Conference rusure::math

Here's a few - I'm not sure about the ratings, but these can all be dealt 
with by ingenuity.

1) Show that every prime > 3 is of the form 6n+-1. (Easy)

2) Which has the larger area, triangle (5,5,6), or triangle (5,5,8)? (Easy,
each is formed by adjoining two (3,4,5) triangles) 

3) Prove that there exists an integer of the form 11...1100...00 (a 
sequence of 1's followed by a sequence of zeros) that is a multiple of 87.
(Harder. Use the pigeonhole principle: consider the 1st 87 numbers of all 1's. 
Either one is divisible by 87, or two leave the same remainder. Subtract 
them.)

4) Which is larger, e^pi or pi^e? (Hard, unless you have a good pocket 
calculator available.)

    
      RE: .1
    
      Spoiler:
    
               2) the same area (I had to draw it out!!) 8^)
    
               4) e^pi > pi^e  (~23,~22)
    
                                                       Shawn L.

    The HARD problem in .0 is not very hard.  Realize that the centers
    of the balls form a regular tetrahedron with edges of length 2.
    This tetrahedron has height sqrt(8/3) (by three applications of
    the Pythagorean theorem), so the center of the top ball is sqrt(8/3)
    inches above the centers of the others balls.  So the bottom of
    the top ball is sqrt(8/3) inches above the bottoms of the other
    balls, and hence the table.

>4) Which is larger, e^pi or pi^e? (Hard, unless you have a good pocket 
>calculator available.)

    This is a pretty good one!  I'd guess that if b > a >= 1, then b^a > a^b
    (based on 3^2 > 2^3), try to prove this conjecture, realize that
    the correct conjecture is that b > a >= e implies a^b > b^a, prove
    this, then assert pi > e, and so pi^e > e^pi.

    Here is a problem that can be added to your list.  You can decide
    how difficult it is.
    
    "There are six points in space. No three points are collinear.
     The points are connected by linear segments. The segments
     are colored at random by two colors.  Prove that there is a
     triangle made up from the segments of the same color."
    
    
    
    
    

    Write pi = e*(1+d).

    We get:

	pi^e = (e*(1+d))^e = e^e * (1+d)^e			[1]

	e^pi = e^(e*(1+d)) = e^e * e^(e*d)			[2]

    expanding (1+d)^e in a series we get

	1+ d*e + d^2/2*e*(e-1) + d^3/6*e*(e-1)*(e-2) + ...	[1a]

    expanding e^(ed) in a series we get

	1 + d*e + d^2/2*e*e + d^3/6*e*e*e + ...			[2a]

    comparing term by term we find that [2a] dominates [1a], so

	e^pi > pi^e

    - Jim

    Very good, Lynn.
    
    But are you quite sure you got ALL the answers?  Every last one?
    
      John

Well, I was too tired yesterday to think about negative radices, but this
morning I see that radix = -13, corresponding to X=6, gives 
	[10] (x+1)(x+6) = x^2 + 7x + 6 = x^2 + 7x -(-13+7) =
	[-13] x^2 + 7x -17,
and there are other negative radices [X=5,4,...] that also work. [X=7 or 8
give radices in which 7 is not a digit.]

Did you have something else in mind?

Lynn 

    Nope, that was it.  -13, -17, and -19 are the three negative radices.
    
    Not every high school or college student has had exposure to negative
    radices.  This may be a poor way to spring it on them.  Perhaps
    there might be a separate problem earlier...
    
      John

I think problem 1 of note 227 is appropriate (moderate difficulty). I still 
don't know of a solution to problem 2 in that note - constructions in 3-d 
geometry are frequently very difficult.

It's pleasant to observe how a number of relatively simple problems in this 
conference have led to generalizations or extended problems of considerable 
difficulty.

Lynn Yarbrough 

    10! = 3628800 has 2 final zeroes.
    How many final zeroes are there in 100! ?

     How many black birds were baked in a pie?
     
     Dan

The number of trailing zeroes in n! is the exponent of 5 in the expansion 
of n! as a product of primes. For any prime p the exponent of p in n! is

	sum([n/p^i], i = 1...)	([x] = integer part of x)

which in the present case is

	[100/5] + [100/25] + [100/125] + ...
	= 20 + 4 + 0 ...

Lynn Yarbrough 

    It's OK to let somebody else solve the easy ones.  Really.
    
      John

	I can't recall what the precise accuracy of Stirling's stuff is,
though I can look it up.   Question is, is it accurate enough to complement
the sort of mod 10^n argument (or perhaps 2^m is most promising, to round
the Stirling guess to the nearest feasible number?

Questing of Valbonne

     re .-1
     
     Is your question whether you can take Stirling's
     approximation for n!, and combine it with the number of
     zeroes in n! (using the formula given earlier), to get an
     exact answer?
     
     I would guess not.
     
     Anyway, here is another problem, not hard.  Prove that for
     positive integers a, b, c, and n, with n > 2, that it is not
     possible to have
     
                     a    b    c
                    n  + n  = n
     
     Dan

>                    -< your proof should fit in the margin >-
>     Anyway, here is another problem, not hard.  Prove that for
>     positive integers a, b, c, and n, with n > 2, that it is not
>     possible to have
>     
>                     a    b    c
>                    n  + n  = n
>     
>     Dan


Simple.  Let n=3, a=1, b=2, c=3.

We therefore have

		3**1 + 3**2 = 3**3

or

		3 + 9 = 27

which is false.

qed

    That's proof that our school system is inadequate.
    
    Dan was asking for a proof that no possible combination (a,b,c,n>2)
    of integers satisfied the equation.
    
      John

BTW, the e^pi problem was discussed back in note 515.