Author Topic: Fun questions :) (Read 114924 times) Tweet Share

kamil9876 · « **Reply #120 on:** April 30, 2009, 07:10:39 pm »

bahaha algebraing on latex fail. saw an extra +a for some reason ;p

TrueTears · « **Reply #121 on:** April 30, 2009, 07:11:23 pm »

Quote from: kamil9876 on April 30, 2009, 07:10:39 pm

bahaha algebraing on latex fail. saw an extra +a for some reason ;p

haha always use the good ol' paper n pen

TrueTears · « **Reply #122 on:** April 30, 2009, 08:45:19 pm »

nice it is correct now kamil

humph · « **Reply #123 on:** May 01, 2009, 04:49:41 am »

gah i want to answer some of the new questions but i r failing at teh sobriety so they are too hard

TrueTears · « **Reply #124 on:** May 01, 2009, 01:15:45 pm »

Quote from: humph on May 01, 2009, 04:49:41 am

gah i want to answer some of the new questions but i r failing at teh sobriety so they are too hard

LOL

TrueTears · « **Reply #125 on:** May 01, 2009, 05:21:45 pm »

First post has reached it maximum of 10000 words. (consists of 50 questions)

New questions will still be posted in the first page, but in my 2nd post on that page. (the one after over9000's answer to Q2)

TrueTears · « **Reply #126 on:** May 01, 2009, 06:18:47 pm »

18 NEW QUESTIONS, FROM Q 42 - Q 60.

ENJOY YOURSELVES ~~~~

GerrySly · « **Reply #127 on:** May 01, 2009, 06:56:50 pm »

Quote from: TrueTears on April 23, 2009, 06:36:54 pm

55. Let $f_0(x) = \frac{1}{1-x}$ and $f_n(x) = f_0(f_{n-1}(x))$ where n = 1, 2, 3, 4... Evaluate $f_{2000}(2000)$

Just taking a massive stab in the dark really...

$\begin{align*} f_0(x)&=\frac{1}{1-x}\\ f_1(x)&=f_0[f_{1-1}(x)]\\ &=\frac{1}{1-\frac{1}{1-x}}\\ &=\frac{x-1}{x}\\ f_2(x)&=f_0[f_{2-1}(x)]\\ &=\frac{1}{1-\frac{x-1}{x}}\\ &=x\\ f_3(x)&=f_0[f_{3-1}(x)]\\ &=\frac{1}{1-x}\\ \end{align*}$
I am assuming that every $4n$ we renew the cycle, and the function is started again. Now if we went 4x499=1996. Now if we start the n counting, it goes 1996=1, 1998=2, 1999=3, 2000=4, therefore we use the function $f_0(x)$ .

Therefore

$\begin{align*} f_{2000}(2000)&=\frac{1}{1-2000}\\ &=-\frac{1}{1999} \end{align*}$

As I said, massive stab in the dark at that haha

TrueTears · « **Reply #128 on:** May 01, 2009, 07:43:07 pm »

Your method is correct, but you are not there yet! Just a few more steps

kamil9876 · « **Reply #129 on:** May 01, 2009, 08:42:56 pm »

54.)

In base n notation:

$10201=n^4+2n^2+1$
$=(n^2+1)^2$
which is a composite number for any natural n.

$10101=n^4+n^2+1$
$=n^4+2n^2+1-n^2$
$=(n^2+1)-n^2$
$=(n^2+1-n)(n^2+1+n)$

both factors are greater than 1 for all n>1. (n cannot equal 1 because 10101 does not exist in base 1 notation)

kamil9876 · « **Reply #130 on:** May 01, 2009, 11:12:46 pm »

51.)
from the centre, O, draw lines OB and OA and consider the triangle OAB. let r be radius of big circle.

It is one of those two sides the same triangle(forgot the name). the height perpendicular to AB is:

$h=\sqrt{r^2 -(\frac{AB}{2})^2}$

However this height is the radius of the smaller circle. Hence the area of the annulus is:

$Area=big circle - small circle$
$=\pi r^2 - \pi (\sqrt{r^2 - (\frac{AB}{2})^2})^2$
$=\pi r^2 - \pi r^2 + (\frac{AB}{2})^2$
$=\frac{AB^2 \pi}{4}$

TrueTears · « **Reply #131 on:** May 02, 2009, 12:40:31 am »

Quote from: kamil9876 on May 01, 2009, 08:42:56 pm

54.)

In base n notation:

$10201=n^4+2n^2+1$
$=(n^2+1)^2$
which is a composite number for any natural n.

$10101=n^4+n^2+1$
$=n^4+2n^2+1-n^2$
$=(n^2+1)-n^2$
$=(n^2+1-n)(n^2+1+n)$

both factors are greater than 1 for all n>1. (n cannot equal 1 because 10101 does not exist in base 1 notation)

correct!

TrueTears · « **Reply #132 on:** May 02, 2009, 12:41:42 am »

and 51 is correct too.

TrueTears · « **Reply #133 on:** May 02, 2009, 01:42:41 pm »

8 more questions, have fun ~!~

kamil9876 · « **Reply #134 on:** May 02, 2009, 10:38:30 pm »

66.)

n, n+1,n+2,n+3 are our numbers. The quantity we're interested in is:

$n(n+1)(n+2)(n+3)+1$
$=n(n^3 + 6n^2 + 11n^2 + 6n +6)+1$
$=n^4 + 6n^3 + 11n^2 +6n + 1$

If this is a square for all n then the above can be written as some quadratic squared.

$(n^2+an+b)^2=n^4+a^2n^2+b^2+2an^3+2bn^2+2abn$

$=n^4+2an^3+n^2(a^2+2b)+2abn+b^2$

This means that the following set of equations must have a solution:

$2a=6$
$a^2+2b=11$
$2ab=6$
$b^2=1$

the first and third equations imply that a=3, b=1. These values are consistent with the second and fourth equation, and are all natural coefficents. Hence such a quadratic exists and so for all natural n this quantity is a square.

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