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hi gmn,

Thanks for your comments. Here's the situation. MathsIsFun made both sites. Some years ago he made me an administrator on the forum. I can do some things here but I cannot alter the puzzle itself.

I have had a go at a rewording of the puzzle. I've tried to make it clear, unambiguous and with a single solution. Please would you have a go at it and see if you agree. If we both think it's ok then I'll email MIF and suggest he changes it.

**Marble Mix Up Puzzle**

Years ago, to puzzle his friends, a scientist gave one of four containers containing blue and/or yellow marbles to each of the friends; Alan, Brian, Colin and Diana.

There were 3 marbles in each container, and the number of blue marbles was different in each one. There was a piece of paper in each container telling which color marbles were in that container, but the papers had been mixed up and were ALL in the wrong containers.

He then told his friends to take 2 marbles out of their container, show the others, read the label without the others seeing, and then tell him, but not each other, the color of the third marble.

Alan took two yellow marbles out of his container and looked at the label. He announced he could not tell the color of his remaining marble.

Brian took 2 blues marbles from his container. After looking at his label he was able to whisper the color of his remaining marble to the scientist.

Colin took 1 blue and 1 yellow marble from his container. He looked at the label in his container and he was also able to whisper the color of his remaining marble.

Diana, without even looking at her marbles or her label, was able to tell the scientist what color her marbles were.

Can you tell what color marbles Diana had?

Can you also tell what color marbles the others had, and what label was in each of their containers?

Bob

Marble Mix Up Puzzle.

I agree; there's something wrong with this puzzle.

I'm using B and Y to indicate marbles; b and y to indicate labels.

TOM draws BB and looks at his label. He says he knows the third marble. This can only be done if his label is bbb or bby.

RICHARD (huh!) draws BY and looks at his label. He says he knows the third marble. This can only be done if his label is bby or byy.

HARRY draws YY and his label doesn't enable him to deduce his marble **. So his label must be bbb or bby.

At this point the three labels bbb, bby and byy have all been seen by these three so

SALLY must have yyy as her label. As she cannot have YYY HARRY must have this.

So why didn't he work this out at ** ? Perhaps his logical ability isn't as good as SALLY's. Maybe TOM, RICHARD and HARRY make their draws, look at their labels and make their decisions but don't announce them until all three have completed all three steps and are then not allowed to change their decision. In that case, HARRY doesn't yet know what the other two have concluded so he cannot deduce his third marble. This is a complicated explanation so a simpler way is to have HARRY going first without knowing the other two's decisions; then TOM, then RICHARD. That way when SALLY comes to make her decision she has more information than HARRY.

Bob

hi Mathegocart

I've never done this before so I had to search for the right commands. I think I've done it ok.

Bob

I think that has done it. Please post to check this.

Bob

hi gmn

Welcome to the forum.

I joined the forum after trying some of the puzzles too. But not those two so I've had a look now.

I tried the Merry Go Round using algebra and got a linear equation with that one solution. So you're right. I suspect that the wording of the solution may be Sam Loyd's own. He liked to set puzzles that could be done just by thinking about them, rather than using 'heavy' maths like algebra. The 'solution' could be improved by adding something like "In fact, this is the only solution." I cannot edit those pages so we'll just have to hope that MIF himself does it for us.

As regards the Nanny algorithm to change certain words, it's a problem faced by all forums. Do you moderate every post before it is published or remove offensive ones after the event. The 'Nanny' let's us do the latter which is easier for the moderators and improves communications generally. I've 'fallen foul' of it myself and I just smiled to myself and found another way to say what I wanted. It's not perfect but it does allow us to function.

I've had a quick go at the marbles puzzle and I couldn't even get a solution. I'm wondering if some crucial words are missing. I'm helping out at my local school today so I'll have a longer look later when I get some time.

Bob

hi

This is two geometric series put together:

16 x (1/3 + 1/27 + …) + 32 x (1/9 + 1/81 + ...)

The common ratio is r = 1/9 for both. The start numbers are a = 1/3 and 1/9.

The formula for the sum to infinity of a GP is a/(1-r)

https://www.mathsisfun.com/algebra/sequ … etric.html

Bob

hi mrpace

You can eliminate the earlier prime factors like this

2? The number is not even so 2 isn't a factor.

