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#1 Re: Help Me ! » I have 3 very similar percentage questions please » Today 05:44:48

hi sybil8464

Your answer for Q2  77.78  is correct.  You need 14+ out of 18.  14/18 is 77.7777.. % so choosing 77.78 does that.

For Q7 you need 18+/21 .  18/21 is 85.71428..... so strictly 85.71 is not enough.  But it's the highest answer so it'll have to do.  These aren't from Compu-High by any chance are they?  Because they use multi-choice and computer marking you cannot say none of these actually work, so try 85.71 and if you can point out to your tutor that it doesn't really work maybe they'll realise they need to take more care setting their questions.

For Q10 you need 15+/18.  15/18 = 83.3333 % so I can see why your sister suggests that, but once again it's not strictly enough to get past 50% of the votes.  So i like your thinking in suggesting answer 4.  However, I suspect the person who set the questions hadn't thought this through properly because of rounding.  I'm not sure how you answer this.  If you can write an explanation for your answer then you deserve extra credit but are their tutors clever enough to spot this?  Please post back what you submit and how it is marked. 

Best wishes,


#2 Re: Help Me ! » Geometry angle of rotation » Yesterday 19:36:53

hi sybil8464

Welcome to the forum.

From the way you have described this, either answer is correct.  The question setter needed to say clockwise or anticlockwise.  Normally one would give the lower value ( 150 ) but, unless the wording makes it clear, 210 would also be good.  In some areas of mathematics anticlockwise is the norm, so I'd suggest you say 150.

Hope that is good,


#3 Re: Puzzles and Games » The Ball-Drop Problem - Solve w/o Knowing the Solution! » 2020-07-29 20:13:28

hi PMH

Go up one from k1, and test the block from there to F, using no more than k-1 drops.
   k2 = k1 + 1

Shouldn't that be k2 = k1 -1  ?

This generates the series k1, k1+1, k1+2, ...
   which has to sum to F.
   Sum:  k*(k+1)/2 + k*k*(k+1)/2
        = (k+1)*(k+1)/2
    so F = (k+1)^2/2
    so k = sqrt(2F)

I think that is Sum = k(k+1)^2 over 2


#4 Re: Puzzles and Games » The Ball-Drop Problem - Solve w/o Knowing the Solution! » 2020-07-28 19:59:49

hi PMH,

In my solution I got to the point where I wanted

b + (b-1) + (b-2) +   ... + (a+3) + (a+2) + (a+1) = 99

If sum(n) means the sum of the natural numbers up to n then my formula can be re-written as
sum(b) - sum(a)

There's a formula for the sum of n natural numbers which can be worked out as follows:

sum(n) = n + (n-1) + (n-2) + ...  + 3       + 2   + 1

sum(n) = 1  +   2    +   3  + ... + (n-2) + (n-1) + n


2 x sum(n) = (n+1) + (n+1) + (n+1) + .... + (n+1)  + (n+1) + (n+1)  = n x (n+1)



#5 Re: Puzzles and Games » The Ball-Drop Problem - Solve w/o Knowing the Solution! » 2020-07-22 09:25:43


There was a command square brackets code that allowed a poster to show their line of LaTex code completely without activating the interpreter.  I don't know why that doesn't now work.  But I've found a way round it.  I've deliberately left out a close bracket.  Then the code command shows the rest so you just have to re-insert the missing bracket.

Here's the first example using that trick.

[math V_{sphere}=\frac{4}{3}\pi r^3[/math]

But there's an easy way to see what the code is.

In the first example I clicked on the LaTex and got this:       V_{sphere}=\frac{4}{3}\pi r^3

You can always do this to see what the code was.

I've decided to re-write post 4 properly in case anyone else is misled by it.


#6 Re: Puzzles and Games » The Ball-Drop Problem - Solve w/o Knowing the Solution! » 2020-07-22 01:22:06

hi PMH,

Apologies for thinking 9 is a prime.  Still it doesn't alter the proof.  I'm wondering whether to make it a bit neater now I've finished developing it.

Also sorry that I've taken so long to reply.  I had forgotten I was still logged in from earlier so I've only just spotted your reply.

The maths characters are done using a code called LaTex.  The FluxBB engine that drives the forum will jump to the interpreter when it meets the command open square bracket math close square bracket.  The interpreter switches off again in a similar way with the command /math

There's a tutorial here: 

The first 20 or so posts will be enough to get you started.  Then you can dip in when you need a specific symbol.

Also useful:

You can create code there to copy and paste into your post.

Furthermore, if you see some LaTex that you want to investigate, just click on it to see how it was done.

