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I think they still use the Elliptic curves ECF or Quadratic Sieves.

Take a look here:

https://www.alpertron.com.ar/ECM.HTM

]]>Write solutions like this.

]]>This is one of the 7 things of the pascals triangle which i find very interesting.

]]>Do not worry about it, it is okay. It is enough that you have it. Post if you get stuck on one or find one interesting.

I will not get stuck since i have the solutions book(yes a solutions book is also available there.....). Yeah i will send you the questions which i find difficult and interesting.

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https://brilliant.org/wiki/converse-of-intermediate-value-theorem/

]]>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

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.]]>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.]]>The formula is too complicated to remember. It can be remembered much more easily by Implementing Faulhaber's formula. It says,

1ˣ +2ˣ +3ˣ +4ˣ ......nˣ =[1/(1+x)](aB₀nˣ⁺¹+bB₁nˣ+cB₂nˣ⁻¹............yBₓn), where Bₙ is the nth Bernoulli no. and a,b,c,.....y are the consecutive terms of (x+1)th row of Pascal's Triangle.

For e.g.

1¹¹+2¹¹+3¹¹+4¹¹.....7¹¹=(1/12)(1B₀n¹²+12B₁n¹¹+66B₂n¹⁰+220B₃n⁹+495B₄n⁸+792B₅n⁷+924B₆n⁶+792B₇n⁵+495B₈n⁴+220B₉n³+66B₁₀n²+12B₁₁n)

(1+B)ⁿ⁺¹-Bₙ₊₁=0

For e.g.

for,B₁ n=1 so

(1+B)²-B₂=0

⇒1+B₂+2B-B₂=0

⇒1=2B=0

⇒B=-0.5

However in this ( the formula for sums of powers )formula B₂=+0.5]]>

Sorry, haven't been keeping up with the "this is cool" feed. But nice job!

]]>URL:http://www.mathsisfun.com/numbers/infinity.html

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