3? If you take any number in the three times table and add its digits you get another (smaller) number that is in the three times table. eg. 48 4+8 = 12

As the digits add to 2 we can eliminate 3 as a factor.

5? Number has to end in 5 or 0, so 5 isn't a factor.

7? This is the hardest so far to test. If you divide by 7 you get

142857152857142857.....14285 remainder 6 so 7 isn't a factor.

As you say 11 does work ( add alternate sets of digits to get a difference of 0) so it's the smallest.

Bob

It's ok to say let a and b be any numbers and define c to be a + b. As c 'cancels' out of the proof you're just left with a = b. That's the mystery bit. Except the proof is flawed at the step where square roots are taken.

Bob

Hi Anthony

Nice 'proof'. When applying a square root you have to remember that two roots are possible.

9 =9 but 3 ≠ -3

To check out any algebra just substitute some numbers to see what is happening.

I chose a = 5; b = 3 and so c = 8.

Here is your 'proof' with those numbers.

8 x 2 = 8 x 2

25 - 9 = 40 - 24

25 - 40 = 9 - 24

25 -40 + 32 = 9 - 24 + 32

(5-4)^2 = (3-4)^2

All so far is correct. But then we have the square root.

(5-4) ≠ (3-4)

But, back to the algebra, if we take a negative root:

a - c/2 = -b + c/2

a + b = c/2 + c/2 = c

Now it's correct but hasn't given anything new.

Bob

hi katoroh

Welcome to the forum.

There is a formula for this but I don't think it will help in a test situation:

https://math.vanderbilt.edu/schectex/courses/cubic/

There are some helpful tips here:

https://www.wikihow.com/Solve-a-Cubic-Equation

If you are doing a math competition, then the questioners are looking for clever applications of math not number crunching. All cubics must have at least one real solution and, once found, the remaining quadratic can be solved with the formula. So the person making up the question should be giving you such a route into the problem. Most likely you may be able to spot a factor using the factor theorem

https://www.mathsisfun.com/algebra/poly … actor.html

Have you got any 'past papers' with such questions?

Bob

hi Lily,

Welcome to the forum.

Your first step is correct:

Therefore

One way to test if a rearrangement is correct is to choose a value for t and calculate V; then use the new formula to find t when you use that value of V.

Let's try that with t = 1

In my rearranged formula with V = 8000

So that helps to confirm the rearrangement.

The formula you end with, v^-1(t) = 4.1 gives 8000^-1 = 1/8000 which is not 4.1

Bob

hi Zeeshan 01

Haven't heard from you for a while. You seem to have made good progress if you are on to topics like these. Well done!

There's no need to put the word IMPORTANT into your titles. I treat all requests for help as equally important and try to help if I can.

I've repeated all your questions here and I'll start with the vector questions because they all ask similar things:

Question 5 : if the two vectors are A=3i+4j+k and B=i-j+k then

a)They are orthogonal b) orthonormal c) antiparallel d) none

Question 9 : The angle between Vectors -2i+3j+k and i+2j-4k is

a) 0 degree b) 90 degree c) 180 degree

Two vectors, A and B are parallel if there is a constant, k, such that A = kB. If k is positive then the vectors go in the same direction so the angle between them is 0. If k is negative then they go in opposite directions, so I suppose you could say the angle between them is 180, although I've never met this in a question before.

If the dot (or scalar) product is 0, then they are at right angles, so the angle between them is 90. Orthogonal is just another way to say this. If the vectors are orthogonal and unit vectors then they are orthonormal. Use Pythagoras to determine whether they are unit vectors.

Question1 : if the function fxx(x0,y0)>0 then f has a ------ at (x0,y0)

a) Relative minima b) relative maxima c) saddle point d) none

I don't understand the notation here 'fxx'. Did you mean f(x,y) ? If so then none of the answers can be chosen since any could be true. If 'fxx' means the double differentiated functions then, as with the 2D case, this would be a local minimum.

Question 8 : If function fxx(x0,y0)=0 then f has a ----- at (X0,y0)

a) relative minima b ) relative maxima c) saddle point

This seems to be the same question again.

Question2 : if phi=2xz^4- (x^2y) , the |delta phi | at point (2,2,-1) is

a) 2sqrt93 b) sqrt 372 c) 2sqrt91 d) both a and b

Sorry, what does ' the |delta phi | ' mean?

