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#1 Re: Exercises » Exponent Slash, #1 » 2019-01-06 10:17:04

Strange result.  Not exactly 'fun' but it may make you laugh.

My wife took one look at my algebra and said "It looks like a sequence."

"Of course", I thought, "a geometric series!"

So I set to work:

Let S = a^6 - a^5 + a^4 - a^3 + a^2 - a + 1

then aS = a^7 - a^6 + a^5 - a^4 + a^3 - a^2 + s

adding : S + aS = a^7 +1

So S =(a^7 + 1)/(a+1)

!!!!!

And to think I did all that complicated binomial expanding.

Ho hum.  Back to the drawing board.

Bob

#2 Re: Exercises » Exponent Slash, #1 » 2019-01-05 12:41:47

Q2.

To make it simpler to type I'll replace 7^7 with A

The problem becomes (A^7+1)/(A+1)

= (A+1)^7/(A+1) - (7A^6 + 21A^5 + 35A^4 + 35A^3 + 21A^2 + 7A^1)/(A+1)

=(A+1)^6 - 7A^5 - 14A^4 - 21A^3 - 14A^2 -7A^1

= A^6 - A^5 + A^4 - A^3 + A^2 - A + 1

I thought I had this factorised but, in typing, I've spotted a mistake so I'll stop for now and sleep on it.

Later edit

I have used WolframAlpha and there is factoisation but it is not simple.  To check the algebra I substituted A = 2 and then subsequently A = 7, at the start and at the end of this simplification.  I got exactly the same value at each end of the algebra.  The chances of this happening when there is, in fact, an error, is very small.

It is of course possible that this changes when A is replaced by 7^7, so I'll keep thinking.

5 minutes later edit.

Replacing A as suggested gives 7 Prime factors, the lowest of which is 197.  So, many numerical solutions are possible.  The question wording IMPLIES that this is not what is expected but rather a non evaluated solution that 'drops' out of the working somehow.  Still thinking.

Bob

#3 Re: Help Me ! » hexagon’s » 2019-01-05 06:17:00

Hi Strangerrr,

Cannot provide a diagram but hopefully you can sketch this.  Starting at the top and going round clockwise, call the points A, B, C, D, E, and F.  Then the height is AE = 14.

Triangle AEF is isosceles with angles 120, 30 and 30.

To make a right angle draw a line across from F to C.  Where FC cuts AE call this point G.  Then triangle AGF is 30, 60, 90 with AG = 7.

So use trig to calculate AF.

Bob

#4 Re: Exercises » Exponent Slash, #1 » 2019-01-04 11:33:25

Q1.

300^3 + 1 = 301^3 - 3x300^2 -3x300 = 301^3 -3x300(300+1)

=301^3 - 30^2 x 301

Divide by 301

301^2 - 30^2 = (301 - 30)(301 + 30) = 271 x 331

Bob

#5 Re: Help Me ! » Vectors and horses » 2019-01-04 11:09:20

Hi Mathegocart,

I haven't got access to my geometry program at the moment but 115 looks ok with a rough sketch. The parallelogram of forces has a long side for the horse so the human has to have a southward component to change the angle to 51.

For the second one the orthogonal vector will have the form (1 , 3).  Form an equation using lambda and mu

(-2 , 5) = lambda(3 , -1) + mu(1 , 3) and solve for the unknowns.

Bob

#6 Re: Exercises » Exponent Slash, #1 » 2019-01-04 06:49:28

Hi Mathegocart

Yes, as in Bob Bundy.  Someone once said that joining twice was against the rules but it isn't.  I wouldn't encourage everyone to do it ... sometimes spammers who've got banned do it so it is suspicious.

If you look up my earliest posts using this name you'll find my reasoning in a problem about a professor of logic. Then, some years later I was asked whether only moderators could do a particular thing and it was helpful to have a non mod account to test it out.

Today I have a reason for using alter ego which I don't want to disclose publicly.  It's to do with forum security.  Please don't speculate.  In a week ask again and I'll explain in a private message.

Q5 has me stumped at present.  Pick any two numbers and they will be in GP.  But the next term may not even be an integer so I'm stuck as to how to proceed.

I'm also not getting anywhere with Q1 and Q2 at the moment.  Lots of pages of algebra but no results yet.  I feel that, in both one needs to start with a binomial expansion.

