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#1 Re: Help Me ! » Find sum of n terms in a geometric series using two other sums » Yesterday 21:21:55


hi zainyusufazam

If u1 and r must be integers then the first equation will take you most of the way.

Write 45 as a product of primes.

By inspection you will see only one value of u1 but two values of r are possible.

By evaluating the sum you'll quickly find out which r value works.


#2 Re: Help Me ! » Find sum of n terms in a geometric series using two other sums » 2021-01-14 21:55:57


hi irspow

While having my breakfast I saw the first post and, keen to have a go, I got out some paper and sketched out a solution.  Then I went to post it and discovered you had already sent exactly the same smile  I suppose my lesson from this is to read everything before starting an answer.  But the 'work-out' was good for me so why not?  I was a bit worried when I got to the quadratic but when it came out as r = 1.5 I was greatly relieved. Not too complicated but unlikely to allow someone to get the answer just by guesswork.

Best wishes,


#3 Re: Help Me ! » Find sum of n terms in a geometric series using two other sums » 2021-01-14 21:50:33


hi zainyusufazam

Welcome to the forum.

Use of Latex:

If you click on one of irspow's Latex lines you will see the underlying code that was used.  That makes it easier to copy useful commands that others have used.  The Sticky thread also has lots of ideas and you can also search on-line for new commands.  Different forums may have different Latex applications so sometimes you may learn about a command that this forum doesn't support, but that only happens rarely for me.

To send your commands to the Latex interpreter you need to begin and end with two bccode commands.  The begin command is the word math inside square brackets [m... here] Sorry I cannot show it correctly without setting it off.  The end command is similar but with /math inside the square brackets.  If you don't have this second command you'll get an error report when you try to post your work.

This site is also useful for creating Latex:

Hope that helps.


#4 Re: Help Me ! » How to right imaginary number with inequalities and in... » 2021-01-11 23:11:27


hi garrisontrantow

Welcome to the forum.

I've never encountered an inequality problem involving complex numbers but I don't see why it shouldn't be allowed.

The 'imaginary' access is not called that because it is somehow 'made up' but rather because it is the image of the other axis.

On the real axis > means to the right of, so on the imaginary axis ** I'll define > to mean higher up the line**.

Then I'll divide the exercise into three parts:

(1) If x is a real number.  x^2 ≥ 0 for all x, so every real number is a solution.

(2) If x is an imaginary number.  You've pretty much solved this already, but consider if x = 2i  Then x^2 = -4

-4 is not ≥ -1 using my definition **  because -4 is not higher up the line.

Consider x = 0.5i    x^2 = -0.25 This is higher up the line than -1. 

And if x -= -0.25i   x^2 = -0.25 again.  So it looks like the solution is -i ≤ x ≤ +i

(3) If x is a fully complex number.  That is to say it has the form a + bi where a and b are both non zero real numbers.

In modulus argument form, let the argument be θ where θ is between 0 and 360 and not equal to 90, 180 and 270.

When you square x the argument is doubled , which means that its argument is also not 0, 90, 180, 270, 360.  Thus the square is also a fully complex number.  The inequality has no meaning in these circumstances so there are no new solutions here.


#5 Re: Help Me ! » Combination » 2021-01-11 22:55:13


hi 2004hdhdf

Welcome to the forum.

More information needed.

Is each tile different so every change of tile is a new possibility or are we just looking at ways of tessellating the tiles.  If that's the problem can I assume that you could use some rectangular tiles and then fill up with squares with no limit about how many of each are available?


#6 Re: Help Me ! » An old problem that I have never solved. » 2021-01-03 05:08:44


hi irspow

You want to solve a cubic equation.  There is a formula for this but it's not easy to use.  Perhaps the following will help.


And I can simplify further by setting r = 1.  This is ok because you can always re-scale so it is true and adjust back once you've got a value for h.

So consider the graph y = 3x^2 - x^3

The graph of y against x is increasing in the range we are interested in .  Translate the graph by -k in the y direction, so it crosses the x axis at (h,0) Our problem is to find h.
This makes the graph

There is a numerical way to find h using the Newton-Raphson method.

If a is an approximation for h then

is a better approximation.

where f' means differentiate once.

so the better 'guess' is found by evaluating

Repeated iterations of this calculation quickly 'home in' on the required value h.  For an initial value x = 0.5 will work quite well.

To test this I chose k = 0.7 and a = 0.5

First iteration:     b = 0.533333         f(x) = 0.00163
second iteration: b = 0.532639,         f(x) smaller than 0.000001
Subsequent iterations left b unchanged to 10 decimal places.

Check in original problem: y evaluates to 0.700000675 which is not bad for just two iterations.