Other square bracket type commands you might find useful:

quote     url      hide     img      b  (for bold)  u (for underline)

Hope that helps,


#7 Re: Puzzles and Games » The Ball-Drop Problem - Solve w/o Knowing the Solution! » 2020-07-21 19:35:51

hi Phro,

Whoops! I feel like such a fool.  I was so pleased I'd found a method, I overlooked this detail.  Which is easier: rework the whole of maths defining 9 to be a prime, or change my earlier post?

Think I'll make the edit.



#8 Re: Puzzles and Games » The Ball-Drop Problem - Solve w/o Knowing the Solution! » 2020-07-20 20:17:47

RE_WRITE of post.

You will see from the next post by Phrontister that I made a careless mistake and said that 9 was a prime.  I've realised that the solution doesn't need primes anyway so here's a complete revision.

We need:

Clearly b - a < b + a + 1 and this pair must be a whole number factor pair multiplyig to give 198, so the possibilities are:

1 x 198, 2 x 99, 3 x 66, 6 x 33, 9 x 22, 11 x 18.

So I'll investigate b - a being each of the first of these pairs:

If b - a = 1, then a = 98 and b = 99

If b - a = 2, then a = 48 and b = 50

If b - a = 3, then a = 31 and b = 34

If b - a = 6, then a = 13 and b =  19

If b - a = 9, then a = 6 and b = 15

If b - a = 11, then a = 3 and b = 14.

You will notice that b gets smaller as b - a gets larger.

This follows because to make b smallest we want b + a + 1 to be smallest, and that will happen when b - a is largest.

So I could have left out all those trials and jumped straight to b - a = 11 immediately.

I'm feeling old and useless.

That's why joining MIF was a good move.  We'll have you back to your prime in no time. smile


#9 Re: Puzzles and Games » The Ball-Drop Problem - Solve w/o Knowing the Solution! » 2020-07-20 04:04:52

hi PMH

Welcome to the forum.

I see that puzzle was provided by JaneFairfax.  She hasn't been active on the forum for a long time now, but, in her day, she was a formidable brain.  And she expected her readers to follow her arguments with the same level of genius.  Not always possible for us mere mortals!

I think there is some logic there but half the steps are missing (Probably she thought they were obvious smile ).

Let's say you try the nth floor and it's the best.  If a ball breaks you can test from 1 to n-1 in that order to establish the level.

If the ball doesn't break you need to try a higher floor in a similar way.  No good trying a number equal to n + another n, because if the ball breaks it will now take another n-1 to find the level so, on top of the two tries you have already had that means n + 1 tries.  But I wanted n to be the maximum number.  So I'll try n + one less than n this time. 

If it breaks I still have n-2 tries to find the level, so n altogether again.

If it doesn't break then by a similar logic I'll try n-2 higher and so on.

I can stop if I reach floor 99, because, if I haven't yet found the level, it must be floor 100.

So I'm looking for n, such that n + (n-1) + (n-2) + (n-3) + ..... comes to 99.  Note: this sequence doesn't have to end in +3 + 2 +1.  It can end sooner, provided I reach 99.

At this stage you can use trial and improvement (a spreadsheet helps) to home in on n = 14 quite quickly. [There may be an algebraic way to get 14 with any trialling.  I'll try that now and post again if I find a way.]


#10 Re: Dark Discussions at Cafe Infinity » Miscellaneous Math Discussion » 2020-07-15 06:23:06

I had assumed that ' is being used to mean the inverse of.  But I cannot find any on-line resources that confirm this.

But, if it is, then

(X' Y' X Y) (Y' X' Y X) = X' ( Y' ( X (YY') X' ) Y ) X = the identity = I

Also (XY) (XY)' = I = X Y Y' X' => (XY)' = Y'X'


#11 Re: Dark Discussions at Cafe Infinity » Miscellaneous Math Discussion » 2020-07-13 20:24:24

hi pi_cubed

I tried X and Y as 2 x 2 matrices, just making up some numbers.  I didn't get the same answer evaluating X' Y' X Y and  Y' X' Y X.  It was a long calculation so I may have slipped up of course.

Did you just randomly make up the relationship, or does it come from something?


#12 Re: Maths Is Fun - Suggestions and Comments » This forum has been really inactive lately... » 2020-07-13 04:33:23

The forum is run for the benefit of the members.  If you want to start a discussion then please do so.  I'm looking forward to your post. smile


#13 Re: Introductions » Hello from Maine, USA » 2020-07-09 03:16:38

hi Justin K. Soto

If you've joined because of an interest in maths, then WELCOME!

If you just joined to advertise  loans then this is NOT welcome.  I've changed your website link to a link to our rules.  Please read them.


