Maxematics

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.

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.