#7 Re: Help Me ! » I'm in trouble of 10 and exponent math. » 2021-01-02 20:52:47


hi don.farr

I'm confused in trying to compare your two methods.

In B


What does this mean?

1.0 x e^-6 is a different thing altogether.  e raised to the power -6 would not be a simple power of ten.

Are you using the 'exp' key on a calculator here.  That would make this 1.0 x 10^-6 which makes this the same as A.


0.02       ((   2        ) / 0.00002)
       0.0002 * 1OOOOO

Here you have changed 0.02 into 0.0002.  I cannot see why.

This physics is not my area but I can substitute values into a formula and evaluate.  Please go back a step and say where the values have come from.  Then I'll try to see what the correct evaluation is.


#8 Re: Help Me ! » Need help » 2020-12-30 20:38:45


hi  karryman,

Welcome to the forum.

You're right.  This poster asked a genuine maths question to which we provided some answers.

Later he edited the post to make it an advert.  I will edit this back for you.  wink


#9 Re: Introductions » Math newbie » 2020-12-23 21:29:07


hi Wayne,

Welcome to the forum.


#10 Re: Help Me ! » Translating quadratic equation language for a mathphobe » 2020-12-23 21:27:36


hi Glenys

stivemorris  joined yesterday and made a post on this thread.  He then deleted it.  Many new members don't join because they want to talk maths but rather just to advertise some product.  Let's wait and see what is posted next.

Best wishes for Christmas and the New Year.


#11 Re: Help Me ! » A man said "I am only four times my son's age » 2020-12-15 04:53:01


hi Pradeepkaslp

Welcome to the forum.

The way I was taught to do these is to let one of the unknowns be x.

Let's say the son is x.

Then the man, (the father) is 4x.

And the grandfather is double that less 10 so 8x - 10

Together the ages add up to 120 so

x + 4x + 8x - 10 = 120

I'll leave you to finish off from there.


#12 Re: Help Me ! » Translating quadratic equation language for a mathphobe » 2020-12-04 21:07:37


hi Glenys,

I like your idea for helping the young lady how to remember a set of instructions using emojis.  smile

Years ago it was possible to upload images to the forum server but it became necessary to switch to a new server and that facility was not included in the new deal.

Another member recommended that I join and upload my images there.  It's what I have used ever since.  imgur has changed its front page a little since I wrote the following instructions but you should be able to follow what to do.  What is needed is the bcc command which can be copied directly into a post. … 06#p352606

post 1686

I think the algebra page you are referring to was written after the other pages so it's not really meant to be page 1 of an algebra course.  One of the strengths* of the MIF material is the cross referencing between pages.  So click around the site and find the pages you need.  If you encounter a question with which you need help, I'm happy if you post it here and I'll try to post what you need.


* I also think the explanations are very clear, with easy to understand language, good diagrams, good use of font changes and colour and some excellent interactive pieces.  I haven't found a better site on the internet.  All the work of the MIF founder MathsIsFun himself.  He is working on a science site so leaves the day to day stuff to the other admin and moderators.

#13 Re: Help Me ! » maths » 2020-12-04 20:48:05


hi mishti

Welcome to the forum.

There are many answers to this (with no loss of generality assume a ≥ b 

LCM = 6     a = 6 b = 6     HCF = 6
LCM = 6     a = 6 b = 3     HCF = 3
LCM = 6     a = 6 b = 2     HCF = 2
LCM = 6     a = 6 b = 1     HCF = 1

LCM = 6     a = 3 b = 2     HCF = 1


#14 Re: Help Me ! » Translating quadratic equation language for a mathphobe » 2020-11-29 20:55:41


hi Glenys,

Welcome to the forum.

Yes, you've come to the right place.  I have taught maths in Uk secondary schools for 37 years so I ought to be able to help.

Here's a start.

Maths is full of equations for all sorts of things.  For example the formula for converting degrees fahrenheit into celsius is this

I've used Latex to make that appear 'nicely'.  I typed C = (F-32) \times \frac{5}{9} and then boxed it in with square brackets math and square brackets /math

If you click on any piece of Latex you'll see the underlying 'code' that made the piece.

Descartes invented the x-y graphing idea so you could make a graph of that formula by using y and x instead:

If you go to this page

and change the equation to (x-32)*5/9 you'll get the graph for changing F into C.  [note that the grapher 'understands' * for multiply and / for divide]

Don't panic when no graph appears.  It doesn't show with the basic default scale. You'll have to adjust the scale using the zoom (drag right) and you can click and drag to see other bits of the graph.

Equations where x occurs like that give rise to straight line graphs.

If an x squared term occurs (usually written as x^2 on this forum) then the equation is called a quadratic, and the graph is always a curve.

While you're looking at the grapher page try this equation:

and you will see a typical quadratic graph.

