disagree with your answer to the cone problem,
rmilson. The answer I obtain is
<br><br>s=3*(1+sqrt(2))*sqrt(3-sqrt(2)) [9.1205, roughly] <br><br>I believe that
you are
mistaken in assuming that two windings allows you to
consider a sector with an included angle of 2*pi/3. If you
extend your argument to six windings, the answer would
turn out to be zero, which is clearly wrong.
<br><br>Without a diagram it is difficult to describe my
approach, but I will try. We agree that the cone can be
unwound into a sector with an included angle of pi/3 (60
deg) and a radius of 6 units. Along one of the radial
edges and the vertex mark off the points P, M, and V as
given. Mark off the point P' where the circular arc
intersects the other radial edge and join P and P'. You now
have an equilateral triangle PVP'. Parallel to the
base PP' and through M construct a line which
intersects the other radial edge at M', making a second
equilateral triangle MVM' whose side length is half that of
the triangle PVP'. Anywhere between M and P mark off
a point Q on the relevant radial edge. Find Q' on
the other radial edge so that QQ' is parallel to PP'
(and MM'), yielding a third equilateral triangle QVQ'.
Let the perpendicular distance between PP' and QQ' be
H1, and that between MM' and QQ' be H2. Let the
straight line distance QP' be S1 and that of MQ' be S2.
You can find S1 in terms of H1 and S2 in terms of H2.
The total distance is then S=S1+S2. <br><br>I won't
do the math here (it's a little tricky), but you
must find H1 (and H2) subject to: <br>(1) H1+H2 must
equal a constant value [3*sin(pi/3)=3/2*sqrt(3)]
<br>(2) QP' is parallel to MQ' <br><br>Another approach
is to use only condition (1) to find S in terms of
H1 (or H2) only, and to minimise S. The answers are
the same. <br><br>P_Jr

a) The four masses are 1, 3, 9 and 27 kg. It is
easy to see how to get 1, 3, 4, 9, 10, 12, 13, 27, 28,
30, 36, 37, 39 and 40 kg from these on one pan of the
balance scale. The remaining masses are obtained by
placing (a) suitable compensating weight(s) on the
opposite pan. E.g. 2 kg is obtained with the 3 and 1 kg
weight; 22 kg is obtained with the 9 kg weight on one pan
and the 1, 3 and 27 (= 31) kg weights on the other;
32 kg is obtained with the 9 and 27 (= 36) kg
weights on one pan and the 1 and 3 (= 4) kg on the other
pan. More abstractly, any whole number between 1 and
40 (inclusive) can be expressed as a combination
(sums and/or differences) of some or all of these base
values. Can you prove that this scheme (of integral
powers of 3 as base values) is indefinitely extendable
to include all whole numbers? <br><br>(b) The first
weighing of any 9 balls against any other 9 will identify
which of the 3 sets of 9 balls contains the lighter
one. From this set take any 6 balls and weigh 3
against 3, which in turn will identify which set of 3
balls contains the lighter one. From this set take any
2 balls and weigh 1 against 1. The last weighing
identifies which one it is. A similar, but somewhat more
difficult, weighing problem with a balance scale which you
may be familiar with is this: you have 12 balls of
which one is either lighter or heavier than the rest.
In 3 weighings find the odd ball (oddball?) and
identify whether it is lighter or heavier. <br><br>(c)
First we note that x^n+1/(x^n) with x&lt;&gt;0 is an
integer for all x, namely 2 (=1+1). Also, x is such that
the expression is an integer for n=1. Thus, it is
true for n=0 and n=1. Suppose now that it is true for
n=k-1 and for n=k, i.e. x^(k-1)+1/(x^(k-1) and
x^k+1/(x^k). Then for n=k+1 the expression becomes
x^(k+1)+1/(x^(k+1))=(x+1/x)*(x^k+1/(x^k))-{x^(k-1)+1/(x^(k-1))}, of which all
three terms are integers (on the
assumption that it is true for n=k and n=k-1), so that the
whole expression is "integer times integer minus
integer", which is clearly integer. Therefore, since it is
true for n=0 and n=1, it is true for n=2. Since it is
true for n=1 and n=2, it is true for n=3, etc. ad
infinituum, and hence true for all positive integers n. In
fact, it is true for all integers (negative, zero and
positive). Can you prove this assertion? It is quite easy.
<br><br>P_Jr

y last problems are 3a,b,c NOT 2a,b,c. I didn't see the previous postings...
sorry!