Bob

#7 Re: Exercises » Exponent Slash, #1 » 2019-01-04 00:39:23

Q4.


If you divide both by n, then the question becomes when is (n-1)! divisible by n.

If n is any prime this cannot happen as none of the factorial factors can contain it because it's Prime and these are all smaller.

So what if n is not Prime?  It must have a decomposition into factors all of which are smaller and so contained somewhere in the factors of the factorial.  But what if such a factor occurs twice?  It can be further factored using numbers we haven't yet used so n will divide the factorial.

This sounds a bit clumsy so I'll illustrate with an example.

Suppose n = 147 = 3 x 7squared

146! = 3 x 7 x 14 x other numbers so 147 divides it.

So all composite numbers have the property.

Bob

#8 Re: Exercises » Exponent Slash, #1 » 2019-01-04 00:12:56

Hi Mathegocart,

At the time I didn't try these.  I can show two now and I'll have a think about the others. No one method for all.

Q3.  You'll need to  know about geometric series and sum to infinity.

The series can be divided up as

5/13 + 5/13squared+ 5/13cubed + ....

        + 50/13squared + 50/13cubed + .....

                                 +500/13cubed +.........

etc

Each line is a geometric series.  I'll abbreviate each by (a,r) where a is the first term and r the common ratio.

(5/13  1/13) + (50/13squared, 1/13) + (500/13cubed 1/13) + .......

Using the sum formula we get

5/13 x 13/12 + 50/13 x 13/12 + 500/13 x 13/12 + ......

= 5/12 x 1/13 x (1 + 10/13 + 100/13squared + ......)

This last part is also a geometric (1 , 10/13) so we get

5/12 x 1/13 x 1 x 13/3 = 65/36

More follows,

Bob.

#9 Re: Help Me ! » Radius » 2017-11-04 23:05:04

Hi Animesh,

Join R to the centre, O, of the circle. Triangles OPR and OQR are congruent as they have a 90 degree angle and one side in common plus equal radii, OP and OQ.

So OR is a line of symmetry .  If OR crosses PQ at S then SP = 3,so RS = 4 by Pythagoras.  If you let OS = x and radius = r you can form two equations in r and x by Pythagoras on OPR and on OPS.

It's easier to eliminate r and find x.  Then find r.

Alter

#10 Re: Help Me ! » Geometry help! » 2017-11-03 23:14:24

Hi sydbernard,

I'm hoping I've understood this.  I'm slightly worried by the term disc rather than circle.

Let''s say the disc has radius r and the required circle has radius s.  The centre of the disc is O, and the line halfway between the parallels is m.

With centre O and radius r + s an arc cutting m at P will give the required circle centre.  So how to 'construct' r + s ?

If m cuts the disc at A, make a line perpendicular to m cutting one parallel at B.  AB = s. Extend OA and with centre A and radius AB make an arc to cut OA produced at C. Note: OC = r + s. With centre O and radius OC make an arc to cut m at P  P Is the required centre .

But what if the disc is too small to be cut by m ?

In that case you can draw a line n, perpendicular to m, through O. Let n cut the disc at E. Also draw any other line, FG, , parallel to n, so that F is on m and G on the parallel. FG = s.

Join F to E and construct GH parallel to FE with H on n. OE + EH = r + s so once again we have the right radius for an arc to cut m.

Alter

#11 Re: Help Me ! » Represent sqrt{2}+sqrt{3} on a number line » 2017-11-01 22:55:10

But what does 'exact' mean?  If you draw a line 1 inch (or cm) long is that exact?  Suppose you look at the line through a magnifying glass.  How do you know it is not actually 0.99999 ?  I think you are creating an unnecessary difficulty.

A

#12 Re: Help Me ! » Can someone please help me with this: 2x + 4cosx = 0 » 2017-11-01 00:48:23

Hi CIV,

There isn't an exact value for this so you'll have to give an approximate value.  On the MathsIsFun main site there is a function plotter.  You can plot -x as one line and 2cosx as a curve.  Zoom in to get your answer where they cross.
Alter

#13 Re: Help Me ! » Represent sqrt{2}+sqrt{3} on a number line » 2017-11-01 00:40:41

Hi Abbas 0000

What are you accepting as a 'representation'?  If you draw a line and Mark points at regular intervals, labelling them 1, 2, 3, etc you have to assume you can do that accurately.  In Euclidean geometry we assume you can, but in the real world there is no such thing as perfect accuracy. Pencil drawn points have size and, by viewing under magnification, you will reveal that drawings are never exact.

