# Math Is Fun Forum

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## #1 Re: This is Cool » Divisibility test for 7 & 13... » 2020-05-08 23:58:57

If the number (y) is odd and divisible by 7........

(y -1)/2=a
If a is odd, b = (a-1) / 2, if a is even b = (a +y-1) / 2
If b is odd, c = (b-1) / 2, if b is even c = (a +y-1) / 2
If y is divisible by 7, c will be divisible by 7..........

Works for any Mersenne Number....

Works because whatever I do to y to get to zero I have to be able to do to a multiple of y, to get to a remainder zero for y! And 7-1/2=3 3-1/2=1 1-1/2=0.

Example:35= 5*7,
35-1/2=17, 17-1/2=8 (8+35-1)/2=21,
21= 3*7

Example: 27=3*9
27-1/2=13 13-1/2=6
6=3*2

3-1/2=1
1-1/2=0

## #2 Help Me ! » Is this Publishable? » 2020-05-08 20:06:44

Replies: 0

I don’t mean to sound stupid. But I got a formula to work out the remainder for mod Mersenne Prime, working backwards on the Lucas Lehmer Primality Test Sequence. The sequence is (s^2) - 2 moving forwards. Is this publishable? If so, where? It seems really obvious to me but I can’t see anything on Wikipedia. And on Wikipedia they’ve put that some guy proved one step going backwards on the sequence = +/- 2^((p+1)/2) but no mention of ANY step????

## #3 Re: Help Me ! » Can someone help me write a proof for my idea? » 2019-01-13 12:35:18

Thanks a lot Grantingriver! I think I understood all that you said. That’s great stuff.

I’ve actually posted on another forum where the idea’s been changed to:

If an integer, 2p + 1, where p is a prime number, is a divisor of the Mersenne number

, then 2p + 1 is a prime number.

My argument is that because divisors of the Mersenne number

can’t be < p if p is a prime number. Therefore if 2p +1 is a divisor of
it has no divisors as p is > the square root of 2p + 1. This will therefore make 2p + 1 a prime number.

What I’m also interested in is how to put the idea that:

An integer minus one, divided by 2, p times equals zero, will be a factor of

.
You may add the integer as necessary if minus one, divided by 2 results in an even number.

## #4 This is Cool » If 2p + 1 is a factor of 2^p - 1 then it is prime, proof. » 2019-01-13 03:29:21

Replies: 0

If an integer, 2p + 1, where p is a prime number, is a divisor of the Mersenne number

, then 2p + 1 is a prime number.

My argument is that because divisors of the Mersenne number

can’t be < p if p is a prime number. Therefore if 2p +1 is a divisor of
it has no divisors as p is > the square root of 2p + 1. This will therefore make 2p + 1 a prime number.

Is this proof correct?

## #5 Help Me ! » Can someone help me write a proof for my idea? » 2019-01-12 11:10:49

Replies: 2

I’ve got two ideas which if proven wrong will hopefully be a learning experience.

The first is a different way to do the Fermat Primality test.

And the second is that 2p^2 + 1 if it passes the Fermat Primality test and is not factorable by 3 will be prime.

Anyone interested? (I don’t know anything about Maths really, apart from GCSE and A-levels and my own research)

## #6 Re: This is Cool » Something ineteresting » 2019-01-10 03:56:41

I was going to put.....

Every integer >5 can be expressed as the sum of 3 primes

Because I find that really beautiful, but then I realised that this was still just a conjecture.

Whoops!

## #7 Re: Puzzles and Games » The Monty Hall Problem » 2019-01-09 17:29:31

Say you choose a goat, it’s in your best interest to change.
Say you choose a car, it’s not in your best interest to change.

You are statistically more likely to choose a goat initially.
Therefore it is statistically better to change.

## #8 This is Cool » Sophie Germain primes and Mersenne number divisors » 2019-01-08 12:36:40

Replies: 0

It is known from Wikipedia that:

“ If a Sophie Germain prime p is congruent to 3 (mod 4), then it’s matching safe prime 2p + 1 will be a divisor of the Mersenne number 2^p - 1.”

And:

“ Fermat's little theorem states that if p is a prime number, then for any integer a, the number a^p − a is an integer multiple of p.”

However if an integer, 2p + 1, where p is a prime number, is a divisor of the Mersenne number 2^p - 1, then 2p + 1 is a safe prime and p it’s matching Sophie Germain prime.

Divisors of the Mersenne number 2^p - 1 can’t be < p if p is a prime number. Therefore if 2p +1 is a divisor of 2^p - 1 it has no divisors as p is > the square root of 2p + 1. This will therefore make 2p + 1 a safe prime and p it’s matching Sophie Germain prime.