.<br>(a) With a set of balance scales and only
four masses it is posible to weigh out any whole
number of kilograms from 1 to 40. What are the values of
the four masses?<br><br>(b) You have 27 steel ball
bearings (don't ask me why!) and a set of balance scales.
One of the ball bearings has a hole in the middle (so
it is lighter than the others). How can you
determine which ball this is, with only three
weighings?<br><br>(c) Suppose that x is a real number such that x+(1/x)
is an integer. Prove by induction on n that
(x^n)+(1/(x^n)) is an integer for all positive integers n.

get the answer 3*sqrt(7)<br><br>A right
circular cone is obtained by taking a sector of a disk and
gluing together the opposite edges. The radius of the
resulting cone is equal to the radius of the original
circle times the proportion used to obtain the sector
(i.e. sector circumference div circumference of the
circle). A tight string on the cone is a shortest path
between the string's endpoints, and therefore corresponds
to a straight line once the cone is unfurled.
<br><br>The givens mandate that the cone's circumference is
1/6th that of the circle, i.e. an angle of Pi/3. Since
a double winding is desired one needs to consider a
sector corresponding to an angle of 2*Pi/3. The path of
the string in question is obtained by taking a radial
midpoint of a circle of radius 6 and joining it by a
straight line to a point on the circumference that is
2*Pi/3 radians away from the direction of the starting
midpoint. Thus one is dealing with a triangle whose sides
are 6,3 with common angle 2*Pi/3. The law of cosines
immediately yields the length of the opposite side: 3*sqrt(7)

his sat. evening I would like for us to get
together in our chat room to discuss any math problems or
any thing for that matter. I will be online about 9
PM EST this SAT. I am going to start putting event
on the calendar do check it every once and a while.
And If you have any times or dates that you want
specifically. Just let me know and I will do what I can to make
an event at that time. Thank
you<br><br>Puzzle_Hacker(Wes)

he radius of the base of a right circular cone
is 1. The vertex of the cone is V, and P is a point
on the circumference of the base. The length of PV
is 6 and the midpoint of PV is M. A piece of string
is attached to M and wound tightly twice round the
cone finishing at P. What is length of the string?

ot so fast next time! I was just posting a
preliminary idea - don't really want an answer until I've had
time to think about it and get really stuck. Thought
it might help others out/be of interest.<br><br>No
worries though.<br><br>I will try and post some problems
when I have time.

f you think about the expression that states the
mean of any set of numbers is greater than or equal to
its geometric mean <br>(a+b)/2 &gt;= sqrt(a*b) true
for all real a,b<br>this has been proven and is
fairly easy to prove if you would like to try.<br>this
is similar to your way al_jabr<br>(1+999)/2=500
&gt;= sqrt(1*999)<br>(2+998)/2=500 &gt;=
sqrt(2*998)<br>(3+997)/2=500 &gt;= sqrt(3*997)<br>...<br>(499+501)/2=500 &gt;=
sqrt(499*501)<br>sqrt(500) &gt;= sqrt(500)<br>now if you muliply both sides
together you get<br>500^498*500^(1/2) &gt;= sqrt(999!)
<br>sqare both sides<br>500^998*500=500^999 &gt;= 999!

ou can't assume anything but here is how you
finish<br>(Y*X)*X<br>((Y*Y)*X)*X, BY PROPERTY 1<br>((Y*X)*Y)*X, BY PROPERTY
2<br>(Y*X)*(Y*X), BY PROPERTY 2<br>Y*X, BY PROPERTY 1<br>you got
most of it though

fairly simple solution(?) - not requiring
Stirling's Formula or the like:<br><br>Consider
999!/(500^999)<br><br>=(1*2*3*...*999)/(500^999)<br><br>or
rearranging<br><br>=[(1*999)/25000]*[(2*998)/25000]*...*[(499*501)/25000]*[500/5\
00]<br><br>Each term (except for the last term (which is just 1)
is of the
form<br>[(500-n)*(500+n)]/25000=(25000-n^2)/25000<br>where n is between 1 and
499. Each of these terms is
obviously less than 1 so the whole expression must be less
than 1 and thus the numerator 999! must be less than
the denominator 500^999.

xpense of being pendatic, it is usual notation to use
just three dots to indicate the continuation of any
list etc. in maths - and elsewhere - it is known as an
ellipsis)<br><br>As can be seen from Prometheus' answer, the average
of ANY even number of terms is -1/2 and any odd
number of terms, say N, is (N+1)/2N.

