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

]]>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

]]>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

]]>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

]]>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

Any idea for 5? thickhead says (sigma of stuff)/(some number choose 3), presumably that is the number of all three-number combinations, though my mind is totally lost on the numerator. Could you explain the numerator? I have to admit, I was spitballing when I saw made these problems and I didn't really have a clue on how to solve these. Bob? As in Bob Bundy?

]]>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

]]>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.

]]>f the 3 numbers are in A.P. what is the probability for first n natural numbers?

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