Maxematics

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

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!

If \(d\) is the common difference between consecutive terms of the arithmetic sequence, then we can rewrite the sequaence as \(a_1, a_1 + d, a_1 + 2d, \dots, a_1+7d, a_1 + 8d, a_1 + 9d\).

We can now werite both equations as

 \(1: a_1 + (a_1 + 2d) = 5\) and \(2: (a_1 + d) + (a_1 + 3d) = 6\) and solve for d and a.

Let's use elimination:

\(2: 2a_1 + 4d = 6\)

\(1: 2a_1 + 2d = 5\)

\(2d = 1\)

\(d = \frac{1}{2}\)

\(1: 2a_1 + 2(\frac{1}{2}) = 5\)

\(2a_1 + 1 = 5\)

\(2a_1 = 4\)

\(a_1 = 2\)

This means that the sequence is 2, 2.5, 3, 3.5, ... and that a_1 = 2

We can wrte down 2 equations for this then use substitution to solve it.

\(6a=8c\)

\(\frac{3}{4}a+280=c+\frac{1}{4}a\)

\(\frac{3}{4}a=c\)

\(c+280=c+\frac{1}{4}a\)

\(\frac{1}{4}a=280\)

\(a=1120\)

\(\frac{3}{4}(1120)=c\)

\(c=840\)

Let's double check:

\(6(1120)=8(840)\)

\(6720=6720\)

\(\frac{3}{4}(1120)+280=840+\frac{1}{4}(1120)\)

\(840+280=840+280\)

\(1120=1120\)

So both equations are true, therefore, Andrew has 1120 stamps and Caden has 840 stamps.

If you consider the cases where the order doesn't matter (so 1,2,3,4,5,6 is the same as 6,4,5,3,1,2), there are 4 possibilities. Find those, then use basic combinatorics to find the number of possibilities.

Let's call m the number of cards Mike has and a  the number of cards Aaron has. We can write the following equations:

\(m = a+56\)

\(a = \frac{3}{5}m+72\)

Let's use elimination to solve this

\(5a = 3m + 360\)

\(-3a=-3m+168\)

\(2a=528\)

\(a=264\)

\(m = (264)+56\)

\(m = 320\)

Let's double-check our answers:

\((320)= (264)+56\)

\(320=320\)

\((264) = \frac{3}{5}(320)+72\)

\(264=192+72\)

\(264=264\)

So both numbers are correct. Therefore, Mike has 264 cards and Aaron has 320 cards.

Let's rewrite the number as \(a18,486.7b\). This number must be both divisible by 9 and 11. For the number to be divisible by 9, the digits must add up to a multile of 9. The sum of all the digits are \(34 + a + b\). The closest multiple of 9 is 36. Therefore, \(a+b \equiv 2\text{ } (\text{mod }9)\), because the sum of any 2 one-digit numbers can't be any greater than 18. For a number to be divisible by 11, the result of alternating adding and subtracting the digits must be divisible by 11 (ex: 121, 1 - 2 + 1 = 0, so 121 is divisible by 11). Therefore, \(a + 1 - 8 + 4 - 8 + 6 - 7 + b = -12+a+b\)

The closest multiple of 11 is -11, so \(a+b\equiv1\text{ }(\text{mod 11} )\). We can now narrow down the possibilities. No two single digit numbers can add up to  more than 18. a + b can now only be 2 and 11 for the first equation, and a + b can now only be 1 and 12 for the second equation. Since none of these are the same, there are no possibilities for the missing digits.

We are starting with this:

\(N \equiv 4 \pmod{6}, \\ N \equiv 4 \pmod{10}, \\ N \equiv 4 \pmod{15}, \\ N \equiv 4 \pmod{24}.\)

We can rewrite this as:

\(N = 6a +4\\N=10b+4\\N=15c+4\\N=24d+4\)

\(N -4 = 6a\\N-4=10b\\N-4=15c\\N-4=24d\)

Therefore, N - 4 is a factor of 6, 10, 15, and 24. To find the minimum value of N-4, we find the least common multiple of 6, 10, 15, 24. We can do this by listing out the factors:

\(6 = 2^13^1\)

\(10=2^15^1\)

\(15=3^15^1\)

\(24=2^33^1\)

\(N-4= \text{lcm}(6, 10, 15, 24) = 2^33^15^1=120\)

Therefore, the least possible value of N is 124.

If a number is written in the form of \(p_1^a p_2^bp_3^c...\), where all p's are prime numners, it will have (a+1)(b+1)(c+1)... factors. Therefore, a terrific number must be writen in the form of either \(p^9\), or \(p_1^1p_2^4\). The smallest possibilities for each respectively are 2^9 or 512 and 3(2^4) or 48. Therefore, 48 is the smallest terrific number.

If you have a number in the form of \(p^n\) where p is a prime number, it will have n + 1 factors. Therefore \(2^9\) or 512 will have 10 factors (1, 2, 4, 8, 16, 32, 64, 128, 256, 512), but 2 will be the only prime number that it is divisible by. This is also true for \(3^9\) (19683), \(5^9\) (1953125), \(7^9\) (40353607), or any number in the form of \(p^9\), where p is a prime number. Therefore, one prime number is the minimum number of factors for a terrific number.

We can use Simon's Favourite Factoring Trick to solve this problem.

\(3x + 2y + xy = 18 + 7x + 5y\)

\(xy-4x-3y-18=0\)

\(xy-4x-3y+12=30\)

\(x(y-4)-3(y-4)=30\)

\((x-3)(y-4)=30\)

The pairs of factors of 30 is (1, 30), (2, 15), (3, 10), (5, 6), (6, 5), (10, 3), (15, 2), and (30, 1). We choose x and y such that the values in the parentheses is equal to one of the two factors of 30. This makes all values of (x, y) equal to (4, 34), (5, 19), (6, 14), (8, 10), (9, 9) (13, 7), (18, 6), or (33, 5). The sum of x and y for each pair respectively is 38, 24, 20, 18, 18, 20, 24, and 38. Therefore, all possible sums of x and y are 18, 20, 24, and 38.