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#1 Re: Dark Discussions at Cafe Infinity » Breathing Trouble while Meditation » Yesterday 12:57:23

George,Y wrote:

Uh it is surprising to find so many mathematicians practice meditation!

People can be spiritual without being religious –

#2 Re: Help Me ! » Arithmetic and Geometric Sequences » 2018-01-13 23:47:00

George,Y wrote:

Note the common ratio is a complex number.

It still works, provided the modulus of the common ratio is less than 1.

#3 Re: Help Me ! » Light bulbs and switches » 2018-01-13 03:36:07

For n ≡ 3 (mod 4), n is odd and the total number of switchings is even, which is impossible if each switch is to be flicked an odd number of times: the sum of an odd number of odd numbers must be odd. This is what you proved yourself in post #1.

For n ≡ 2 (mod 4), n is even and the total number of switchings is odd, again impossible as the sum of an even number of odd numbers must be even.

#4 Re: Help Me ! » Light bulbs and switches » 2018-01-12 18:03:18

So if n ≡ 2 or 3 (mod 4) it is never possible to have all lights on at the end. I will show that you can have them all on if n ≡ 0 or 1 (mod 4).

For n = 4, the process is simple:


using bob bundy’s notation. Let us give this an easy-to-remember name and call it the “four-switch process”.

Now label the switches 1, 2, …, n. Suppose n is a multiple of 4. The first four people perform the “four-switch process” to the first four switches. The next four people then toggle the first four switches and perform the “four-switch process” to switches 5 to 8. Then switches 5–8 will all be on; since switches 1–4 are already on, toggling them an even number of times will leave them on. Hence after the first 8 people, switches 1–8 will be on. Then we can continue the process, persons 9–12 toggling switches 1–8 and doing the “four-switch process” to swtiches 9–12, and so on. Hence when n ≡ 0 (mod 4) it is always possible to leave all the lights on.

Bob was scratching his head over the case n = 5; let me put him out of his misery. smile


The first person turns the first switch on, persons 2–5 toggle the first switch and perform the four-switch process on switches 2–5. So if n ≡ 1 (mod 4) proceed as follows: persons 6–9 toggle the first 5 switches and do the four-switch process on switches 6–9, persons 10–13 toggle the first 9 switches and do the four-switch process on switches 10–13, and so on – leaving all lights on at the end.

#5 Re: Help Me ! » Light bulbs and switches » 2018-01-12 16:11:36

chen.aavaz wrote:

Total number of switchings by the n people: 1+2+…+n = n*(n+1)/2 = always even
For all lights to be on after the whole process, each of them must have been switched by an odd number of times.
So this is a sum of odd numbers, and if n:odd then the total is odd, which is contradictory with the above.

The total number of switchings is even if either n or n+1 is divisible by 4, otherwise (as in bob bundy’s example) it is odd.

However you have managed to show that if n ≡ 3 (mod 4) then it is never possible to have all lights on at the end. You still need to determine whether it’s still the case for other values of n.

PS: You can also argue in similar fashion that if n ≡ 2 (mod 4) then again it’s not possible. This time the total number of switchings is odd and n is even.

#6 Re: Help Me ! » Exponential problem » 2018-01-11 05:06:46

bob bundy wrote:

Can anyone assist with the following problem to find a where x = 0.85, k=1 and y = 11.2;


So 11.2 = a^(-0.85)

#7 Re: Help Me ! » Problem around modular arithmatic » 2018-01-01 06:42:55

Let me put this in that language of group theory. We have

because 58 is the order of G, the multiplicative group of nonzero integers modulo 59 (a prime). If

then m must be a multiple of r, the order of 10 in the group G. Now r must be a divisor of the group order |G| = 58, i.e. its possible values are 1, 2, 29, 58. In my posts above, I eliminated 1, 2, 29. Therefore r = 58 (i.e. 10 is generator of the cyclic group G).

#9 Re: Help Me ! » Problem around modular arithmatic » 2018-01-01 02:22:49

Since 59 and 9 = 10 − 1 are coprime, any solution to

is also a solution to

i.e. to

and vice versa. By Fermat’s little theorem, as 59 is prime,

Therefore the smallest positive value of n+1 must divide 58. Clearly 10 ≢ 1 (mod 59) and 10² = 100 ≢ 1 (mod 59). If you can show that

then n+1 will be a multiple of 58 and the general solution will be n = 58k − 1, k = 1, 2, 3, ….