#14 Re: This is Cool » Po-Shen Loh discovers a new way to solve quadratic equations » 2020-07-09 02:49:44

Quadratic Equations

There has been a lot of internet ‘fuss’ generated by Po-Shen Loh’s ‘new’ method for solving quadratic equations.  One thing to say about that.  It’s definitely not new.  I spent a few minutes looking at the comments from his Utube and found the following:
“In Indian schools this is taught to students in grade 8 as a standard method”
“I was taught this method by my high-school math teacher in former Soviet Union in early nineties”
“My geometry teacher taught me this 3 years ago”
“I already known that I learned that in middle school”
“This is the thing i had already been taught in 9th class”

So, not new then.

I’m not sure if he is also claiming the sum and product of the roots bit but that’s certainly not new.  I was taught that in 1967 as part of my A level and it was in the text that had been around for years.

I thought I would write up the standard quadratic theory and compare.  Apologises to those for whom this is ‘old hat’ but there will be some readers who may find this useful.

A quadratic is an equation of the form ax^2 + bx + c = 0 for values of a, b and c. Apart from in mathematics they are used in physics, economics, astronomy, engineering, agriculture, military and law enforcement to mention a few.

If we draw the graph of y = ax^2 + bx + c we get something like this:


Using calculus dy/dx = 2ax + b and so the stationary point is when x = -b/(2a)

The gradient graph is linear and crosses the x axis at that point.  If we differentiate again we get d2y/dx^2 = 2a.  When a is positive this is positive so the gradient line goes from negative to positive values as we pass from left to right through the stationary point.  So it is a minimum.  If a is negative we get a maximum.

With k>0, if we consider a point to the right of the minimum where x = -b/(2a) + k then

And to the left, with x = -b/(2a) - k

These y values are the same.  As this is true for all values of k this makes the vertical line x = -b/(2a) a line of symmetry for the graph.  All quadratics are symmetrical in this way.

So what value(s) of x will fit the equation?  The usual thing we do to solve an equation is to get x on the left hand side and numbers on the right and then calculate the right.  But x occurs twice in a quadratic once as a square and that messes up simple manipulation of the equation.  x^2 = expression with x in and x = expression with x^2 in are no help so something else is needed. 

One way is to complete the square.  I’ll illustrate with 3x^2 + 2x -20 = 0

Step 1.  Divide through by 3                x^2 + 2x/3 -20/3 = 0

Step 2. Move the constant term over to the right and add (half the coefficient of x) squared to both sides.

x^2 + 2x/3 + 1/9 = 20/3 + 1/9

Step 3.
This makes the LHS a ‘perfect square’ so it will factorise with one repeated bracket.

(x + 1/3)^2 = 20/3 + 1/9

Step 4.  Note that x only occurs once now so we can proceed to make it the subject of the equation.
Take the square root of both sides, remembering that a number has two square roots (+ and -)

x + 1/3 = +/- root(20/3 + 1/9)

Step 5. Subtract 1/3 from each side to finish the task.

If you apply this same process to the form with a, b and c in it you get:

That’s called the quadratic formula.  You can still use ‘completing the square’ if you prefer.  It’s essentially the same steps and the same amount of calculation … just means you do several small steps instead of plugging the whole thing into a formula, which you have to learn.

In Po-Shen Loh’s original article ( he uses an example where a=1.  This ‘cherry picking’ of a favourable example is hardly a rigorous way to introduce his method.

If we put a = 1 in the formula we get:

Splitting this into 2 terms and moving the ‘over2’ inside the square root we get

which is his formula.  He says his method makes it unnecessary to memorize any formula at all,  but here it is in a screen shot I’ve taken from his article:

Is it a quicker and simpler way to solve a quadratic?  Well given that you can only use his method by dividing through by ‘a’, and you still have to do a calculation with b^2 and c, and square root and do the +/- bit, it seems pretty much the same to me.  It’ll work, so use it if you want.  Computationally it comes to exactly the same thing.  Well of course it does.  It would be rather worrying if you got a different answer wouldn’t it?

Best wishes, keep safe!


#16 Re: This is Cool » Po-Shen Loh discovers a new way to solve quadratic equations » 2020-07-05 19:55:34

If you 'complete the square' or use the formula you have to do some calculations involving c and b squared, do a square root, add in -b and divide by a term with a in it.

This 'new' method involves eliminating a from the expression first, then a calculation involving b squared and  c, then square rooting and then adding in -b/2.

So in what sense is this quicker, or different from what we've all been doing all these years?  If you learn an algorithm and learn to do it quickly, then that's fine for you.  But is this really worth all the hype?

later edit:

quadratic formula when a = 1 :

So this 'new' method is just the old one with a = 1.


#17 Re: Exercises » puzzle by Xodus » 2020-07-03 23:40:34

hi Phro,

Great to hear from you.