For example 11 which is prime, (11*2) + 1 = 23. 2^11 - 1 is divisible by 23 making 11 a Sophie Germain prime and 23 it’s matching safe prime.

## #9 Re: Dark Discussions at Cafe Infinity » The Meaning Of Life Is In The Dictionary! :) » 2018-08-27 09:03:30

Yes. I would say it’s culturally biased, just being a normal/good citizen of that society. I don’t think there is an ultimate meaning of life, just personal meanings of life. It’s a question I’ve been most concerned about.

Common themes might include not being on drugs, getting a fairly good job, looking attractive to the opposite gender, having a tidy house, not breaking the law of that country etc. I also think soldiers would be particularly proud of themselves for serving their country, but that’s down to the individual. Contrarily, being an unemployed drug addict, not having good personal hygiene, living in a very messy house and having a criminal record might leave you feeling none too proud of yourself. I don’t think your family would be too proud of you either, but anyone can change!

The revelation to me was that they don’t actually want to live like that, they want to be proud of themselves. I always thought drug addicts etc. don’t want to change sometimes, but if they’re truly honest with themselves they actually do. (According to the dictionary).

Why you get pleasure and satisfaction from doing what makes you proud is another question entirely. Who cares? If you know what you want to do in life the only problem is how you’re going to do it.

## #10 Re: Dark Discussions at Cafe Infinity » Logical thinking of Humans » 2018-08-27 08:34:08

Oh! So you use abstract concepts to solve problems but you’d get to the same answer if you did it the long way round. I see. I would still say logic is a universal tool though. Just because some other people use it differently doesn’t mean the actual tool is different. ???

## #11 Re: Dark Discussions at Cafe Infinity » Logical thinking of Humans » 2018-08-26 07:59:13

I thought the tool of logic was universal?

What are you saying then? You use the existence of things that don’t exist to prove things that do exist?

How does that work?

If A then B
No A
Therefore B

Wouldn’t your reasoning just be false then?

.........It’s very interesting.

## #12 Dark Discussions at Cafe Infinity » The Meaning Of Life Is In The Dictionary! :) » 2018-08-26 06:07:14

Replies: 3

According to the Cambridge English Dictionary:

Proud means feeling pleasure and satisfaction because you or people connected with you have done or got something good

Satisfaction means a pleasant feeling that you get when you receive something you wanted, or when you have done something you wanted to do

Happy means feeling, showing, or causing pleasure or satisfaction

Enjoy means to get pleasure from something

Let:
P=Proud
S=Satisfaction
L=Pleasure
W=Getting/Doing what you want
H=Happy
E=Enjoy

If P then L and S
If S then W
If L or S then H
Therefore If P then W
Therefore if P then H
Therefore If P then W and H

If E then L
If L then H
Therefore If E then H

So, if you’re proud of yourself you will have got what you wanted and you will be happy. As opposed to if you’re just enjoying yourself you will be happy but it’s not what you want to do.

So, the meaning of life according to the dictionary is to do good things, not necessarily good things like helping people or being a good person, but good things that make you feel proud of yourself, like achieving things or giving up drugs or getting a good job etc. Whatever makes you feel proud of yourself! It will make you happy and it’s what you truly want! It will also make people around you happy too because they’ll be proud of you too. And it’s what they truly want for you.

## #13 This is Cool » Which Mersenne Number is this number a factor of? Proof » 2018-08-01 21:19:42

Replies: 0

Whatever done to a Mersenne number to get to zero must be able to be done to it’s factors.

Mersenne Number= 2^11 -1
Factors = 23 and 89
2^11 -1 takes eleven goes to get to zero if I continually minus one and divide by 2. Watch what happens to 23 and 89 acting as remainders as I do the same to them.

2^11 -1
2^10 -1           First go
2^9 -1            Second go
2^8 -1            Third go
2^7 -1             Fourth go
2^6 -1             Fifth go
2^5 -1             Sixth go
2^4 -1             Seventh go
2^3 -1             Eighth go
2^2 -1             Ninth go
2-1                  Tenth go
1-1=0              Eleventh go

23
(23-1)/2 = 11                                First go
(11-1)/2 = 5                                  Second go
(5-1)/2 = 2                                   Third go
23+2 = 25, (25-1)/2 = 12           Fourth go
23+12 = 35, (35-1)/2 = 17         Fifth go
(17-1)/2 = 8                                  Sixth go
23+8 = 31, (31-1)/2 = 15            Seventh go
(15-1)/2 = 7                                  Eighth go
(7-1)/2 = 3                                    Ninth go
(3-1)/2 = 1                                    Tenth go
(1-1)/2 = 0                                     Eleventh go

