Rom

15 items must be kept in stock.

there's probably some very clever way of attacking this but by the time you figure out what that is you'll have solved it by brute force and be off savoring a margarita.

First let's see what multiples of 99 are 4 digits.  There can't be all that many of them.

99 x 11 = 1089 is the first one, 99 x 101 = 9999 is the last one, so 90 all together.

In this day and age of fast software I'd just code up a quick little program to see which of these are palindromes.

I come up with

{1881, 2772, 3663, 4554, 5445, 6336, 7227, 8118, 9009, 9999}

Now how many 4 digit palindromes are there.  Well you're basically choosing 2 numbers.

The first digit can be 1-9, and the 2nd can be 0-9, so we've got 9x10 = 90 total palindromes.  Thus

p = 10/90 = 1/9

2)

I guess the easiest way to do this is check the Pythagorean theorem for a right triangle.

For each value of c find the length squared of each of the 3 sides of the triangle and sort them.

Then if the triangle is a right triangle the sum of the two smaller squared sides will equal

the largest squared side.

I'm going to let you slog through the arithmetic.  I suggest using a spreadsheet or something.

1)

\(-3x+4y=4 \\ y = \dfrac 3 4 x + 1 \\ slope = \dfrac 3 4 \\ \text{now we construct a line with slope }\dfrac 3 4 \text{ passing through }(4,0)\\ \text{using point slope form}\\ (y-0) = \dfrac 3 4 (x-4) \\ y = \dfrac 3 4 x - 3 \\ -3x + 4y = -12\)

\(\text{here may be some clever way to attack this but if so I'm not aware of it}\\ p(x) = a x^3 + b x^2 + c x + d\\ \text{using the function values give we can form the equations}\\ -a+b-c+d=15\\ d=0\\ a+b+c+d=5\\ 8a+4b+2c+d=12\\ \text{I would dump these into a matrix and solve it like that}\\ \begin{pmatrix}-1 &1 &-1 &1\\0 &0 &0 &1 \\1 &1 &1 &1 \\8 &4 &2 &1\end{pmatrix} \begin{pmatrix}a \\ b\\ c \\ d\end{pmatrix} = \begin{pmatrix}15 \\0 \\5 \\12 \end{pmatrix}\)

Solving this using the method of you choice  we find

\(a=-3,~b=10,~c=-2,~d=0 \\ p(x) = -3x^3 + 10x^2 -2x\)

The x-intercepts occur where p(x)=0, factoring we get

\(p(x)=-3x^3+10x^2-2x = \\ -x(3x^2 - 10x+2)\\ \text{and this has zeros at } x=0 \text{ and}\\ x = \dfrac{10\pm \sqrt{100-24}}{6} = \dfrac 1 3\left(5 \pm \sqrt{19}\right)\)

ok so basically the zeros of this are

\(x =\dfrac 1 2 e^{\dfrac{i(\pi + 2\pi k)}{4}},~k=0,1,2,3\)

and then what we do is pair up the conjugates and form

\(\left(x - \dfrac 1 2 e^{i\pi/4}\right)\left(x - \dfrac 1 2 e^{-i\pi/4}\right) \left(x - \dfrac 1 2 e^{3\pi/4}\right)\left(x - \dfrac 1 2 e^{-3i\pi/4}\right)\)

which becomes

\(\left(x^2 - \dfrac{x}{\sqrt{2}}+\dfrac 1 4\right)\left(x^2+\dfrac{x}{\sqrt{2}}+\dfrac{1}{4}\right)\)

what do you mean?

out of the 10 dice you can find (1,2,3,4,5,6)?

you won't get (1,2,3,4,5,6,7,8,9,10) unless they are 10 sided dice

you subsets in the morning

then you subsets in the evening

ain't we got fun?

\(\text{A long straight can appear }6! \text{ different ways}\\ \text{There are }6^6 \text{ total possible different dice rolls (dice are distinguishable)} \\ P[\text{long straight}]=\dfrac{6!}{6^6} = \dfrac{720}{45656} = \dfrac{5}{324}\)

Use the quadratic formula to get the roots in terms of p and q.

Being integers and primes and non-negative put constraints on the resulting discriminant (what's under the radical sign)

Solving for values that meet those constraints gets you the two pairs.