So I think what you have described is acceptable.  Draw a unit line from zero to point A and then a vertical line up of one unit to point B.  With compass point at zero and radius OB make an arc to cut the number line at C.  OC is root 2.

This point is as accurate as A, so if the point for 1 is acceptable so is C.

Draw a line vertically up from C one unit, to point D. Compass set to radius OD and make an arc to cut the number line at E. This is root 3.

Repeat the root 2 construction from point E and F will be at root 2 + root 3.

Alter

#14 Re: Help Me ! » Sangaku Problem 525 » 2017-10-30 23:37:19

EOdash = r2

AOdash = d - r2

KJ = r3

AJ = 2r1 - r3

Alter

#15 Re: Help Me ! » what is Brute-force search in numerical analysis, » 2017-10-29 18:52:06

Say you have an equation in n, where n is a positive integer less than 10000.

A brute force method would be to try every possible value for n, and see which values work.

A

#16 Re: Help Me ! » Geometry help » 2017-10-29 18:49:12

Again no diagram for the first three.

You have correctly found the angle sum for the polygons. In a regular polygon all the angles will be equal, so now calculate one of them.

Your scale factor is correct for small shape as a fraction of the large shape.  I think the second answer is for large as a multiple of the small.

In the last a minus sign is needed.  I'm not seeing that.

What course are you studying?

A

#17 Re: Help Me ! » Geometry work need help! » 2017-10-29 18:41:44

I don't think it is possible to help without a diagram.  Sorry.

A

#18 Re: Help Me ! » Fraction » 2017-10-29 01:00:12

Hi Animesh,

Write 1000n/810 = d25.d25d25d.......

Subtracting gives 999n/810 = d25

This will greatly simplify and you'll find only one value of d is possible.  The value of n follows easily from this

Alter

#19 Re: Puzzles and Games » The Professor and the Student » 2013-12-16 20:47:40

I thought this was totally sorted by Bob Bundy.  Maybe he wasn’t clear enough so I’ll have a go for him.

Let’s call the test availability period,  TAP, measured in days. 

The professor has declared:  Statement   = “…but you won't know in advance which”

So on Sunday evening,

 

As we reach the evening of each day without a test,  TAP reduces by 1 and if we count back one day TAP increases by 1.

Deduction one:  So by Thursday evening

   and so the student deduces that the statement implies  the test cannot take place on Friday, so
 

So far, so good.  But then the student’s logic goes wrong.  The student’s logic, not the professor’s.  Which is at it should be; he is the professor after all.  :)

On Wednesday evening

  .  But the student thinks
  because of deduction one and counting back one day.   And hence the test cannot take place on Wednesday, making
 

But  you cannot use deduction one on Wednesday evening , because deduction one requires that it is Thursday evening  and it isn’t.  You cannot use logic like this.

In post 56 Bob set up a parallel situation involving his next post.  In post 65 he made this post.  No one successfully predicted when he would post, but the post occurred.  He clearly wasn’t lying!

He also said he was giving up on the thread thereafter.  That’s why he has asked me to remind everyone about what he said; so he can maintain the position that he was not lying.  

I am, of course, happy to assist him.  smile

Alter ego

#20 Re: Puzzles and Games » The Professor and the Student » 2013-05-04 05:51:35

Agnishom wrote:

Ai do not mean to be rude, but I doubt your identity…

What do you mean?  I am who I am.  I recommend you re-read posts 57 and 66.

Alter

#22 Help Me ! » what does this become? » 2012-12-15 09:59:59

alter ego
Replies: 1

Can somebody help me do

Thanks

#23 Re: Puzzles and Games » The Professor and the Student » 2012-03-30 09:34:33

hi guys,

You seem to be missing something.

In post 57 bob said he would post during the week but you wouldn't know when.

In post 66 he posted.  It was a Tuesday.  Nobody predicted that.

He also said "So, I'll have one last go and then I'm giving up on this thread."

So he is clearly not going to come back to this thread again.

Case closed.

alter ego

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