Rom

It must be that there are 2 of one color and then 1 each of the other two colors.

\(P[\text{we pick the two like colored balls in two choices}]=\dfrac 1 2 \dfrac 1 3 = \dfrac 1 6\)

hey... where are my popsicles?

You can convert all this to base 10 and solve it but to me that goes against the grain of the problem.

All the following is assumed to be in base 6

AB+C = C0

B<6, C<6, so B+C=6, This produces a carry of 1 to the left so

A+1 = C

AB+BA=CC so B+A = C mod 6

B+A=A+1 mod 6

B=1

C=6-B = 5

A=C-1 = 4

ABC = 415₆

\(a=900+0.1s \\ b=1200+0.15(s-8000)\\ \text{find s such that }b>a\\ 1200+0.15(s-8000) > 900 + 0.1 s\\ 1200 +0.15s - 1200 > 900 + 0.1s \\ 0.05s > 900 \\ s> 18000\)

I'm not seeing any simple way of doing this though the answer is simple enough it makes me suspect there is one.

You need to know the basic equation of ballistic motion

\(h(t) = -\dfrac{g t^2}{2} + v_0 t + h_0\)

here they give you \(h_0 = 60\)

they also seem to be using \(g = -10 ~m/s^2\) which isn't quite accurate but close enough for government work.

You have two equations

\(h(2)= 80,~h(6) = 0\\ \text{you really only need one of them, plugging in values we have}\\ 80 = -5(2^2) + 2v_0 + 60\\ 20 = 2v_0 -20\\ v_0 = 20 \\ \text{checking with the other equation we have}\\ h(6) = -5(6^2) + 20(6) + 60 = \\ -180 + 120 + 60 = 0 \text{ as expected. So our equation of motion is}\\ h(t) = -5t^2 +20t + 60 \\ \text{we can rewrite this as}\\ h(t) = -5(t^2 - 4t - 12) = \\ -5((t-2)^2-4-12)=\\ -5(t-2)^2 + 80 \\ \text{and this last is the standard form of the parabola }\)

If the expression (x-a) is a factor of f(x), then f(a) = 0.

So just plug each of them into f(x)

f(5) = 160

f(-3) = 0

f(15) = 4320

\(\dfrac{y_2}{y_1}=\dfrac{1}{\sqrt{\dfrac{x_2}{x_1}}}\\y_2 = \dfrac{1}{\sqrt{\dfrac{x_2}{x_1}}}y_1 \\ 1 = \dfrac{1}{\sqrt{\dfrac x 2}} 4 \\ 1 = 4 \sqrt{\dfrac 2 x}\\ 1 = 16 \dfrac 2 x \\ x = 32 \)

\(BBA_6+41B_6+A15_6=A152_6 \\ \text{looking at the 1's digit we have, (all numbers will be base 6)} \\ A+B+5 =2 \mod 6 \\ \text{given the possible ranges of A and B we have}\\ A+B = 9 \text{ with a carry of 2 or }\\ A+B = 3 \text{ with a carry of 1}\)

\(\text{let's suppose the first case}\\ \text{then looking at the 6's digit we have}\\ B+1+1+2 = 5 \mod 6\\ B+4 = 5 \mod 6 \\ B = 7 \text{ or }B=1 \\ \text{well B=7 is clearly out as B<6, and A+B=9, which would make A=8, which is also out}\)

\(\text{so let's assume the 2nd case}\\ \text{looking at the 6's digit we have}\\ B+1+1+1 = 5 \mod 6\\ B+3 = 5 \mod 6\\ B=8 \text{ or }B=2 \\ \text{again B=8 is clearly out. If B=2 then A=1 and this seems like a solution}\)

\(\text{checking we have }\\ 221 + 412+115 = 1152 \text{ all in base 6}\)

\(|A-B| = |1-2|=1\)

two formulas are of use

\(\sum \limits_{r=1}^n~r^2 = \dfrac{1}{6} n (n+1) (2 n+1) \\ \sum \limits_{r=1}^n~r = \dfrac 1 2 n(n+1) \\ \text{applying these we get, after some algebra}\\ \sum \limits_{r=1}^n~6r^2 + 32r = n (n+1) (2 n+17) \\ \text{plugging }n=1000 \text{ into the rightmost factor we see that }\\ 2(1000)+17 = 2017 \\ \text{and thus the overall sum is a multiple of 2017}\)