Author Topic: TT's Maths Thread (Read 145816 times) Tweet Share

TrueTears · « **Reply #570 on:** December 31, 2009, 02:15:50 am »

Quote from: kamil9876 on December 31, 2009, 01:39:02 am

something feels sticky

lolololol

Prove that if $m$ is a square then $m$ leaves a remainder $0$ or $1$ when divided by $4$ .

kamil9876 · « **Reply #571 on:** December 31, 2009, 02:27:57 am »

if m is even. let $m=2k: m^2=4k^2$ which is a multiple of 4, hence remainder 0.

if m is odd. let $m=2k+1: m^2=4(k^2+k)+1$ which leaves remainder 1 when divided by 4.

TrueTears · « **Reply #572 on:** December 31, 2009, 02:33:58 am »

something feels sticky

TrueTears · « **Reply #573 on:** December 31, 2009, 07:11:32 pm »

Calculate the least non-negative residue of $3^8$ $\mod (13)$ .

/0 · « **Reply #574 on:** December 31, 2009, 07:15:53 pm »

TrueTears · « **Reply #575 on:** December 31, 2009, 08:02:59 pm »

Quote from: /0 on December 31, 2009, 07:15:53 pm

$3^3 \equiv 27 \equiv 1 \pmod {13}$

$\Rightarrow (3^3)^2 \equiv 1^2 \equiv 1 \pmod {13}$

$\therefore 3^8 \equiv (3^3)^2\times 3^2 \equiv 1 \times 3^2 \equiv 9 \pmod {13}$ xD

Cool. Thanks

TrueTears · « **Reply #576 on:** December 31, 2009, 08:04:48 pm »

Find the final decimal digit of $1! + 2! + 3! + ... + 10!$ .

kamil9876 · « **Reply #577 on:** December 31, 2009, 10:12:38 pm »

$\equiv 1!+2!+3!+4! mod 10$

since all the other terms are multiples of 10. This is because they contain a 5 and a 2 in the product, hence a 10.

$1!\equiv 1$

$2!\equiv 2$

$3!\equiv 6$

$4!\equiv 24=4+20\equiv 4$

therefore the sum $\equiv 1 + 2 + 6 + 4 \equiv 13 \equiv 3$

hence 3 is the last digit.

TrueTears · « **Reply #578 on:** December 31, 2009, 10:32:50 pm »

Quote from: kamil9876 on December 31, 2009, 10:12:38 pm

$\equiv 1!+2!+3!+4! mod 10$

since all the other terms are multiples of 10. This is because they contain a 5 and a 2 in the product, hence a 10.

$1!\equiv 1$

$2!\equiv 2$

$3!\equiv 6$

$4!\equiv 24=4+20\equiv 4$

therefore the sum $\equiv 1 + 2 + 6 + 4 \equiv 13 \equiv 3$

hence 3 is the last digit.

But you could also have $1!+2!+3!+4! \equiv a+b+c+d \pmod {10}$

So $1! \equiv 1 \pmod {10}$

$2! \equiv -8 \pmod {10}$

$3! \equiv -4 \pmod {10}$

$4! \equiv 24 \pmod {10}$

So $1!+2!+3!+4! \equiv 1-8-4+24 \equiv 13 \pmod {10}$

In other words how do you know what values to pick for $a,b,c,d$ ?

kamil9876 · « **Reply #579 on:** December 31, 2009, 10:53:24 pm »

still get mod 3 in the end

TrueTears · « **Reply #580 on:** December 31, 2009, 10:54:30 pm »

Quote from: kamil9876 on December 31, 2009, 10:53:24 pm

still get mod 3 in the end

But is there a way I can minimise the work at the end so I get 3 at the end?

kamil9876 · « **Reply #581 on:** December 31, 2009, 10:59:27 pm »

just delete all the digits from each summand, except the last one.

TrueTears · « **Reply #582 on:** January 01, 2010, 04:28:45 pm »

~~1. Prove that the polynomial $f(x) = x^5 - x^2 + x - 3$ has no integer roots.~~

2. Prove that there is no non-constant polynomial $f(x)$ , with integer coefficients, such that $f(x)$ is prime for all integers $x$ .

TrueTears · « **Reply #583 on:** January 01, 2010, 05:13:09 pm »

Quote from: TrueTears on January 01, 2010, 04:28:45 pm

2. Prove that there is no non-constant polynomial $f(x)$ , with integer coefficients, such that $f(x)$ is prime for all integers $x$ .

Assume $f(x)$ is a non-constant polynomial with integer coefficients and is prime for all integers $x$ .

This means for any integer $a \implies f(a) = p$ for any prime $p$ .

$f(a) \equiv 0 \pmod {p}$ since $f(a) = p$

But $a+pk \equiv a \pmod {p}$ for some constant $k$ .

$f(a+pk) \equiv f(a) \equiv 0 \pmod {p}$

Which means $f(a+pk) = p$ since $p$ is prime and the only thing $p$ can divide is itself.

This means $f(x) - p = 0$ has infinitely many solutions since if you plug in any number in the form of $a+pk$ then $f(x)-p=0$ is satisfied.

However $f(x)$ can only have at most $d$ roots where $d$ is the leading degree.

Contradiction!

brightsky · « **Reply #584 on:** January 01, 2010, 08:03:25 pm »

An interesting question:-

"There are only about 20 people in Victoria that can prove this." - Dr. Jianming He

Prove that 1 + 1 = 2.

Are the 20 people among the VN community?

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