Melody

Thanks Alan.    

The positive integers from 1 to 150 inclusive are placed in a 10 by 15 grid so that each cell contains exactly one integer. Then the multiples of 3 are given a red mark, the multiples of 5 are given a blue mark, and the multiples of 7 are given a green mark. How many cells have more than 0 marks?

Here is how to start it.

Draw a Venn diagram

The bit that overlaps the most has the number of multiples of all of them in it. 

3*5*7=105   There is only one multiple of 105 in that set of numbers.

Now work out from there.

How many multiples of 3 and 5 are there?   3*5=15       150/15=10      10-1=9

So where are you going to put the 9?

Thanks Alan,

 So that means there is a polynomial of degree 5 that fits?  

Good one Ginger :)

The function f(x) = ax^r satisfies f(2) = 1 and f(32) = 16. Find r.

\(1=a*(2^r) \qquad 16=a*(32^r)\\ 1=a*(2^r) \qquad 2^4=a*(2^{5r})\\ 2{-r}=a \qquad \qquad 2^{4-5r}=a\\ 2-r=4-5r\\ etc\)

LaTex:

1=a*(2^r) \qquad 16=a*(32^r)\\
1=a*(2^r) \qquad 2^4=a*(2^{5r})\\
2{-r}=a \qquad \qquad 2^{4-5r}=a\\
2-r=4-5r\\
etc

She has 12 moves.   

So that is 2^12 possible paths

Let her take r moves clockwise  (positive)  and 12-r moves anti-clockwise.  (negative)

So she will end up at   r - (12-r) = 2r-12 in a positive directions which is

2r-12 (mod6) = 2r (mod6) 

2r(mod6) = 0 when r = 0,3,6,9,12

If r=0 then she goes anticlockwise every time so there is only one way she can do that      1

If r=12 then she goes clockwise every time so there is only one way she can do that          1

If r= 3 there are  12C3 = 220 possible paths

If r = 9  there are also   220 possible paths

If r=6 there are 12C6 = 924 possible paths

924+440+2 =  1366 possible paths where she will end up back at A

Probability that she will end up back at A

\(=\frac{1366}{2^{12}}\\ =\frac{683}{2^{11}}\\ \approx 0.335 \qquad\text{Correct to 3 decimal places}\)

**    I am not claiming to be positive that this is completely correct.

Thanks Ginger.

What calculator/program  did you use to get those values Gino? 

hint:

(n+3)! = n!(n+1)(n+2)(n+3)

sqrt(2*sqrt(2 - t)) = 7 - t

\(\sqrt{2*\sqrt{2 - t}} = 7 - t\\ 2*\sqrt{2 - t} = t^2-14t+49\\ \sqrt{2 - t} =\frac{ t^2-14t+49}{2}\\ 2 - t =\frac{( t^2-14t+49)( t^2-14t+49)}{4}\\ 8 - 4t =( t^2-14t+49)( t^2-14t+49)\\ 0 =( t^2-14t+49)( t^2-14t+49)+4t-8 \)

\(0=2401 - 1372 t + 294 t^2 - 28 t^3 + t^4+4t-8\\ 0=t^4 - 28 t^3 + 294 t^2 - 1368 t + 2393\)

Wolfram alpha says no real solutions.

You do need to check my algebra though. (for careless errors)

LaTex:

\sqrt{2*\sqrt{2 - t}} = 7 - t\\

2*\sqrt{2 - t} = t^2-14t+49\\
\sqrt{2 - t} =\frac{ t^2-14t+49}{2}\\

2 - t =\frac{( t^2-14t+49)( t^2-14t+49)}{4}\\
8 - 4t =( t^2-14t+49)( t^2-14t+49)\\

0 =( t^2-14t+49)( t^2-14t+49)+4t-8