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#101 Re: Puzzles and Games » square numbeers forever. » 2012-12-28 22:51:23

scientia

You didn't expect that, did you? tongue

#102 Puzzles and Games » Mandy and Sandy » 2012-12-28 22:34:38

scientia
Replies: 12

This is my adaptation of a math problem I came across elsewhere on the Internet. smile


Mandy has just done her shopping at the shopping center; now she is hungry and wants to go to the restaurant for lunch. Sandy has just had something to eat at the restaurant; now she wants to go to the shopping center and buy some clothes. A park is situated between the shopping center and the restaurant. Mandy sets off from the shopping center at 2:00pm, walks through the park, and arrives at the restaurant at 2:30pm. Sandy leaves the restaurant at 1:43pm, takes the same route through the park (but in the opposite direction) and gets to the shopping center at 2:23pm. Both girls enter the park at the same time and Mandy leaves the park 4 minutes earlier than Sandy.

First question (easy): Find the time (to the nearest minute) at which both girls meet.

#103 Re: Help Me ! » abstract algebra » 2012-12-28 21:57:07

scientia

Hint: Any set containing an uncountable subset is uncountable.

#104 Re: Euler Avenue » Wilson's Theorem » 2012-12-27 23:35:59

scientia

Yes.

#105 Re: Euler Avenue » Wilson's Theorem » 2012-12-27 22:56:14

scientia

Right. A corollary of Fermat's little theorem is that if p is a prime and a is an integer coprime with p, then there is an integer b such that ab ≡ 1 (mod p).

Proof: Take b = a[sup]p−2[/sup].

Moreover, by the division algorithm, we are always guaranteed to find a unique b in the range 1 ≤ b ≤ p−1 such that a[sup]p−2[/sup] ≡ b (mod p).

In the language of group theory, we say that b is a multiplicative inverse of a modulo p. This is the result I was using in my proof. smile

#106 Re: Euler Avenue » Wilson's Theorem » 2012-12-27 22:03:12

scientia

Sorry, I thought you understood group theory. sad


Are you familiar with Fermat's little theorem? This states that if p is a prime and a is coprime with p, then a[sup]p−1[/sup] ≡ 1 (mod p).

#107 Re: Puzzles and Games » square numbeers forever. » 2012-12-27 12:18:59

scientia

#108 Re: This is Cool » What is the point (ordered pairs) of Graph of a function? » 2012-12-27 07:30:59

scientia
21122012 wrote:
bob bundy wrote:

Next question?

Bob

You wrote everything correctly but a question didn't answer.

Is that the way you want to see your "answer" being written?

#109 Re: Help Me ! » Mixing Word Problems & Distance Word Problems » 2012-12-26 21:30:49

scientia
demha wrote:

Hi Scientine:
I was taught how to do it using d = rt but wasn't understanding it right.

Now for what you shows me, am I supposed to substitute C for the current (which I got as 4) and then continue to solve for B?

Yes, you need to solve for b because the question is asking for the boat's speed in still water: in other words, what boat's speed would be if there were no current at all. The question isn't asking for the speed of the current (so your answer was not correct).

Bob bundy has created two tables for you for these two questions. Have a look. I think you might understand the questions better from the tables. smile

#110 Re: Help Me ! » Mixing Word Problems & Distance Word Problems » 2012-12-26 18:26:41

scientia
demha wrote:

Hi Scientina:
I'm not too sure how t continue. Could you please help me a little on that?

What are you not too sure of? The two problems are similar, and I did almost all of question 4 for you. Just apply the concept speed = distance / time.

#111 Re: Help Me ! » a challenging problem for all » 2012-12-26 15:39:50

scientia

I found some more errors and corrected them. And now I'm sure everything up there is perfect.

The challenge now is to compute the actual integral. Not for the faint-hearted. faint

#112 Re: Help Me ! » a challenging problem for all » 2012-12-26 14:20:34

scientia

I made some careless mistakes in my calculations and got some horrendously frightening values for the elevating angle.

I have re-computed them. They should be

and

And you want to find the following integral for the solid angle:

#113 Re: Help Me ! » Mixing Word Problems & Distance Word Problems » 2012-12-26 00:37:37

scientia
demha wrote:

4.
Q. A boat travels 60 km upstream (against the current) in 5 hours. The boat travels the same distance downstream in 3 hours. What is the rate of the boat in still water?

A. The rate is 4 current.

Let b be the velocity of the boat in still water and c be the speed of the current. Then you have

Solve the two simultaneous equations for b.


demha wrote:

5.
Q. A duck can fly 2400 m in 10 minutes with the wind. Against the wind, it can fly only 2/3 of this distance in 10 minutes. What is the rate of the wind?

