Math Is Fun Forum

hi Superlynx

In post 272 I just took your line

and inserted multiplication signs.

aei means 'a' times 'e' times 'i'

So it was just that you had misunderstood the meaning of the formula.

Hope that clears it up for you.

By all means post a geometry question, but can I suggest you start a new thread with a title that fits that problem?  This will make it easier for me and others to pick it up and help.

Bob

hi Lazernugget

If you want to combine maths with astronomy here's a suggestion that shouldn't cost you too much.

"A field guide to the stars and planets" by Donald Menzel. 

I bought this book 1966 edition but it's been re-printed lots of times and seems to be available second hand for a few cents.

It has accurate diagrams of the visible planetary orbits along with the data you need to calculate which one(s) will be in the night sky and in what constellation.  I find it very satisfying to work out what I will be able to see on a given date, and then go out and confirm it by making the observation.  A wonderful combination of mathematical modelling and science in action.  You can also automate the calculations by setting up the formulas in a spreadsheet program like Excel.

Every year Nortons Star Atlas is published.  It too is full of useful stuff for the star gazer.  Couldn't find a 2011 edition on-line (surely it's out there somewhere?) but 2010 certainly exists.  Worth a search for the latest I think.

later edit:  Looks like it isn't published every year.  My mistake.  So 2010 is the latest.

http://www.amazon.com/Nortons-Star-Atlas-Reference-Handbook/dp/0131451642/ref=cm_cr_pr_pb_t/183-5354459-7109922

This will cost you a bit.

Bob

hi rcwitt

To add or subtract matrices, they must be the same 'shape';  ie. same number of rows and same number of columns.

So the two you give in your post cannot be added.  (first is 3 by 2; second is 2 by 3)

To multiply the number of columns in the first matrix must be the same as the number of rows in the second.

So if first is m by n  and the second is p by q then they may only be multiplied if n = p.

And, because of the way matrix multiplication works, the answer is m by q.

Your example ( 2 by 2 and 2 by 3 ) is possible with the result 2 by 3.

Bob

hi 1a2b3c2212

I did trial and error and got no solutions either.

A few possibilities can be immediately eliminated because of prime factor type rules and the rest by slog through the alternatives.

Bob

hi bobbym,

How are you?  I agree with your answer. 

Bob

hi jmusbach,

I've not used the 83+

Don't know if this will help ... it's for the 86

http://www.ticalc.org/archives/files/fileinfo/145/14573.html

Bob

hi Anakin,

Yes, I got 5 too.  And no other solution.

The way I see it is this.

Given some vectors, you can make new vectors by making linear combinations from them.  Not necessarily every random vector can be made as a linear combination, so those that can are 'special'.  They are the ones that are 'in the span' of the start vectors.

So if you are given a new vector how do you know whether it is in the span?

Well just try to form it as a linear combination and see if you can.  It may not be possible in which case it isn't in the span.  For instance, just pick a = 100 in the example and you cannot find r, s, and t at all.

But the question says it is in the span so plough (plow) through the equations and see what drops out.  Only if a = 5 does a combination become possible so that's the value of 'a' and it's unique.

I was going to add that linear combinations are always unique but then I realised that's not true.

eg. v1 = ( 1, 2, 3 )  v2 = ( 2, 4, 6 )  then y = ( 3, 6, 9 ) = 3 x v1 = 1.5 x v2, so that's two different combinations for a start. 

You may also be able to see that y2 = ( 3 , 6, 1001 ) cannot be made as a linear combination .... only vectors parallel to v1 or v2 can be, because of the components I chose for these two base vectors.  (I made them parallel).

I expect you are about to learn about vectors that can span the whole set of vectors and are therefore called a basis for the
space.  In the example, v1, v2 and v3 are not a basis.  You should  be learning soon why that is and what to look for to guarantee a basis.

Bob

hi shocamefromebay

Suppose you have found all the ways of using A, B, C, D, and E in some order.

There are 5! = 5x4x3x2x1 = 120 of these.

But only one will be in the correct ( alphabetical ) order so dis-allow 119 possibillities.

But this will be true for any five letters.

eg B, Z, Y, T, and P can only be allowed if the order is BPTYZ, again that is 119 ways, with these letters, dis-allowed.

So to get your answer, take the number of ways (without order considerations) and divide by 120; that should give you the result you want.

Bob

Got some.

Just added "200 digits" ( in quotes ) to my search.

Site at bottom of post so you can find it yourself if you want.

But you'll also have to prove the product has 400 digits, not 399 or 401.

(note.  100 x 100 is 5 digits, but 900 x 900 is 6 digits, so you cannot just assume it's ok to add the digits 200 + 200 = 400 )

Bob

.......................spoiler below ...........................

http://primes.utm.edu/lists/small/small2.html

ps.  These are considered 'small' !!!

And if you Google 'list prime numbers' you will find there are sites that do exactly that although I didn't find 200 digits that easy, because there seem to be sites that give the first n primes and sites that give very large primes (over 10000 digits, say) and not much in between.  You may need to tweak your search and that's a problem solving skill ... hence the nature of the task.

Best of luck,

Bob

hi Charlie,

So what are you allowed to 'know'?

eg.  If you can find two 200-digit primes those would probably meet the criteria, so are you allowed to look at a list of big primes, or use one from a known subset of the primes?

And are you hoping to do the multiplication?,  because that will need  mathematical software.

Or, is this a question requiring an analytical, number theory, type answer?

Bob

hi AISSAISHAK

By 'solve' do you mean 'simplify'?

Extra brackets not needed

Or did you mean:

Let's try both

(i)

so

or

(ii)

so

Oh! That was a surprise!

Bob