To All You BFM Geniuses

[Titanium»X«]

2, 3, and 4 seem pretty trivial to me, as does 6 if you know how to fit polynomials to specific points.

I think three360 was talking about triviality based on the perception of the average Joe Schmoe IMO.  Why is the rest not trivial to you?


I haven't done #6 yet, but can the solution be obtained by solving this 3x4 matrix below using the Jordan-Gaussian algorithm to put it in reduced row echelon form?

1 -1 1 14
1 1 1 4
9 3 1 10

[jim360]

Everything is trivial when you know how to do it.

For #6 I think you start by setting up 3 simultaneous equations then doing with them how you please. I imagine the matrix form suggested is neatest but I'd probably pick the Ax=b form and look at it like:

1 -1 1 a = 14
1  1 1 b      4
9  3 1 c      10

And try Gaussian elimination or something like it. SO I'm solving a 3*3 matrix which I believe is slightly easier.

[BFM_Edison]

2, 3, and 4 seem pretty trivial to me, as does 6 if you know how to fit polynomials to specific points.

I think three360 was talking about triviality based on the perception of the average Joe Schmoe IMO.  Why is the rest not trivial to you?

I was going with the ones that I think Joe Schmoe should know and consider trivial the most, of course the key word is should but >.>

[BFM_Hydra]

Use the reasonable person test!

[Titanium»X«]

Personally, I think it's easier to just RREF a a 3x4 matrix than solving for Ax=b

I RREF'ed

1 -1 1 14
1 1 1 4
9 3 1 10

and got

1 0 0 2
0 1 0 -5
0 0 1 7

I think it's right.

[BFM_Edison]

I refereed it and checked the points for my original answer, which I think is the same as yours.

[Titanium»X«]

I refereed it and checked the points for my original answer, which I think is the same as yours.

Yup, it appears so.

[conrad9600]

what is a differential equation???????!!!!!!!!

[jim360]

A differential equation is pretty much exactly what it says it is - an equation invloving differentials. So I expect you're really asking what a differential is. I'll try to answer that with a practical example rather than mathematically.

Suppose you have a hot cup of tea and you're interested in how long it will take to cool down. For that information you need to know how hot it is, how hot the surrounding environment is and how fast heat moves from one to the other. But it's not unreasonable to assume that the hotter the tea is, the faster heat moves from it to the room. Hence not so much the time taken to cool is important but the rate at which the tea cools down. This depends on the difference between the tea's temperature, T, and the room's temperature, R, and importantly it is not a constant because hot things cool down more rapidly that colder things. Since at all times the quantity (T-R) is changing, the rate of cooling is given by d(T-R)/dt where d/dt is a differential and expresses the fact that conditions are always in a state of change. It's then possible to solve the resulting equation (time taken to cool is proportiional to d(T-R)/dt ) using the rules of calculus.

Differentials come from the problem of trying to solve problems that involve curves when it's not possible to use straight lines, rectangles and so on. Then the first step was to approximate to a straight line and then find better and better approximations, when the exact answer comes from considering a sraight line with no length. Kind of confusing I know, but that's sort of what happens although it's a little more complicated than that.

[TUR80]

1) How many unique 10-character password exist, if valid characters consist of only lowercase letters and numbers, and at least one character in the password must be a number?

ok thats a math c question
26+10 is 36 possible characters taking into account the number
35x34x33x32x31x30x29x28x27

[BFM_Kiwi]

I don't think that's right.  35x34x33... assumes you only use each letter/number once.  But you could have aaaaabbbb9 where the a's and b's repeat.

I think it's more like

10 (all the numbers) * 36 (letters/numbers) to the ninth power  = 10*36*36*36*36*36*36*36*36*36

or does the answer involve factorials...hmmm...

[BFM_Crimson]

1) How many unique 10-character password exist, if valid characters consist of only lowercase letters and numbers, and at least one character in the password must be a number?

I believe that it would be 10x36x36x36x36x36x36x36x36x36
The x's mean multiply (I never liked *).
Eeach number there (there are 10) represents the sample space (don't quote me on correct use of terms and definitions).

The password must contain at least one number, meaning there are 10 to choose from (0,1,2,3,4,5,6,7,8,9).

Then the rest can be either a lower case letter or a number, meaning there are 36 to choose from (q,w,e,r,t,y,u,i,o,p,a,s,d,f,g,h,j,k,l,z,x,c,v,b,n,m,1,2,3,4,5,6,7,8,9,0).

Therefore, the answer is 1015599566684160, or one quadrillion, fifteen trillion, five hundred and ninety-nine billion, five hundred and sixty-six million, six hundred and eighty-four thousand, one hundred and sixty

[TUR80]

I don't think that's right.  35x34x33... assumes you only use each letter/number once.  But you could have aaaaabbbb9 where the a's and b's repeat.

I think it's more like

10 (all the numbers) * 36 (letters/numbers) to the ninth power  = 10*36*36*36*36*36*36*36*36*36

or does the answer involve factorials...hmmm...

ur right
i was thinking non replacement

[BFM_Edison]

This topic is old, and I am far too lazy to go back and look for what answers were posted. The best way to go about this is to find all without the restriction on at least one number and then take away all those that don't have a number. That is, you have 36^10-26^10 = (36^5-26^5)(36^5+26^5) = 48,584,800x72,347,552 = 3,514,991,344,409,600. The factoring was an attempt to not get errors due to tenth powers being too large, but I needed Mathematica anyways. The problem with just trying 10x35^9 is that you restrict where you are placing the number. This would seem to multiply your value by 10 again for each of the places it can be placed, but then you're counting some possibilities more than once. For example, 12b1 changed to 2b11 gives you something you already would have counted. Thus in cases where you have "at least", you want to start with total cases and subtract those away that don't satisfy the "at least" factor.

[BFM_Crimson]

This topic is old, and I am far too lazy to go back and look for what answers were posted. The best way to go about this is to find all without the restriction on at least one number and then take away all those that don't have a number. That is, you have 36^10-26^10 = (36^5-26^5)(36^5+26^5) = 48,584,800x72,347,552 = 3,514,991,344,409,600. The factoring was an attempt to not get errors due to tenth powers being too large, but I needed Mathematica anyways. The problem with just trying 10x35^9 is that you restrict where you are placing the number. This would seem to multiply your value by 10 again for each of the places it can be placed, but then you're counting some possibilities more than once. For example, 12b1 changed to 2b11 gives you something you already would have counted. Thus in cases where you have "at least", you want to start with total cases and subtract those away that don't satisfy the "at least" factor.

I was waiting for you to say that.