A.
How would I do this?

As above you must set up two simultaneous equations and solve them. Let d be the velocity of the duck in still air and w the speed of the wind. The first equation is

I leave you to derive the second equation and solve the them for w.

#114 Re: Euler Avenue » Wilson's Theorem » 2012-12-25 21:05:01

scientia


The set
, where
is prime, forms a group under multiplication modulo
.

#115 Re: Euler Avenue » Infinite Descent » 2012-12-25 21:00:58

scientia

Do you understand it?

#116 Re: Euler Avenue » Wilson's Theorem » 2012-12-25 17:53:55

scientia

Okay, Wilson's theorem.


Suppose
is prime. We first note that
.

Consider the multiplicative group of

modulo
, where
is odd. We have
and
. Conversely, if
and
, then
divides
divides
or
or
. Hence only 1 and
are their own multiplicative inverses modulo
.

It follows that

. (Each number in the list
multiplies by another number in the list to ≡ 1 (mod p).)

Conversely suppose

and
. Then
for some integer
. Let
be a prime divisor of
. If
were not prime,
would be smaller than
and so would divide
and hence would divide
, a contradiction. Thus
must be prime.

#117 Re: Puzzles and Games » What words can you make with julianthemath? » 2012-12-25 13:43:12

scientia

Go to http://www.letterpresscheat.net/.

Type "julianthemath" in the box and click SEARCH.

Note that it doesn't give proper names.

#118 Re: Help Me ! » a challenging problem for all » 2012-12-25 10:30:27

scientia


I think you mean
and also you want
.

Supposed the trajectory is angled

horizontally from the vertical plane through the y-axis, taking
to be positive to the right and negative to the left of the y-axis. The straight line on the ground through the origin making this angle with the y-axis intersects the circle at two points. The distances from the origin to these two points can be calculated; these are the minimum and maximum ranges respectively at the angle
. Let
be the vertical angle from the ground at which the particle must be projected to cover the minimum range, and
be the vertical angle to reach the maximum range. (This can be computed from the formula
where
is the acceleration due to gravity and
the desired range.) The maximum value of the horizontal angle
is
.

Then the solid angle you want is:

#119 Re: Euler Avenue » Infinite Descent » 2012-12-25 06:36:33

scientia
Agnishom wrote:

Suppose, w and v are relatively prime to each other
and


The theorem claims that there exists integers x and y such that
and

Example, 10^2 = 25*4 where (25, 4) are relatively prime to each other now see that 25 = 5^2 and 4 = 2^2


If
, take
and
. Otherwise, since
and
are coprime, each prime divisor of
must divide
exactly as many times as it divides
, namely a multiple of
times. Thus
, being a product of primes of power a multiple of
, is a power of
. Same for
.

#120 Re: Help Me ! » Triple integration » 2012-12-25 01:13:15

scientia

Write the equation as

This is an ellipse with semimajor axis

and semiminor axis
. Its area is
. Integrate this from
to
to get the volume of the solid.

The surface area is a bit more complicated …

#121 Re: Help Me ! » solid angle » 2012-12-22 14:42:25

scientia

Well, the polar co-ordinates of the ellipse is

so you're looking at a right-angled triangle with height

and base
. The solid angle is

where

.

#122 Re: Help Me ! » Geometric progression » 2012-12-22 11:27:09

scientia

Something is wrong here. 1, 1, 1 are in geometric progression (r = 1) but

#123 Re: Euler Avenue » Wilson's Theorem » 2012-12-22 10:50:49

scientia


The way to prove that
is divisible by
for composite
is as follows.

If

is composite and greater than 4, then
can be written as a product of an integer greater or equal to 3 and an integer greater or equal to 2, i.e.
where
and
.

Now I claim that

. Proof.

+ rearranging gives
as claimed.

This means that the factors of

contain a sequence of
consecutive integers and a nonoverlapping sequence of
integers. The first sequence contains a factor of
and the second sequence contains of factor of
. So the product of those integers is divisible by
. But the two sequences of integers are nonoverlapping and so
is divisible by the product of those integers. Hence
is divisible by
.

#124 Re: Computer Math » 2,3,4 triangle. » 2012-12-18 02:17:02

scientia

http://www.bmoc.maths.org/solutions/bmo1-2013/

Problem no. 5 there.

#125 Re: Puzzles and Games » Sum of 3 digits and its compliment to 1000 » 2012-12-17 22:10:14

scientia

In general, if you have an n-digit number

and

is the n- or (n−1)-digit number that when added to
makes
, then:

If

is the smallest integer such that
,
, then
,
, and
,
.

Thus the sum of digits is

.

Example:

This is equal to

(8 is the no. of digits of N and 4 is the lowest-placed nonzero digit of N).

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