Math Is Fun Forum

Hi! Welcome to the forum.

Here is my procedure for solving it.. which is probably not too good but it is easy to follow:

let N = number of comp disks.

N < 100

By the descriptions:

11A + 10 = N   .. eqn 1

10B + 7 = N     .. eqn 2

6C + 3 = N       .. eqn 3

tweak these starting with maximums for A, B, C:

A = 8  -> N = 98
B = 9  -> N = 97
C = 16 -> N = 99

They are not equal, try again.
There is no way to make eqn 2 produce 98, so lower A to 7.

A = 7 -> N = 77 + 10 = 87

Try to make eqn 2 produce 87:

B = 8 -> N = 80 + 7 = 87

Yay it worked! Now try to produce 87 with eqn 3

C = 15 -> N = 93 ,
C = 14 -> N = 87 ; works!

So it seems 87 is the highest number we can get out of all 3 equations at once.

But perhaps there are other combinations that work?

I don't really like this question, it seems to be all about trial and error.

I guess Devanté got it by just reasoning it out.. impressive!

I assume you mean:

and not:

If you mean the first one, then the answer is:

If you mean the second one, there is no further way to simplify it.

If you want to know why this is so, just ask.

I hope u are still awake since I did finally prove this! :

let A be nxn

We need to use these facts:

(1) (A^-1)^-1 = A
(2) (k A)^-1 = (1/k) A^-1
(3) det(adj A) = (det(A))^(n-1) where n is the dimension
(4) A^-1 = 1/det(A) adj(A)

begin with (4):

A^-1 = 1/det(A) adj(A)

(A^-1)^-1 = (1/det(A) adj(A))^-1

A = (det A)(adj A)^-1

A = (det A) [ 1/(det (adj A)) adj(adj A) ]

A = (det A)/(det (adj A)) adj(adj A)

A = (det A)/(det A)^(n-1) adj(adj A)

A = (det A)^(2-n) adj(adj A)

adj(adj A) = (det A)^(-(2-n)) A

adj(adj A) = (det A)^(n-2) A

and clearly as a special case for 3x3 matrices:

det(A) = 1 -> adj(adj A) = A

Finally !

Well, writing out the definition of adj(A) for A nxn is long and annoying, and then taking the adjoint of that, would be painful. I'm not sure what the point of the question is, but perhaps this is an acceptable answer (since this is an expression for ajd(adj(A)):

A^(-1) = (1/det(A)) * adj(A)

det(A) * A^(-1) = adj(A)

adj(adj(A)) = adj(det(A) * A^(-1))

Though this may not be what is required.

Well, you are not solving ONLY that equation -- that equation has *2* unknowns, so you cant solve it by itself.
you are solving *both equations simultaneously*,  6J + 5L = 63 AND J = 3L - 1. 

So here is what you do:

We know that  6J + 5L = 63

But we know  also that J = 3L - 1

So substitute this second fact about J into the first equation:

6(3L - 1) + 5L = 63

18L - 6 + 5L = 63

23L = 69 

L = 3

Now you can go back to either equation and get J.

Thanks for this great tutorial, hope its ok to 'practice' on this thread.

But i wonder, im getting some odd behaviour: if i add lots of newlines inside the math tag, the interpreter gives me something like 'math?' in latex font. e.g. this does not work because of the newlines inside the math tag:

what could be wrong?

Here is my understanding:
It has to do with the fact that we can never measure a circumference to an infinite number of decimal places -- so whatever we do measure is a finite decimal expansion, and therefore rational. So calculating pi as a ratio of circumference to diameter is just a rational approximation to pi. The actual irrational number pi can be defined as various infinite sums or products.

There is a good Dr Math article about this:
http://mathforum.org/library/drmath/view/54660.html

Recall that the base-4 logarithm with argument y is the function that gives you "what number 4 needs to be raised to, in order to get y", and the equation says: "this number which 4 needs to be raised to, is x".   

Thus, this is the same as saying:  4^x = y.  Which is in exponential form.

If you like a more algebraic method:

x = log4(y)

apply the change of base formula logb(t) = ln(t)/ln(b):

x = ln(y)/ln(4)

x*ln(4) = ln(y)

recall that k*ln(t) = ln(t^k):

ln(4^x) = ln(y)

exponentiate both sides:

e^(ln(4^x)) = e^(ln(y))

4^x = y

Same thing.

The second method is better if you remember those formulas and it requires less thinking.

another way is to take the 'Antilog base 4 ' of both sides:

antilog4(x) = antilog4(log4(y))

4^x = y

(b)

Here I think you need the formula:

P[a < X <= b] = F(b) - F(a)

so

P[2 < X <= 4] = F(4) - F(2) = 1 - 11/12 = 1/12

notice that:

f(x) = cot x = 1/tan x

set f'(x) = 0

ie:

d/dx f(x) = -csc^2 x

-1/sin^2 x = 0 -> no solution.

But at the boundaries:

lim as x->0 f(x) = lim x->0 1/tan x -> infinity

So we have a vertical asymptote at x=0.

lim as x->pi f(x) = lim as x->pi  1/tan x -> infinity  [since tan x is 0 at x = pi]

So we have a vertical asymptote at x=pi.

Also note that tanget of pi/2 goes off to infinity. So, for 1/tangent:

lim as x->(pi/2) 1/tan x = 0

This is an important point since concavity changes on opposite sides of this.

Now we must check for concavity on both sides of this point using: 

d^2/dx^2 f(x) = -csc x * -(csc x)(cot x) = (csc^2 x)(cot x)

Shall I go on?

oops! sorry about that, i thought that post was lost, but i must have just used the wrong button