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#201 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:

[list=1]
[*]Drop the last digit 5.[/*]
[*]Multiply the number without the 5 by the number that’s one greater.[/*]
[*]Tack on the digits 25 at the end.[/*]
[/list]

For example, to find 85²:

[list=1]
[*]85 “−” 5 → 8.[/*]
[*]8 × 9 = 72.[/*]
[*]72 “+” 25 → 7225.[/*]
[/list]

∴ 85² = 7225.

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

[list=*]
[*]135² = [13×14]25 = 18225.[/*]
[*]12345² = [1234×1235]25 = 152399025.[/*]
[/list]

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

[list=*]
[*]

[/*]
[/list]

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

Solve
[list=*]
[*]

[/*]
[/list]
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.

#203 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.

#204 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.

#205 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:

[list=*]
[*]

[/*]
[/list]

#206 Re: Help Me ! » Equation » 2017-11-24 21:30:14

Brijesh wrote:

[list=*]
[*]

[/*]
[/list]

#207 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:

[list=*]
[*]

[/*]
[/list]

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.

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

Nehushtan wrote:

Here is an interesting example:

#209 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).

#210 Re: Help Me ! » Prime numbers importance » 2017-11-24 06:27:18

A prime number is a number (positive integer) that is divisible only by 1 and itself (excluding the number 1 itself). Examples of prime numbers are 2, 3, 5, 7, 11, 13, ….

IMHO the most important practical use of prime numbers in this digital age is their use in encrypting passwords. In the RSA cryptosystem, for example, there is asymmetry between the public encryption key and the private decryption key, based on the practical difficulty of factorizing the product of two large prime numbers. The method is as follows:

A user of RSA creates and then publishes a public key based on two large prime numbers, along with an auxiliary value. The prime numbers must be kept secret. Anyone can use the public key to encrypt a message, but with currently published methods, and if the public key is large enough, only someone with knowledge of the prime numbers can decode the message feasibly.

Prime numbers crop up in many areas of mathematics. The fundamental theorem of arithmetic states that every integer greater than 1 either is a prime or can be factorized uniquely (apart from the order of factors) into prime numbers. In abstract algebra, the integers are thought of as forming a subring of the field of rational numbers, and algebraic-number theory tries to generalize the unique-factorization property of the integers to algebraic integers, which are subrings of algebraic-number fields. The study of such UFDs (unique-factorization domains) was motivated by the search for a proof of Fermat’s last theorem. A complete proof proved elusive – it was not until the last decade of the 2oth century that Andrew Wiles succeeded where all others had failed by proving the Taniyama–Shimura conjecture for semistable elliptic curves – and the best that could be achieved before the 20th century was Kummer’s proof using the theory of ideals that the equation

[list=*]
[*]

[/*]
[/list]
has no nonzero integer solutions in x, y, z when n is an odd regular prime (or a multiple thereof).

#211 Re: Help Me ! » Common Multiples of 9, 10, 11, and 12? » 2017-11-24 05:18:56

lorynjane wrote:

I have a homework that i am really stuck with! dunno
Does anyone know what all of the common multiples of 9, 10, 11 and 12 please??

The common multiples of 9, 10, 11, 12 are the multiples of their lcm (lowest common multiple).

To work out their lcm, first note that 9, 10, 11 are coprime so their lcm is simply their product: lcm(9,10,11) = 9 × 10 × 11 = 990.

Now we work out the lcm of 990 and 12. This is equal to their product divided by their gcd (greatest common divisor). We have gcd(990,12) = 6. Hence lcm(990,12) = (990×12)/6 = 1980.

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