

\(4+\frac{1}{\frac{24}{5}}\)

\(4+\frac{5}{24}\)

\(\mathbf{\frac{101}{24}}\)

\(a + b = c\)

\(b - a = d\)

\(a < b\)

\(a \neq 0\)

(If a was 0, then b and d would be the same)

If a was 1, b could be any number from 3 - 8 (6 numbers)

If a was 2, b could be 3, 5, 6, 7 (4 numbers)

If a was 3, b could be 4, 5 (2 numbers)

If a was 4, b could only be 5 (1 number)

There are no other possible values for a

Therefore there are 6 + 4 + 2 + 1 = 13 possibilities.

We can rewrite this as \(b^2 + 1\). This is one of Landhau's 4 problems, all of which are unsolved to this day. Therefore, this problem cannot be solved (yet).

\(\angle{P} = \angle{TUV}\)

\(\angle{PUV} = \angle{UTR}\)

\(\angle{PQV} = \angle{UVR}\)

\(\angle{QVU} = \angle{VRT}\)

Therefore, trapezoid PUVQ is similar to trapezoid UTRV

\(\overline{TR} = \frac{2}{7}(20) = \frac{40}{7}\)

Since \(PUVQ \sim UTRV\):

\(\frac{20}{UV} = \frac{UV}{\frac{40}{7}}\)

\(\frac{800}{7} = \overline{UV}^2\)

\(\textbf{UV = }\frac{\textbf{20}\sqrt{\textbf{2}}}{\sqrt{\textbf{7}}}\)

\(\textbf{UV = }\frac{\textbf{20}\sqrt{\textbf{14}}}{{\textbf{7}}}\)

Let me know if I made any mistakes!