There's several examples of quadratics here: … world.html

Hopefully that will give you a starting point.  Post back when you want more smile


#16 Re: Help Me ! » Similar Polygons !? » 2020-11-05 20:15:16


hi cooperblu

Welcome to the forum.

This same question has been asked before here:


#17 Re: Help Me ! » Sports draw problem » 2020-11-02 03:12:11



I haven't forgotten you.  I'm working on a couple of approaches which may help.  Is it possible for you to post the working you tried ... don't worry if it's muddly ... I'll try to work it through.


#18 Re: Help Me ! » Sports draw problem » 2020-10-31 23:13:08


hi blob123

Welcome to the forum.

One caution.  I'm a bit shaky on questions like this so I may be wrong but here's how I'd tackle these.

Make the question easier by simplifying the numbers.

Let's say you have just 6 teams and you're trying to make up three matches.  I'll call them C1 C2 C3 C4 M1 and M2

How many ways can we select the matches?

Pick any team.  There are 5 opponents to choose from.  Pick another team.  There are 3 opponents to choose from.  Pick a final team.  There is just 1 opponent to 'choose' from.

So that makes 5 x 3 x 1 = 15 possible selections.

EDIT:  I've had a further think and the next bit isn't correct yet.  I still think it's ok to try and simplify the problem. I'm working on a modification that will yield the correct version.  Hope to get there later today. LATER EDIT. This is now corrected.

Now require that just two Cs are together.  There are so few possibilities I can easily list them:

{C1C2 C3M1 C4M2} {C1C2 C3M2 C4M1} {C1C3 C2M1 C4M2} {C1C3 C2M2 C4M1} {C1C4 C2M1 C3M2} {C1C4 C2M2 C3M1} {C1M1 C2C3 C4M2} {C1M1 C2C4 C3M2} {C1M1 C3C4 C3M2} {C1M2 C2C3 C4M1} {C1M1 C2C4 C3M1} {C1M2 C3C4 C2M1}

So I have 12 possibilities making a probability of having just2 Cs of 12/15 = 4/5

Is there a way to get this without listing all the possibilities?

Start with a C (4 choices) Choose another (3 choices) Divide by 2 as I'm counting eg C1C2 and C2C1.

The remaining Cs must each be paired with an M; there's just two ways to do this. (eg C3M1 or C3M2 ... last pair is fixed).

So we have 4 x 3 / 2 x 2 = 12.

That approach can now be scaled up for the 20 teams.


#19 Re: Help Me ! » Calculate the probability of election result from 1% of votes counted » 2020-10-19 21:12:54


hi mrpace

You need to create a mathematical model to 'describe' this situation.  With only two parties the binomial distribution might work.

Probability in this 'constituency' of a voter supporting A is 0.65 (p) and for B is 0.35 (q).

Now you have to assume that these probabilities will remain for the rest of the country and that's where this comes unstuck.

Expected number of votes for A =  np = 1200 000 x 0.65 = 780 000

Standard deviation is √ (npq) = 522.

You can use the normal distribution approximation for binomial results, which I have tried.  I don't think it's worth showing the calculation because it makes it virtually certain that A will win overall.

The problem is with the underlined assumption.  Think about it.  0.65 is pretty high and if that's the probability in every constituency then A will keep winning.  In the UK that doesn't happen.  There are big regional variations between support for the parties so the 'pundits' know they cannot assume general results from a single announcement. The pre-election opinion polls carefully select their samples by spreading across areas with so many people of each gender and age group.

To do what you want, you'd have to know the probabilities for each constituency; then you could calculate on an area by area basis.

The underlying theory is here: … ution.html

The method works ok for bags of sugar if it's ok to assume consistency during the production run.  Voters are more complicated.

In the UK they do an 'exit poll' asking a sample of people how they voted after they have voted.  Providing the sample is across all areas like the opinion polls this does usually give a good prediction of the final outcome.


#20 Re: Help Me ! » Derivatives » 2020-10-07 20:48:08


hi 666bro,

I'll use the example from above.

We know there's a turning point at √(1/3)  ≈  0.57735

Investigate the gradient just left and just right of the turning point.  I've chosen x = 0 and x = 1 to make the calculation easy.


The top line in the table shows the x values chosen and the second line shows the gradient at those points.

Below that I've sketched those gradients and the sketch shows it's a minimum.


#21 Re: Help Me ! » Maths » 2020-10-06 22:27:02


hi Pema,

Welcome to the forum.

Thanks to zetafunc for the initial hint.

You might also find the following useful:

w^2 + w + 1 = 0  =>  w + 1 = - w^2

You can prove this equation either by substituting the complex values for each root (the working is identical) or like this:

Let w^2 + w + 1 = z ; then w^3 + w^2 + w = wz

As w^3 = 1 we end up with z = wz.  So either z = 0 or w = 1.  As we know that second isn't the case z must be 0.