*Y=(X*X)*Y, by property 1<br>=(X*Y)*X, by
property 2<br>=(Y*X)*X, by property 2<br>=Y*(X*X), by
associativity<br><br>[You need to assume associativity, or have I missed
something?]<br><br>=Y*X, by property 1.

eturns taking 1min<br>5&10 cross taking 10mins<br>2
returns taking 2mins<br>1&2 cross taking 2mins<br><br>and
now all are across the bridge in just 17mins.

think we should start numbering our problems
from now on so if we want to refer to a problem It
will make it easier.<br><br>Let S be a set and * be a
binary operation on S satisfying the two laws:<br><br>X
* X = X <br>For all X in S,<br>(X * Y) * Z = (Y *
Z) * X <br>For all,X,Y,Z in S<br><br>show that X * Y
= Y * X for all X,Y in S.

n solving the number problem I just simply
labeled letters for the six digits<br> abcdef and fabcde
is one third of it then <br>fabcde<br>
x3<br>abcdef<br>I knew the digit f had to be 1,2, or 3 inorder for
the the answer to be the same number of digits. So
what I did was substitute a digit in for e so that f
could be only 1,2, or 3. those digits are 1,4,7 and
then I just worked the rest out and 1 did not work out
but 4 and 7 did.<br><br>answer:<br>I am assuming that
they had to walk together with the torch. In that case
the minimum amount of time is 19 minutes. but if they
didn't have to walk togegher side by side than the
minimum amount of time would just be 10 minutes but that
makes the problem way to easy so, I don't think that is
the way. Here is how they did it:<br>rates<br>person
1 = 1 min<br>person 2 = 2 min<br>person 3 = 5
min<br>person 4 = 10 min<br><br>lets let person 1 have the
torch at all times<br><br>2,3 ---10
mins-----&gt;1,4<br>1,2,3 &lt;--1 min------- 4<br>2 ---5
mins------&gt;1,3,4<br>1,2 &lt;--1 min------- 3,4<br> ---2
mins------&gt;1,2,3,4 <br>10+1+5+1+2=19 mins total<br>this could be
done in any order as long as person 1 is always
crossing the bridge.<br>I am not sure if this is the
asnwer you are looking for but that is what I can figure
from the problem

our answers to the number problem are correct.
Would you care to show us how you arrived at them?
<br><br>The nuts problem: there are only two ways in which
all three bags can be mislabelled - <br><br>1) "P" =
{w}, "W" = (p,w} and "PW" = {p} <br>2) "P" = {p,w},
"W" = (p} and "PW" = {w}. <br><br>In both cases the
bag labelled "PW" only contains one type of nut, viz.
p or w. So if we look at a nut taken from the bag
labelled "PW" will identify it as either p or w, and in
turn identify which of the above two choices it is.
Thus, if we find a p in "PW", the correct labelling is
"P" = "W", "W" = "PW" and "PW" = "P". If we find a w
in "PW", the correct labelling is "P" = "PW", "W" =
"P" and "PW" = "W". <br><br>Another problem: four
people are walking along a road at night and come to a
bridge. The bridge can only take a maximum of two people
at a time, and they have only one torch. The torch
must always be used when crossing the bridge. The
fastest walker can cross the bridge in 1 minute, the
second fastest in 2 minutes, the third fastest in 5
minutes and the slowest in 10 minutes. What is the
shortest possible time in which all four can cross the
bridge, subject to the rules (torch and max two people)?
How do they do it? <br><br>P_Jr

57142 and 428571 are two answers.<br><br> If you
have three bags, the first one is labeled peanuts, the
second one is labeled peanuts and<br> walnuts, and the
third one is labeled walnuts. Now you find out that all
three bags are mislabeled<br> so you now have to label
them correctly. All you are allowed to do is open one
bag and blindly<br> pull<br> out just one nut. By
doing just this can you correctly label all three bags
of nuts. If so explain how.

..here's another, rather simple, problem to
which there are two answers. Find both: <br><br>A whole
number consists of six digits (base 10). If the last
digit is removed and placed in front of the first
digit, the new number is exactly one-third of the
original number. What is the original number? <br><br>P_Jr