Question 3 : if y=f(x) has continuous derivatives on [a,b] and s denotes the length of arc between the lines x a and x b then

Question incomplete.

Question 4 : arc length s=integral a to b sqrt(r^2+(dr/d thetha)^2 ) d thetha is called

a) Rectangular Coordinates b) polar coordinates c ) spherical coordinates d) none

Is thetha meant to be θ ? That looks like polar coordinates r and θ to me. I don't think spherical as another angle would be required.

Question 6: If phi =1/(r^2) then (delta)^2 phi is

What does '(delta)^2 ' mean?

a) 1 b) -1 c) 0 d)1/(r^2)

Question 7: If f(x,y)=1 throughout the region D then I=integral integral dx dy represents

a) Volume b) Arc C) bounded below d) bounded above

Imagine a 3D graph with x,y in the horizontal plane and z values above. If f is constant then the graph of f is a plane parallel to the x,y plane, one unit higher in the xz direction. If you integrated in 2 directions you would get the volume below the graph.

Please post back clarifying questions 2,3 and 6.

Bob

I'm still unclear.

a must be an event for which a probability can be estimated or determined.

b similarly.

In that case a & b exists as a measurable event. It means both a and b happen.

Look here: https://www.mathsisfun.com/data/probabi … types.html

Bob

hi 666 bro

I'm not quite sure what you're asking so here's some general help with this sort of probability. I'm re-using an old diagram to save making a fresh one as it will work for this. Let's say that a=bread and b=milk.

There are 100 houses in an area. Some have bread delivered; some have milk; some have both; and some have neither. If you pick a house at random what's the probability of …..

p(a) ie. probability of picking a house that has bread delivered = 60/100

p(b) = ie. the probablilty of picking a house that has milk delivered = 50/100

p(a OR b) = ie. the probability of picking a house that has either bread or milk or both = 80/100

p(a AND b) = ie. the probability of picking a house that has both delivered = 30/100

You can see that p(a OR b) = p(a) + p(b) - p(a AND b) [80/100 = 60/100 + 50/100 - 30/100]

This will always be true because when you add p(a) to p(b) you get all of a, all of b but you count (a AND b) twice.

As (a AND b) is contained in both a and b it will never be greater than their sum. It could be equal if, say, the region of b that doesn't overlap a is empty.

Bob

hi MKNB

Welcome to the forum.

This is not going to be easy.

Start with

When you expand the brackets the first two terms have the presence of a function and its derivative and so are directly integrable; the third is too.

That gives you a way to do the inner integral.

Then substitute y = u^2 and switch the outer integral to one involving du rather than dy. It can be done in a similar way, converting the even powers of sine into cosines and using the function and its derivative method.

Worth switching the outer limits to u limits as the numbers convert easily too.

Good luck. Let me know how you get on.

Later edit:

I gave this a try and the first part worked ok. But the substitution didn't. I've been out all day so haven't had a chance to try anything else. I'll make a fresh post if I do.

Bob

Also, once a person is banned, all their posts, including ones from before the ban was made, will show as member banned. So it is possible he initially had PM privileges before the ban date. The forum was on a different server at that time and the rules governing member privileges had to change following the switch.

ARB has re-joined more than once and had his new membership banned also.

Bob

hi 666 bro

If you say "sin x to the power n" some people might not be clear if you mean find sin(x) and raise that to the power n or raise x to the power n and then find the sine of that .

This arises because, in English maths, where to put the function in an expression is inconsistent. Sometimes the function comes after the input such as x^2 and sometimes the function comes first such as √x.

If

To differentiate use the chain rule;

If

This time the chain rule gives

Hope that helps,

Bob

hi greg1313

Good point, thanks. I'll try not to make this error in future.

Also using math tags you can do this:

Bob

hi Janet,

I've had a further thought about this.

All hyperbolas have two lines of symmetry. This one is a translation (and possibly an enlargement depending on k) of y = 1/x

That standard hyperbola has axes of symmetry y = x and y = -x with gradients of 1 and -1. Thus our hyp will have the same. So if we know that BCDE is such an axis then we can see it has gradient 1 and all the answers follow fairly easily (but not without some calculation!) This would explain the earlier help you received.

But the wording of the question does not say that this line is a line of symmetry (or the major axis which amounts to the same thing). So why did your helper think it is? The diagram doesn't show the line as a line of symmetry either.