89
8(89-1)/2 = 44                                  First go
89+44 = 133, (133-1/2) = 66       Second go
89+66 = 155, (155-1)/2 = 77       Third go
(77-1)/2 = 38                                  Fourth go
89+38 = 127, (127-1)/2 = 63        Fifth go
(63-1)/2 = 31                                  Sixth go
(31-1)/2 = 15                                   Seventh go
(15-1)/2 = 7                                     Eighth go
(7-1)/2 = 3                                       Ninth go
(3-1)/2 = 1                                       Tenth go
(1-1)/2 = 0                                        Eleventh go

They both take eleven goes to get to zero too.

Working backwards it can be seen that this must be the case. Starting with zero I multiply by 2 and add one continually until I get to 2^11 -1. The  same must be done to the remainders which will then become 23 and 89 or zero, the factors of 2^11 -1.

Let the number of goes an odd number, y, takes to get to zero by continually minusing one and dividing by 2, (adding y when even) =z.

The z for 2^11 -1, 23 and 89 =11.

Whatever z equals for y, 2^z -1 must be factorable by y. When we use the method for 2^z -1, we never add 2^z -1 as it never equals an even number. This could potentially alter remainders for potential factors, but for Mersenne Numbers, this is not the case.

## #14 Re: This is Cool » Proving Primality Using Minus 1 Divided By 2 » 2018-07-31 10:22:56

If z is the smallest Mersenne number that an odd number, y, will be a factor of.

Does anyone know what “z” a composite can’t be?

## #15 Re: This is Cool » Proving Primality Using Minus 1 Divided By 2 » 2018-07-22 05:47:43

If z= a factor of (y-1), we know that 2^(y-1) -1 is factorable by y, as the process starts again for a multiple as seen above. So y is either a prime or a Fermat pseudoprime, which according to Wikipedia are rather rare.

Another way is to take any odd number, y, you want to know if it's prime. So count the number of times it takes you to get to zero by minusing 1 and dividing by 2, adding y if even, this number of times =z. If z= a factor of (y-1) then y is either a prime or a Fermat pseudoprime. Otherwise it is composite.

## #16 Re: This is Cool » Proving Primality Using Minus 1 Divided By 2 » 2018-06-30 06:29:22

Thanks Bob

Yes you are right. 11 has z=10 which is not prime although is above (square root 11), so we could try using the method again, but we can't because 10 is even. So we can't determine if 11 is prime, and in turn 23 using this method. But if we know 11 is prime and above square root 23, we know 23 is prime, using this method.

The method does not prove that all primes are prime. You seem to have the method down, so I'll expand a little on the proof.

I forgot to mention that for mersenne Numbers, 2^p -1, z=p as you can see below;

2^11 -1
2^10 -1           First go
2^9 -1            Second go
2^8 -1            Third go
2^7 -1             Fourth go
2^6 -1             Fifth go
2^5 -1             Sixth go
2^4 -1             Seventh go
2^3 -1             Eighth go
2^2 -1             Ninth go
2-1                  Tenth go
1-1=0              Eleventh go

So, as 2^10 -1 is factorable by 11 according to Fermats Little Theorem, I know that the z for 11 must be 10 or smaller. So as z's can never be bigger than their corresponding y's, if z is bigger than the square root I know that that exact number did not occur in any potential Factors below square root y. And if y has no factors below it's square root, I know it's prime. HOWEVER, if z is a multiple, y could be a factor of something with a smaller z. If you can imagine z being prime and when you get to z goes you turn it into zero and then it takes 2z goes to get to zero again, then 3z, 4z, 5z etc.
Example:

7 is a factor of 49
49 takes 21 goes to get to zero
7 takes 3 goes to get to zero
49's z= a multiple of 3
Therefore 49 has the potential to be factorable by other y's with z=3
7 has z=3, 7 has the potential to be a factor of 49

(0x2) +1=1           First go
(1x2) +1=3           Second go
(3x2) +1=7           Third go
7-7=0                   Process starts again
(0x2) +1=1           Fourth go
(1x2) +1=3           Fifth go
(3x2)+1=7            Sixth go
7-7=0                   Process starts again
(0x2) +1=1           Seventh go
.................
(3x2) +1=7           Twenty first go

They are only ever potential factors, which is why 11 can have z=10, which has the potential to be factorable by z=2 or z=5, the factors of 10.

Hope this helps......?