he point of my solution to the f(x)*f(f(x)) = 1
is that this problem, as phrased with the condition
that f(1000) = 999, does NOT have a solution since the
only functions f(x) which satisfy f(x)*f(f(x)) = 1 are
those given - i.e. f(x) = 1/x, f(x) = 1 and f(x) = -1.
Moreover, f(x) = 1/x is continuous at all real x, except x
= 0, where there is an infinite jump discontinuity.
<br><br>The condition f(1000) = 999 is thus in conflict with
the premise that f(x)*f(f(x)) = 1. <br><br>As far as
pi^e vs e^pi is concerned, it escapes me why you wish
to complicate the matter by avoiding calculators -
pi and e are numbers (albeit "special" ones with
their own symbols) just as any other. How would you
"prove" that 2.5^3 &gt; 3^2.5 without in some way
evaluating both sides of the inequality, possibly after some
transformation? Expressing 2.5 as (5/2) merely shifts the problem
into the domain of rational numbers and thus integers:
<br><br>To illustrate: <br>2.5^3 ? 3^2.5 <br>2.5^6 ? 3^5
<br>(5^6)/(2^6) ? 3^5 <br>5^6 ? (2^6)*(3^5) = 2*(6^5) <br>15625 ?
15552 <br>Since 15625 &gt; 15552, 2.5^3 &gt; 3^2.5 -
QED. <br><br>The point is that you still need to
evaluate both sides to establish which is the greater. The
fact that e and pi are transcendental (irrational)
numbers makes no difference to the principle. Thus a
properly functioning calculator is proof that e^pi &gt;
pi^e. <br><br>P_Jr

/x is not a continous function so that is not an
option. The problem is not to find the function itself
but to find f(500) there is a way to show what f(500)
by knowing f(1000)=999 without knowing the what the
function actually is. I have yet to solve it yet either.
Here is what came up with so far. if f(1000)=999
then<br>if we put 1000 into f(x)*f(f(x))=1 then we can see
that f(999)=1/999 and if we put in 999 then we see
that f(1/999)=999 with this I can kind of see two
possible solutions for f(500) one is 1/500 and another
could be 499 but I can not figure out how to prove
neither.<br><br>With the problem Pi^e and e^Pi anyone one can punch it
into a calculator but that is not what it is about I
am still tring to find some way to prove it.
<br><br>But with the rest of the problems I agree with your
answers. good job

tirling's factorial approximation looks like
this: <br><br>n! ~ sqrt(2*pi*n)*(n^n)*exp(-n)
<br><br>which gets asymptotically closer to n! with increasing
n. A better approximation is <br><br>n! ~
sqrt(2*pi*n)*(n^n)*exp(-n+1/(12n)). <br><br>This comes from the Gamma function
(calculus). Need more info? <br><br>A different way of
establishing that 500^999 &gt; 999! is by realising that both
are positive numbers. Take the logs of both
expressions. Then log(999!) =
log(1)+log(2)+log(3)+...+log(999) and log(500^999) = 999*log(500). Using 10 as
the
base for the logs, we find (using a spreadsheet) that
log(999!) ~ 2564.605 and 999*log(500) ~ 2696.271, so that
500^999 is about 4.64*10^131 times as big as 999! is.
<br><br>Any scientific calculator will establish that e^pi ~
23.14, while pi^e ~ 22.46, so that e^pi &gt; pi^e.
<br><br>If f(x)*f(f(x)) = 1, then substituting y = f(x)
yields y*f(y) = 1, whence f(y) = 1/y. Thus, f(x) = 1/x.
Two other solutions exist for f(x), namely f(x) = 1
and f(x) = -1. None of the three solutions can
satisfy f(1000) = 999, so that the problem has no
solution. Am I missing something? <br><br>The average of
the first 200 terms of the sequence 1, -2, 3, -4, 5,
-6, ... is the sum of the first 200 terms divided by
200. Thus, <br><br>Ave =
(1-2+3-4+5-6+...-...+199-200)/200 <br>= ((1-2)+(3-4)+(5-6)+...+(199-200))/200
<br>=(-1-1-1-...-1)/200 <br><br>There are 100 terms in the bracket of the
last line above. Thus, the answer is -1/2.
<br><br>P_Jr

don't have solutions to all of this problems.  That is why I started this club,
to learn.  And to compare solutions with other people.

onsider the sequence<br>  1,-2,3,-4,5,-6,......<br>whose Nth term is
((-1)^(n+1))*n.  What is the average of the first 200 terms of the sequence?

et f be a continuous function that satisfies f(x)*f(f(x))=1 for all real x.  It
is known that f(1000)=999.  find f(500).

am not familar with Stirling aprox for factorials.

hich is bigger, e^pi or pi^e ?

00^999 surprisingly is bigger.<br><br>Using Stirling's asymptotic approx. to the
factorial, we get approx. 500^950 for 999!

hich is bigger<br><br>999!  or  500^999<br><br>explain your answer

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