Easy math puzzles

At time t=T min, salt=0.1 gal, water=100 gal.

At time t=x, salt=y. We have dy/dx=-5y/100; therefore, the general solution to the differential equation is

𝑦=A* exp(-x/20)=2*exp(-x/20).

Where we have used: at time t=0 min, salt=2 gal, water=100 gal. Therefore

x=-20 ln(y/2)= -20 ln (0.1/2)‎ = 59.915 min.

13

?

44 4/7

Let T(r,c) be the number of turns it will take to get from (r,c) to (3,3); the solution is T(0,0)

Because of symmetry, T(r,c) = T(c,r), which reduces the number of equations involved

T(3,3) = 0
T(3,2) = 1 + 1/3 T(2,2) + 1/3 T(3,1) {+ 1/3 T(3,3)}
T(3,1) = 1 + 1/3 T(3,2) + 1/3 T(2,1) + 1/3 T(3,0)
T(3,0) = 1 + 1/2 T(3,1) + 1/2 T(2,0)

T(2,2) = 1 + 1/2 T(3,2) + 1/2 T(2,1)
T(2,1) = 1 + 1/4 T(3,1) + 1/4 T(2,2) + 1/4 T(1,1) + 1/4 T(2,0)
T(2,0) = 1 + 1/3 T(3,0) + 1/3 T(2,1) + 1/3 T(1,0)

T(1,1) = 1 + 1/2 T(2,1) + 1/2 T(1,0)

T(1,0) = 1 + 1/3 T(2,0) + 1/3 T(1,1) + 1/3 T(0,0)
T(0,0) = 1 + T(1,0)
T(1,0) = 1 + 1/3 T(2,0) + 1/3 T(1,1) + 1/3 (1 + T(1,0))
2/3 T(1,0) = 4/3 + 1/3 T(2,0) + 1/3 T(1,1)
T(1,0) = 2 + 1/2 T(2,0) + 1/2 T(1,1)

This is 8 equations in 8 unknowns; I won't bother with the details of solving them here, but T(1,0) = 305/7, so T(0,0) = 23 + 1 = 312/7

18

18

For a 3x3 puzzle there is some symmetry so some of the junctions would have the same values as the ones. Also the middle square has to go to A or C, so has the same value as B. The puzzle therefore simplifies to $-A=B=C-X where
(i) $tart has to go to A. Hence EV(A)=EV($tart)+1.
(ii) A has one chance to go to $tart or two chances to go to B. Thus the expected number of turns from A to exit is 1 + average{ $,B,B }.
(iii) B has two chances to go to A, or two chances to go to C (i.e. has equal chances). (ev=1 + average{A,C} )
(iv) C has one chance to eXit or two chances to return to B. (ev

=1 + average{0, B, B} )
(v) X is the exit, so has zero more moves!

$ A B
a B C
b c X

Using the above assuming the original value EV($)=x. Then EV[A]=x-1.

 x  x-1 -?-
... -?- ...
... ... ...

Now EV[A]= average ($ B B) + 1; hence $+B+B=3(x-2) : this means B=x-3.

 x  x-1 x-3
... x-3 ...
... ... ...

Using similar logic B is average(A C)+1, giving C = x-7

 x  x-1 x-3
... x-3 x-7.
... ... ...

Now we know is the average of (x-3, x-3 -?-)+1. This gives it as x-18.

 x  x-1 x-3
... x-3 x-7
... ...x-18

But we know that's the exit. Hence x=18.