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This has come up before here:

http://www.mathisfunforum.com/viewtopic.php?id=21327

Post 2 has the problem and post 7 has an outline of my solution.

Bob

**bob bundy**- Replies: 0

hi LaughingMan_dD

Welcome to the forum. Thanks for your comment.

I've just tried this puzzle.

I understand your point. I think you're meant to assume that each (male) friend takes his two marbles, looks at his label and makes his deduction in isolation. Only Sally has access to all three statements. Maybe it's because women are better listeners.

Here's my reasoning:

There are four possible set ups and four possible labels: BBB, BBY, BYY, YYY.

If someone looks at their label and is able to say what the hidden colour is, their label must have two colours in common with what they picked.

eg. For Tom, he knows he has BBx. If the label is BYY or YYY, then he cannot say what his third colour is because there are two possibilities. But if his label says BBY then he knows he has BBB, and if his label says BBB, then he knows he has BBY.

So from his statement we can deduce Tom's label is either BBB or BBY.

From Richard we can deduce his label is either BBY or BYY.

Harry does not know his third colour so he isn't looking at a label that is either BYY or YYY, so he must be looking at a label saying BBB or BBY.

Thus, between them Tom and Harry have the BBB and BBY labels. This means that Richard must have the label BYY.

As he can deduce his third colour, he must have BBY. So Tom must have BBB.

Sally can deduce her label is YYY just from the other three labels.

So she doesn't have YYY. Therefore Harry must have YYY.

So, finally, Sally has BYY.

Note to MIF. This puzzle could be worded to cover this point.

eg.

Years ago, to puzzle his friends, a scientist gave one of four containers containing blue and/or yellow marbles to each of the friends; Tom, Richard, Harry, and Sally.

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 all of his friends to take 2 marbles out of their container, read the label, and then tell him the color of the third marble.

Harry went first. He took 2 yellow marbles from his container. He looked at the label in his container, but could not tell what color the remaining marble was.

Then Tom took two blue marbles out of his container and looked at the label. He was able to tell the color of the third marble immediately.

Then Richard took 1 blue marble and 1 yellow marble from his container. After looking at his label he was able to tell the color of his remaining marble.

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

Bob

hi SPARKS_CHAN, and Phro,

There are lots of similar triangles here, so I dare say you could do this with lots of ratio calculations, and D will be a corner by symmetry.

If you want me to do the calcs, post again.

Bob

hi Andre3Thousand

Welcome to the forum.

Is this what you were trying to draw?

As you know 5 out of the 6 measurements in this triangle, there are many ways you could do this. I think the simplest is to use the sine rule:

Bob

Oh, thanks. No one is more surprised than me. And I'd like to say that I don't know what the Euclidean algorithm is (my failure to bother to learn such things is legendary ) so I claim double credit for thinking it up myself.

Bob

I'd be grateful for any comments about the validity of this:

Say 7n+4 and 5n+ 3 have a common divisor, d.

Then the result of subtracting the smaller from the larger must also be divisible by d.

So 5n + 3 and 2n + 1 must be divisible by d.

Continuing

3n + 2 and 2n + 1 must be divisible by d.

And again

2n + 1 and n + 1 must be divisible by d.

And, one last time

n+ 1 and n must be divisible by d.

Only d = 1 is therefore possible.

Bob

hi Heron,

Welcome to the forum.

This is my explanation.

P(JJQ) = 4/8 x 3/7 x 4/6

P(JQJ) = 4/8 x 4/7 x 3/6

P(QJJ) = 4/8 x 4/7 x 3/6

This P(JJQ in any order+ = 3 x 4/8 x 4/7 x 3/6 = 3/7

Bob

I get that value too. You might be able to put together an analytic proof of existence like this.

Let y = cos(tan(sin(x))).

Establish the following properties:

(1) y is an even function.

(2) y = 1 when x = 0

(3) y is periodic.

(4) y > 0 for all x.

(5) There exists an x = k so that y is symmetrical in that line. (There are many but find the lowest > 0 )

(6) y is decreasing between x = 0 and x = k

(7) The line y = x cuts the curve once between x = 0 and x = k

(8) This intersection gives the value of the constant. The other properties should be sufficient to show it is unique.

Hint: Use the MIF function grapher to see what the graph of y looks like.

Bob

So do you mean:

or maybe

Bob

hi PatternMan

I really should learn to use Latex.

Yes, that would help.

Put this as a single power of x

Now use the rule for differentiating powers (multiply by the old power and lower the power by 1)

which is what you have I can now see.

But the 'book' answer is the same:

hi Agnishom,

I am unclear about what the 'pattern' of this function is. What goes in the ....s space.

Bob

hi Chris,

In puzzle books they give you grids into which you can put crosses for combinations that are impossible. They always give just enough information so that you can eventually eliminate everything except the one solution. I think that's how Phrontister does them. If you do a search through his posts you might find an example (sorry I couldn't but I think there's one there).

I don't like to use that method as it seems to take away a lot of the thinking, so I've devised my own method:

I start a spreadsheet with columns for all the attributes {name, car, book owned, ...etc)

Then I make one row for each given fact. Sometimes an attribute comes up again in which case I put the second instance into the row where it first came up. eg the book Mulatka Gabriela comes up more than once so you can amalgamate those rows.

You end up with lots of rows (too many) with lots of empty cells. But you can spot some rows that must go with others just from the empty spaces, so you can start to amalgamate more rows. Every time you do that, the number of possibilities goes down and you start to home in on the unique solution.

Good luck with solving

Bob

hi Ayushya

Welcome to the forum.

Bob

hi Ping,

12 x 95 will give the total for all 12 students.

If 11 of them got full marks, how much remains for the low mark student ?

Bob

Post 1787 ???

Bob

hi SPARKS_CHAN

Have you forgotten the area of the top of the frustrum ?

Bob

I have seen inverse trig equations that have been constructed to simplify nicely but I don't think this is one of them. If your friend has just thrown together a load of functions then it is very unlikely it will have a 'nice' solution. If he wants some problems that do make use of trig and inverse trig I can probably find / invent some.

Meanwhile, what's to stop you just using a numerical solver for this. Or graph the RHS and intersect with y = x.

Bob

Did you just use a random function generator for this one ?

Got to get on my new rowing machine. Maybe by the time I've done 3.6Km I'll have some inspiration.

Bob

Cover the solid in approximate cm squares and count them:

Bob

By being taught by someone else in school, who actually teaches things incorrectly.

Bob

Hopefully, that's what the questioner had in mind. The graph cost against V will have a minimum so you should be able to solve this now.

Bob