(I must admit I'm not very good at explaining things.)

## #17 This is Cool » Proving Primality Using Minus 1 Divided By 2 » 2018-06-27 01:25:24

Replies: 4

THEORY
Take any odd number, y, count the number of times it takes you to get to zero by minusing 1 and dividing by 2, this equals, z. If this action results in an even number, simply add y. If z is a prime number >(square root of y), then y is prime. Example:

y=23
(23-1)/2 = 11                                First go
(11-1)/2 = 5                                  Second go
(5-1)/2 = 2                                   Third go
23+2 = 25, (25-1)/2 = 12           Fourth go
23+12 = 35, (35-1)/2 = 17         Fifth go
(17-1)/2 = 8                                  Sixth go
23+8 = 31, (31-1)/2 = 15            Seventh go
(15-1)/2 = 7                                  Eighth go
(7-1)/2 = 3                                    Ninth go
(3-1)/2 = 1                                    Tenth go
(1-1)/2 = 0                                     Eleventh go
Therefore z=11
z>(square root 23) AND z is prime THEREFORE 23 is prime.

Also if you don't know if z is prime , you can repeat the process!

PROOF
Take any odd number, y, minus 1 and divide by 2 until you get to zero, adding y as necessary when it results in an even number. The factor on the other hand will operate the same way. You can see that this must be the case if you work backwards. What is the remainder for n for 0/n? That's correct, zero. What have I done to y after that? I've multiplied it by 2 and added 1, so I must do the same to the remainder to get the new remainder. What happens when I add y? Well, you will effectively be adding a multiple of n, so you would end up just misusing them again to get the remainder. Eventually on the last number of goes I will end up with y and the remainder n, for y/n where n is a factor.

If y is a prime number the number of goes, z, will never be greater than (y-1). Proof of this is in the fact that, according to Fermats Little Theorem, if y is prime, 2^(y-1) -1 will be factorable by y. So y will always be a factor of 2^(y-1) -1 making z=y-1 or smaller, having occurred beforehand in a smaller z.

Therefore if z= a prime >(square root of y) no potential factor, ( primes below the square root of y) can possibly have the same z because they can't be that big.

## #18 This is Cool » Mersenne Number Factors Using Minus 1 Divided By 2 » 2018-06-22 04:58:25

Replies: 0

Mersenne numbers are of the form 2^p -1.
If I minus 1 and divide by 2 it will take p goes to get to zero.
Example:

2^11 -1
2^10 -1           First go
2^9 -1            Second go
2^8 -1            Third go
2^7 -1             Fourth go
2^6 -1             Fifth go
2^5 -1             Sixth go
2^4 -1             Seventh go
2^3 -1             Eighth go
2^2 -1             Ninth go
2-1                  Tenth go
1-1=0              Eleventh go

Similarly if 2^p -1 has a factor of the form, n, it should take p goes to get to zero. Only, if minus 1 divided by 2 results in an even number, I should be allowed to add n to the number to continue. As this is what happens to n, as a remainder for 2^p -1, as each go takes place.
Examples:

23
(23-1)/2 = 11                                First go
(11-1)/2 = 5                                  Second go
(5-1)/2 = 2                                   Third go
23+2 = 25, (25-1)/2 = 12           Fourth go
23+12 = 35, (35-1)/2 = 17         Fifth go
(17-1)/2 = 8                                  Sixth go
23+8 = 31, (31-1)/2 = 15            Seventh go
(15-1)/2 = 7                                  Eighth go
(7-1)/2 = 3                                    Ninth go
(3-1)/2 = 1                                    Tenth go
(1-1)/2 = 0                                     Eleventh go

89
(89-1)/2 = 44                                  First go
89+44 = 133, (133-1/2) = 66       Second go
89+66 = 155, (155-1)/2 = 77       Third go
(77-1)/2 = 38                                  Fourth go
89+38 = 127, (127-1)/2 = 63        Fifth go
(63-1)/2 = 31                                  Sixth go
(31-1)/2 = 15                                   Seventh go
(15-1)/2 = 7                                     Eighth go
(7-1)/2 = 3                                       Ninth go
(3-1)/2 = 1                                       Tenth go
(1-1)/2 = 0                                        Eleventh go

Both 23 and 89 take eleven goes to get to zero, as they must do, as they are factors of 2^11 -1, and 2^11 -1 takes eleven goes to get to zero also.