I have wondered all along why the questioner used k for one parameter. Why that letter? It makes sense for f(x) to have m and c. That's normal for straight lines. There are no equivalent letters to use for a hyperbola. So why not a, b and c? Why use k rather than a. It doesn't matter in terms of doing the question, but it strikes me as odd.

On the maths is fun teaching site there is a graph plotter (https://www.mathsisfun.com/data/function-grapher.php) that has an additional variable 'a'. This allows you to explore the effect of changing one variable on the resulting graph, easily just by moving the slider.

Perhaps the questioner had a similar grapher but one that used k rather than a. So perhaps they set up the hyperbola and experimented with different k values to get one that would make a decent question. Then forgot to change k to a number when the book went to print. You get a good question if you make k = 1 so that's my suggestion.

Bob

hi Janet,

Thanks, I've got it now. I was close with my guessed version. I've tried a couple of values for k ( =1 and = 0.5) and here's the comparison:

Either could be the graph for g(x).

You're right. Not enough information; too many unknowns.

From the asymptote information we can conclude

If I let the y coordinate of C be p then

Eliminating p gives

and there doesn't appear to be any other clues to determine m and k.

k=1 looks most like the given graph, and then p = -2. But, as it stands, not enough to even answer the first part of the question.

Bob

hi Janet,

Welcome to the forum.

Tough to help without the diagram.

I have an on-line account at www.imgur.com and I can upload images there and then point to them in my posts using BBCode. The BBCode here is

`[img]https://i.imgur.com/xSC6dDg.gif[/img]`

but what you see is my diagram.

Hopefully it looks a bit like yours but I've had to guess some of the values on the graph.

The function g(x) has one vertical and one horizontal asymptote and this will be true whatever the values of k, b and c. Here's why:

If x is just bigger than b then the denominator of the fraction will be small and hence y will be large. As x tends towards the value b from the right y will tend to infinity. If x is just smaller than b, then the same thing happens except the fraction is now negative so as x approaches the value b from the left y tends to minus infinity. So one asymptote is the line x = b

As x tends to infinity the fraction reduces to zero so without the +c term the x axis is the asymptote. The curve tends to zero from above when x is positive (assuming k is also positive) and tends to zero from below when x is negative. So that would make the x axis the other asymptote (without the +c)

But we are told that the asymptotes cross at D (-1, -3) so I chose k = 7 (just a guess) then b = +1 (that is forced by the D information) and c = 4 (another guess)

So then I had to make a straight line. All I know is it goes through the same y intercept as g(x), which is 4 on my diagram. So I chose an arbitrary gradient to finish the diagram.

So where do we go from here? Hopefully you've got a similar diagram, but not quite the same. If you post back the changes I'll have a proper go at the diagram and then I can help you finish the question.

the gradients of a hyperbolic function are -1 and 1

No idea what that means. Hyperbolas have continuously changing gradients. And it's not the gradients of the asymptotes.

f(x) = mx + c and g(x) = [k/(x-b)] + c

Question uses 'c' for both functions. Tried that and it yields k = 0 which makes the hyperbola disappear altogether so I don't think the 'c' value can be the same for both.

Bob

Sorry for the confusion. Someone emailed me with two questions and it was easier to post my answers here so I could get math formatting.

The questions were:

Q1. What is the smallest number that has 30 positive factors?

I've given an answer but I'm not certain I've fully proved it yet. It helped that the email had a suggested answer but asked how do we prove it's the lowest?

Q2. (x^2 - y^2) = P a prime >= 2. I'm unsure if this required the lowest values or just any values which is why I explored several.

Bob

That looks like:

[center]This post is just to test

whether we have the center tags or not.

Sorry for the inconvenience.[/center]

for me. That's what happens when the BBcode parser cannot interpret the code.

Bob

Are you sure this is the exact question?

From the RHS I got an expression close to the LHS but not the same. So I tried to spreadsheet some values and did not get LHS = RHS.

eg. If alpha = 30, beta = 60 and gamma = 90 then

LHS = 2x0.5 + 2 x root(3)/2 + 2x1 = 3 + root(3)

RHS = 4 x 0.5 x root(3)/2 x 1 = root(3)

Bob

hi

Not all BBCode is implemented on the forum. I cannot even make it work in the body of a post. Can you?

Bobn