Math Is Fun Forum

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

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

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

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.

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

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

EOdash = r2

AOdash = d - r2

KJ = r3

AJ = 2r1 - r3

Alter

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

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

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

Alter