#22 Re: Help Me ! » Please help with the final question (n. 20)! Thank you. » 2020-09-28 20:40:22


hi juney06

Welcome to the forum.  Thanks for posting your answers to the earlier parts of the problem.  It's a great help in knowing what you do understand about the problem.

I started with a diagram and marked on the height of the cliff and the initial velocity upwards.  This is 'up' but the cliff is down and so is the direction of gravity.  So you need to decide at the start which way you will take as the positive direction and then use + or - correctly for the info given.

You've got the right formula to use but the signs aren't consistent.  If you take 'up' as positive then u = 24.5, g = - 9.8, and s = - 117.6 (that's taking the cliff top as zero).

So the quadratic comes out as

If you take 'down' as positive then it is

As you can see these amount to the same thing so it doesn't matter which you use; but be consistent.

I tried this and got 8 seconds but not with your quadratic. I've just tried to solve your version with a quadratic solver and the square root part comes out negative so it cannot be evaluated. So did you make another sign error which 'cancelled out' the first?

For the last part the ball is to be 49m above the ground. But we are both measuring from the cliff top so you need to work out the distance down from the top to the 49 position ie 117.6 - 49.  Then you can use that as your new 's' value to once again get a quadratic.

Hope that helps,


#23 Re: Help Me ! » Derivatives » 2020-09-17 01:02:51


hi 666bro,

It's great to hear that things are going well for you.

I'll illustrate what's happening with an example.

I've chosen the function

When we differentiate we get

and second derivative


Here are the graphs of (black) the function; (red) the first derivative; (green) the second derivative.

I've put the three graphs one after the other, lined up in the x direction but one on top of the other in the y direction.  For this reason don't worry about the numbering on the axes; you don't need it.  But the changes in gradient for the function are aligned correctly with the first and second derivatives.

The function is a cubic.  When you differentiate it you get a quadratic.  The quadratic has two zeros and these line up with the turning points on the cubic.  When you differentiate again you are finding the gradient function of dy/dx, and you get a linear function. 

On the (black) function graph you can see the two turning points; the first a maximum; the second a minimum; but let's say we don't know this yet.  Just to the left of a maximum the function has positive gradient; just to the right it is negative.  At the turning point the gradient is zero of course.

You can see this on the red graph.  Just to the left of the function's turning point the red graph is above the axis so it's values are positive.  The red graph crosses its axis at the turning point and after that it has negative values.  In a similar way the next turning point (a minimum) has gradient negative to the left of the turning point; zero at the turning point; then positive.

So you can tell if a turning point is a maximum by looking to see if the first derivative goes from positive, through zero, to negative.  And a minimum will have first derivative that goes from negative, through zero, to positive.

One way to investigate this is to draw the graphs; another is to calculate gradients just left and right of the turning point.  But the second derivative gives a quick way to find out without needing to do any graph drawing.  Here's how:

Function has a  maximum.  Gradient function goes from positive, through zero to negative.  So its gradient function must be negative at that point as dy/dx has a reducing gradient.

Function has a minimum. Gradient function goes from negative, through zero, to positive.  So its gradient function must be positive at that point as dy/dx has an increasing gradient.

In my example

  so the turning points are at x = -   √ (1/3) and +  √ (1/3)

  At  x = -   √ (1/3) the second derivative is negative => this turning point is a minimum.

At x = +  √ (1/3) the second derivative is postive => this turning point is a maximum.

Note: I don't even have to do the full calculation here; I just need to know if the second derivative is positive or negative at each turning point; so it's a quick way to tell.


#24 Re: Help Me ! » Slicing method Integration Calculus 2. » 2020-09-14 19:24:49


hi Double

Welcome to the forum.

Sketch the graph and draw two horizontal lines close together from the y axis to the curve, so you have drawn a typical strip.  This strip has area x across and dy width. If you were asked for the area of the part of the curve between y = 0 and y = 1 you would add up the strips like this:

You cannot integrate a function of x with respect to y, so you need to substitute with

The resulting function is directly integrable.

But you ask for a volume, so I'll assume this means that the area is rotated around the y axis to create a solid 'volume of revolution'.

Go back to the original strip and imagine it rotates around the y axis so as to generate a thin disc.  The area of this disc is π x^2 ( ie. pi radius squared ) and so its volume is π x^2 .dy  Here the new integral:

Once again you need to replace the function of x with one involving y .......    x^2 = 1 - y

So we then end up with this integral:

Can you finish from here?


#25 Re: Formulas » Primality test and easy factorization » 2020-09-11 00:07:02


hi Hugo28036

Ok. So what is the algorithm?


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