## #19 This is Cool » To find out if x is prime using only addition! » 2017-09-04 14:57:58

Replies: 0

Let x= ANY ODD "+" INTEGER >3

Let y=(x/3) Rounded down

IF any of the below are TRUE then x is composite OTHERWISE x is prime.

x+1=EVEN triangular number
x+1+2= ODD triangular number
x+1+2+3=EVEN triangular number
x+1+2+3+4= ODD triangular number
x+1+2+3+4+5=EVEN triangular number
x+1+2+3+4+5+6= ODD triangular number
........etc.
Until x+1+2+3......+y

EXAMPLE

x=35
x/3=11+2/3 rounded down =11
Therefore y=11
x+1=35+1=36 36 is an EVEN triangular number
Therefore 35 is composite

EXAMPLE

x=17
x/3=5+2/3 rounded down =5
Therefore y=5
x+1=17+1=18
x+1+2=18+2=20
x+1+2+3=20+3=23
x+1+2+3+4=23+4=27
x+1+2+3+4+5=27+5=32
Therefore 17 is prime

## #20 Re: This is Cool » More accurate than the Prime Number Theorem. » 2017-07-23 08:00:50

Oh, I see. I just thought it was impossible to get an accurate approximate of the number of primes without actually counting them.......... .

## #21 This is Cool » More accurate than the Prime Number Theorem. » 2017-07-22 04:06:27

Replies: 2

The primes there are in a certain range can be estimated because there are;
1 No. not factorable by 2 in (2)
There are 2 No.'s not factorable by 2 or 3 in (6)
There are 8 No.'s not factorable by 2 or 3 or 5 in (30)
There are 48 No.'s not factorable by 2 or 3 or 5 or 7 or in (210)
times 48 by (prime -1) to get the next number of no.'s. i.e. =480 no.'s in (2310) not factorable by 2,3,5,7, or 11.....and so on.

So just work out the percentages and apply the correct one to your number x. For example you will not be concerned with No.'s not factorable by 13 if x<13 squared=169. This should be a far more accurate method than the Prime Number Theorem that approximates the number of primes upto x as x/log(x). You will need a computer to do this.

## #22 This is Cool » Triangular numbers and composites » 2017-04-20 14:17:37

Replies: 1

Triangular numbers are the sum of 1+2+3+4+5+6+7+8..........
a and b= triangular numbers where a>b
Any odd composite= a or (a-b)

This is because the sum of any group of numbers all separated by +1, I.e. 3,4,5,6,7 will be an odd composite with factors of the middle number and the length.

Examples:

2+3+4+5+6+7+8
Middle number=5
Length=7
Therefore=5*7=35

14+15+16+17+18+19+20
Middle number=17
Length=7
Therefore=7*17=119

116+117+118
Middle number=117
Length=3
Therefore=3*117=351

Therefore I would have thought we would be able to compute prime numbers faster by minusing the potential prime, p, off triangular numbers <p to see if they equal another triangular number. If they don't p is prime. This surely must be faster than seeing if p is factorable by all possible factors......................?

Does anyone know how people are using computers to test if very, very, VERY large numbers are prime?

## #23 This is Cool » An easy way to find out if a number is prime. » 2017-01-14 22:08:44

Replies: 0

p=Any number
A=2x3x5x7x11x13x17........ up to

m=Any multiple

A-pm= A number factorable by a factor of p

if p has one.

To get A-pm, go through the primes, subtracting p as many times as you like.

Example:
p=129

Rd. Dwn. to nrst. prime=11
A=2x3x5x7x11
2x3x5x7=210 210-129=81
81x11=891 891-(129x6)=117
A-129m=117
117 will have a factor the same as p if it has any.
117/129=39/43 129 is NOT prime (Common denominator =3)

In this way we can find out if p is composite without ever having to use a number

. Might be useful for computers.

## #24 This is Cool » Largest prime gap theorem » 2017-01-13 02:22:37

Replies: 0

Basically I take my number p, square root it, Rd. Up to the nearest prime, then the next prime greater than that -1= largest prime gap below p.

Example:
p=130

Rd. Up to nearest prime= 13
Next prime after that = 17
17-1=16
Largest prime gap  <130 = 16 (Correct)

This works because the greatest number of composites between two primes occurs when factors are not combined. So what could have been two composites is actually just one, like 15=3x5. To create the greatest possible number of composites I start at 2 not 0. 0 has an infinite number of prime factors, and so the greatest gap between the next repeat will occur after 0. Starting with the smallest composite which is NOT combined factors, I move up. Deleting all numbers factorable by primes less than the square root. The first time I attain TWO primes is when I reach the second prime after the square root. So this -1 is the gap required to create two non-composites with greatest possible occurrence of composites.

Largest prime gap <p =

Rd. Up to second nearest prime -1.