#12 Re: Maths Is Fun - Suggestions and Comments » Fourier Series » 2017-12-26 21:53:12

I think it looks great!

I'm not as au fait with Fourier series (and analysis topics in general) as I am with other, more algebraic areas of mathematics, so it's great to learn something new.

#13 Re: Help Me ! » Please Help! » 2017-12-19 17:16:54

What is the domain of the function f? It certainly cannot include 0, otherwise

Vedanti wrote:

f(3x)=f(x)+2 for all x

would imply 0 = 2 upon substituting x = 0. Please state the question more fully.

#16 Re: Guestbook » Correct Answer to 12 Days of Christmas Puzzle » 2017-12-11 21:46:32 … ution.html

It’s a partridge IN a pear tree, not partridge AND a pear tree. The tree provides a setting for the bird to be in and is not counted as a separate gift.

The solution is involves a sum of triangular numbers


#17 This is Cool » Numbers ending in 5 » 2017-12-03 21:18:00

Alg Num Theory
Replies: 1

To find the square of a number ending in 5:

  1. Drop the last digit 5.

  2. Multiply the number without the 5 by the number that’s one greater.

  3. Tack on the digits 25 at the end.

For example, to find 85²:

  1. 85 “−” 5 → 8.

  2. 8 × 9 = 72.

  3. 72 “+” 25 → 7225.

∴ 85² = 7225.

It works for numbers with more than two digits as well:

  • 135² = [13×14]25 = 18225.

  • 12345² = [1234×1235]25 = 152399025.

Why does it work? Because for any integer n we have

#18 Re: Help Me ! » Cubic Equation Solve for Infelction Point » 2017-12-01 19:49:43


This is a linear equation involving b only and so should be solvable for all values of b. The cubic function itself should also have inflection points for all values of the constant c since this only determines the height of the curve relative to the x-axis, not its shape.

#19 Re: Help Me ! » Group of 30 Table Rotations » 2017-11-25 06:28:15

When I saw the thread title, I thought it was going to be about the dihedral group of degree 30 (order 60)! lol

Anyway, I see you what you mean. I don’t think it’s possible. With more people than tables, a rotation will mean that at least two people sitting at one table will have to sit at the same table and see each other again.

#20 Re: Euler Avenue » Fermats Last Theorem » 2017-11-25 05:57:33

Ricky wrote:
Identity wrote:
JaneFairfax wrote:

Is this the most difficult proof ever?

No.  That would probably be, "Classify all finite simple groups."  That spanned thousands of papers and hundreds of mathematicians.  Another good one is "Prove that any group of odd order is solvable."  That one was 255 pages.

I believe we have a new contender for the record:
Baffling ABC maths proof now has impenetrable 300-page ‘summary’.

A summary of a massive mathematical proof that has baffled mathematicians for the past five years may help a few more people get to get grips with the key ideas. How long is the explainer? A mere 300 pages.

And that is only the summary: the original work – Shinichi Mochizuki’s proof of the ABC conjecture published in 2012, using a radical new theory developed over two decades – contained over 500 pages.

#21 Re: Help Me ! » Square of any determinant is symmetric. » 2017-11-24 21:55:49

Perhaps the book isn’t putting things very clearly. IMO what it’s trying to say is that the square of a determinant is the determinant of a symmetric matrix. For instance:

#23 Re: Exercises » What is 0^0 » 2017-11-24 11:38:11

Whether 0⁰ is defined or not, it is convenient in many mathematical formulas to treat it as equal to 1. For instance, consider the cosine power series:

If you put x = 0, you’ll find that the first term involves the expression 0⁰. Rather than saying the formula is invalid because 0⁰ is undefined, we happily let 0⁰ = 1 anyway.

#24 Re: Euler Avenue » Closures, interiors, and boundaries » 2017-11-24 10:58:27

Nehushtan wrote:

Here is an interesting example:

#25 Re: Help Me ! » A hard modular arithmetic problem » 2017-11-24 10:44:16

By direct calculation,

12! = 479001600 ≡ 736 (mod 2012),
20! = 2432902008176640000 ≡ 344 (mod 2012);

20!×12! ≡ 344×736 = 253184 ≡ 1684 (mod 2012).

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