Math Is Fun Forum

Hi sam123

I think you could assume g^i = g^j where i and j are not equal.  Say i > j.  Times g^i by enough 'g's to make it into the identity and the other by the same.  You've got a contradiction as you now have an power of 'g' below 'n' that is the identity.  So no two are the same.

That means you've got 'n' distinct elements so you got them all => g has generated them all.

Bob

Hi MathsIsFun

I think the picture below is a major strength of the front page.

It's colourful, full of children who look like they're having fun, the pictures relate well to what they link to and it says 'Welcome'!

I think that picture says so much;  I'd say keep it.

Bob

hi

I've not explained myself well enough.  I had that proof but was just surprised that those initial conditions would lead to such a result.  I'm still trying to do (iv)

Bob

Is this the one you meant?  Shave off the beard and put him in modern clothes and ..........

B

I think 'some humans are greedy, some are stupid, some are greedy and stupid and some are none of these (thankfully!)'

Hi NESIC,

Let's call the teachers A, B, C, D, E, and F.  Is the arrangement ABCD different from BDCA, for example?

I'd say YES because, from a teacher's point of view being in the same group of four, but teaching a different section is different.  So this is more like a permutation than a combination.

If you were picking different sweets from a bag, picking ABCD would be the same as picking BDCA if you can eat them in any order, so it would be a combination problem. 

(i)   Choose the first teacher.  6 choices.
      Choose the next.             5 choices.
      Choose the next.             4 choices.
      Choose the last.              3 choices.

So there are 6 x 5 x 4 x 3 ways of doing this.

(ii)    Choose the first.             6 choices.
        Choose the next.     Still  6 choices.
        Choose the next      Still  6 choices.
        Choose the last.      Still  6 choices.

So there are 6 x 6 x 6 x 6 ways this time.

Bob

Ok. I'm still stuck on part (iv) so I'll throw in my attempts at parts (i) to (iii) and maybe that'll inspire someone else to complete the question.

(i)  Consider triangles ATE and ART (vertices in that order)

Angle A is common to both.
Angle ATE = angle ERT (circle theorem .. angle between chord ET and tangent = angle subtended by chord ET)

=> ATE and ART are similar triangles (2 angles the same).

(ii)  note: 3RE = 2RA => 2AE = RA.  The three items of information all => triangle ERD is similar to triangle ARM with scale factor of enlargement 3/2.

proof
angle R is common to both
angle RED = angle RAM (AR cuts parallels in same angle)
=> RC = 2CT and perpendicular height R to AM = 3 times height C to AM
=> area ATE : area ART = 0.5 x AT x height C to AM : 0.5 x AT x height R to AM = 1 : 3

(iii)  I have used a lesser known circle theorem so I'll prove it first.  Refer to diagram below.

A, B, C, and D are any points on the circumference of any circle.
E is the point where AC and BD intersect.

angle ADB = angle ACB (angles subtended by the same chord)
angle DAC = angle DBC (ditto)

=> triangles AED and BEC are similar

=> AE/DE = BE/CE => AE.EC = BE.ED

Now back to the problem, where the letters are as given in the original diagram

EC.CD = RC.CT  => 2/3 AT . 2/3 TM = 2/3 RT. 1/3 RT

=> AT.TM = 1/2 RT.RT

(iv) thoughts

That theorem also works if E is extrenal to the circle and when two points coincide so maybe a starting point is

MT.MT = MD.MR    and AT.AT = AE.AR

Also it is possible to show that angle ART = angle TRM
which makes for a lot of similar triangles

ETC, TMD, RDC, RMT and RTE are all similar

ATE, ART, TDC, ERC and TRD are all similar

AMR and EDR are similar.

that's as far as I've got with this.

Bob

hi Laughsingrun,

Ok, but that's not enough information.  The volume of a prism is base area x height but you only have the hypotenuse. (??)

Now if it is an isosceles triangle, we can deduce that the other sides are each x, [by Pythag]

so  (½ times x times x) times height =

If not isosceles, you'll have to check what else you have been given to find the sides.

Incidentally, bobbym has many great powers to solve problems, but we have no evidence that mind reading is one of them.  It may be that he keeps this skill a secret. 

Bob

hi C25

The method for doing this is to 'force' it into the following format:

Asin(x-y) = Asinxcosy -Acosxsiny (use of compound angle formula; slight complication as you have a + sign between the elements))

so

sinx + 8cosx = 1sinx + 8cosx

= A(1/A  sinx  + 8/A cosx) where A =  sqr root (1 + 64) 

[This use of Pythagoras means there is a 'right angled triangle' with angle y where cosy = 1/A and siny = -8/A. Strictly no such triangle exists because of the minus 8; more usually there is no sign change when using the compound angle formula; but it won't matter as long as you get a 'y' that's over 180 degrees.] 

So, find A by the above Pythag. calculation and y by y = inv tan (8/1); this will be an acute angle; so now find the reflex angle that has cos positive and sin negative; so an angle between 270 and 360 is expected.

Hope that makes sense; I've got some jet lag today; post again if you need more clarification.

Bob

hi giacat,

So hard.  23 ?  If the balance is like the picture below, I can do 12 coins. 

Do we know the weight of a real coin?

Bob

Hi bobbym,

I'm well thanks.  I'm expecting to visit Niagara, then Yellowstone and Grand Canyon soon.  Big adventure for me. 

I'm beginning to think he has difficulty with some of the more subtle nuances of English so he doesn't realise the difference between 'help me',  'puzzles and games' and 'exercises'.  Or the usefulness of starting a new thread with its own topic title.

Oh well, I guess we're all on a learning curve here.  Can you send him an email asking for his native language?  Then we could put out a request for someone who could translate for us.   Just a thought.

bob