Just knocked together a short prog. to crunch this.  Couldn't do better than 131.

Is there a way to get this analytically I wonder?


#18 Re: Exercises » puzzle by Xodus » 2020-07-03 21:13:38

401/491= 0.816700....

I am not yet claiming that is the lowest 'n'.  Still working on it.


#19 Re: Exercises » puzzle by Xodus » 2020-07-03 20:50:00

hi pi_cubed

Remember this isn't my problem.

It says '3 digits are'; it's unclear if that means the first three or any three.  It doesn't say the decimal is just 0.167  There may be more digits after that.

Assuming it's the first three, I have found an 'n' that is lower than 1000.  I got 0.167989....

If the answer may be 0.dddd167dddd then I think I could get an even lower n.


#20 Exercises » puzzle by Xodus » 2020-07-03 01:41:18

bob bundy
Replies: 7

Let m,n be positive integers. When m/n is expressed as a decimal, three consecutive digits after the decimal point are 1,6, 7 in order. Find the smallest possible value of n.

#21 Re: Maths Is Fun - Suggestions and Comments » This forum has been really inactive lately... » 2020-07-03 01:40:09

I've looked at the admin controls and we don't seem to have a 'move post to another topic' option. I think it exists in FLUXBB but you have to download extra modules.  So I'll re-post it in Exercises.

As regards to underused areas of the forum and users who seems to have given up posting, in both respects that's up to folk.  A topic gets as much action as members want.


#22 Re: Euler Avenue » Would certain 3D objects be viewed as impossible in 4D? » 2020-07-03 00:52:07

I have had a loooong think and come up with this:

Firstly, I'll examine an impossible cube.  Then try to repeat this with a 4-D hypercube.

Why a cube?  Just that it's easier to write the coordinates of the corners.

Let's start with a 0-D point.  This has 1 vertex only.

In 1-D I'll consider a line.  This has 2 vertices and 1 line.  Each vertex has 1 line leading from it.

In 2-D I'll consider a square.  This has 4 vertices, and 4 lines.  Each vertex has 2 lines leading from it. These lines are at right angles.

In 3-D I'll consider a cube. This has 8 vertices, and 12 lines.  Each vertex has 3 lines leading from it.  Pairs of these lines are at right angles.  There are 6 square faces.

In 4-D I'll consider a hypercube.  By continuing the pattern this will have 16 vertices and 32 lines.  Each vertex has 4 lines leading from it.  Pairs of these lines are at right angles. There are 8 cubes.

Here's an impossible cube:


The line HD is in front of the line EF.  This is impossible.  By choosing coordinates for the position of the viewer such as (1.5, -1.5, 1.5) you can compute the distance to EF and to HD and hence show that HD is further away from the viewer.

To show a hypercube on a 2-D surface is not easy.  The usual way to to draw one cube inside another and then join corresponding vertices.  Here's my attempt:


I have tried to show the cube ADHEILPM in front of everything else.  Best I could manage so far.


#23 Re: Introductions » hello from australia » 2020-07-01 23:00:41

hi airporttcmelb

If you have joined because of an interest in maths then welcome.  If you just wanted some free advertising then no!

I've made your website link into a link to our rules so you can study them.


#24 Re: Euler Avenue » Would certain 3D objects be viewed as impossible in 4D? » 2020-07-01 03:04:32

I'm not sure humans are equipped to answer this question.  When you 'see' an object, let's say a chair, what happens at your retina is certain light receptive cells are activated and send a message along neurons to the brain.  The retina is slightly curved but, we can say it is 2 dimensional.  So a set of colours are received by the part of the brain that interprets visual information and our brain works out from experience that it is looking at a chair.  It's our perception that tells us it's a 3-D object. 

The same happens if you look at a painting.  There may be a small amount of relief on the surface but not enough to tell us it's a picture of the 3-D world.  Again, our brains do that.

Illusions like the Penrose Triangle (for more look here rely on the way our brains work.  We're so used to seeing solid objects that we'd rather believe an impossible object exists than see it as just a clever bit of 2-D drawing.   

Because our retina only gets a 'flat image', all 3-D world pictures are a construct of our brain.  So how can we perceive a 3-D object in all its 3 dimensions in the way you want.

Nice question though smile and I shall continue to ponder it.  If I come up with anything, I'll let you know.


#25 Re: Help Me ! » geometry » 2020-06-30 19:25:21

hi JunPark

Welcome to the forum.

First two parts look good to me.

Then this:

168 times 2 + 240 = 576

That should be 166 not 168.  I would suggest you go back to the un-rounded value, then x2, then round off again, to preserve maximum accuracy.


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