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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,411

***1

Two numbers x and y are chosen at random from the set [1,2,3,4,..........,3n]. What is the probability that x² -y² is divisible by 3?

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**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

This sounded like a trick question at first, but having spent a little time on it, the pattern is somewhat surprising and gives a remarkably high number of results divisible by 3!

The answer I get seems to tend towards

, but it does vary depending on the value of n. (eg, if n=1, then it is 33.33%, as I had expected).Having worked on this the hard way, let's now try to get a formula.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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I'm afraid not. If n = 1, then there are 9 possible combinations for the two numbers.

1-1=0 --> Yes

4-1=3 --> Yes

9-1=8 --> No

1-4=-3 --> Yes

4-4=0 --> Yes

4-9=-5 --> No

9-1=-8 --> No

9-4=-5 --> No

9-9=0 --> Yes

So there we have 5 out of 9 combinations, which is 55.55%.

But actually, it doesn't matter what the number is, all that matters is whether the numbers you pick are of the form 3k, 3k-1 or 3k-2. And as the set ends in 3n, there is always an equal chance of picking any of these. So if the above list was generalised, it could be used to show that the chance is always 55.55% for any value of n.

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**ashwil****Member**- Registered: 2006-02-27
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Mathsy, I agree with your logic, but the question asked for two numbers to be selected at random, not the same number to be selected twice. Hence, I excluded the possibility of x=y. This has the effect of removing 3n from both numerator and denominator. Progressing through the series, up to very high values of n will still tend towards 55.55%, but always slightly less than this and, at very low values of n, significantly less.

A technicality, maybe, but if I am playing poker, I like to know how many aces are in the deck!

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**mathsyperson****Moderator**- Registered: 2005-06-22
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Ah. True point. Well, in that case, you're completely correct. If n = 1, then the probability is 1/3 and as n increases the probability will tend to 5/9 without ever reaching it.

It'll probably be extremely difficult to calculate the formula though. I'll have a go, but not right now.

Why did the vector cross the road?

It wanted to be normal.

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**ashwil****Member**- Registered: 2006-02-27
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Try this:

[((3n)^2 x 5/9) -3n] / [3n x (3n-1)]

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**ganesh****Moderator**- Registered: 2005-06-28
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Good going, mathsyperson and ashwil! I shall wait for more responses. In a day or two, I shall post the solution.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,411

***2

Solve for x.

(x-1)³+(x-2)³+(x-3)³+(x-4)³+(x-5)³=0

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**ganesh****Moderator**- Registered: 2005-06-28
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***3

Given

find the sum

1² + 3²/1! + 5²/3! + 7²/5! + ............∞

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

***2

do you want complex numbers?

Eq=5(x-3)(x^2-6x+15)

so x=3 or x=3+I√6 or x=3-I√6

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

***3

cool. Very interesting question. I get 1+5e but my solution is stinky. I want to see better.

IPBLE: Increasing Performance By Lowering Expectations.

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**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

***2

Well, the answer is

, but my algebra is so rusty that I am struggling to come up with the elegant solution rather than the trial and error version! Give me time and I WILL get there!Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

If you ractorize what you got as equation you'll get the above equation.

From the first term x=3 but the second term is quadratic equation:

x^2-6x+15=0

The discriminant is <0 So there are 2 complex answers too.

IPBLE: Increasing Performance By Lowering Expectations.

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**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

Thanks, Krassi.

You were obviously posting your answer while I was still working on mine, so I saw yours after posting mine! I had got as far as multiplying out and simplifying, but I just hadn't got as far as the final factorisation. It used to be so easy when I was 18, as I was doing it all the time, but you don't have much need for this kind of maths as a hotelier, so it takes me a bit longer to get my thinking in gear!!

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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You mathz is (the moderators won't allow me to say this ) GOOD.

Do you know how many people know a 1/10 of what you know?

I would say 5-10% MAX!

So... Get back to work!!!

(If you understand me )

Can you give some solution of ***3?

IPBLE: Increasing Performance By Lowering Expectations.

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**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

What a lovely compliment! I wish that my teacher had agreed with you in 1978 (or, even the examination board!!). I was far from the perfect student and my maths was only average at best. But what I have done in life since then is to learn from mistakes, work from first principles wherever possible and always try to understand WHY things are the way they are. It may take a lot longer than remembering formulae, but it is much more sound in the long-term.

Alas, question ***3 is currently beyond me and I really need to dust away a lot more cobwebs before I can suggest a properly worked solution, but I do agree with 1 + 5e being the correct answer, based on observation rather than on algebraic proof.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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I used ugly algebra so I prefer you to post your proof.

For the first: Your teacher drags!!!

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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I wanted to say another thing from drags but the word blocker from the forum didn't allowed me to do it.

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,411

krassi_holmz and ashwil, I shall post the solutions (with steps) to ***2 and ***3 by the weekend! You both are on the right track, Well done!

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,411

***4

How many number of n digits can be made with nonzero digits in which no two consecutive digits are the same?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,411

Excellent, mathsyperson!

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,411

***5

Prove that

is greater than or equal to where a, b, and c are positive real numbers.

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**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

ganesh, I am seriously struggling with the proof of ****3. I can work through the logical structure of the 4 given equations and I can see an obvious similarity with the expression to be solved. However, for the life of me, I just cannot get an expression that absolutely links this expression to those funtions of e. The problem is that the numerator's progression is a funtion of (1+2n)^2 rather than x^n, as given in the examples, which is destined to diverge greatly as n increases.

My solution was simply based around common sense, in that the answer HAD to have a relationship to e, otherwise the examples are pointless!. Doing a sum of the first few values of the expression soon led to an answer of 14.5914, which is clearly 1 + 5e. Equally, the progression follows a very similar pattern to the formula for e, so it was clear that it is not necessary to bring in high values of x as their effect is negligable.

The worked example that gives this missing link would be greatly appreciated!

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**ganesh****Moderator**- Registered: 2005-06-28
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ashwil,

I shall post the proof soon.

I shall try to do it this evening.

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