Author Topic: Fun questions :) (Read 114280 times) Tweet Share

TrueTears · « **Reply #75 on:** April 27, 2009, 06:49:48 pm »

nicely done. hah

kamil9876 · « **Reply #76 on:** April 27, 2009, 09:15:33 pm »

38.)

Basically you have to find a number that is of the form $k*5^{88}$ . Let's focus on how to get those 88 5's. First of all when doing n! you will get some multipled of 5. some of them contribute a single factor of 5.(5,10,15) some contribute two (25,50,75) etc...

the pattern is that every 5 multiples of 5 is of the form 5^2. Every 5 multiples of 5^2 is a multiple of 5^3. etc.**

$5^{88}=5*5*5*5*5*5....$ 88 times

6 goes into 88 14 times. Hence in the above expression there are 14 strings of the form 5*5*5*5*5^{2} and so

$5^{88}=(5*5*5*5*5^2)^{14} *5^{88-14*6}$
$=(5*5*5*5*5^2)^{14} *5^{4}$
which is looking more similair to our sequence of multiples of 5. Reread ** if unsure.

However, every 5 of those $5^2$ is a $5^3$ . 5 goes into 14 twice. Hence the 5th and 10th 5^2 must by donated an extra 5 so that it turns into a 5^3 so that we have something that looks more similair to our pattern of multiples of 5. To do this well will steal two 5's from the 5^4 sitting at the end. SO now our expression looks like this:

$5^2=(5*5*5*5*5^2.... 5^3... 5^3....) 5*5$ (1)

there are only two $5^3$ so we won't worry about any $5^4$ . So now we have the pattern of multiple of 5's in an appropriate form in which the numbers of terms can be easily counted. there are 14*4 single 5's in the brackets. 12 lots of $5^2$ and 2 lots of $5^3$ and two lots of single 5's outside the brackets.

That's a total of 72 numbers being multiplied. hence (72*5)! gives u that pattern found in (1). Hence n=360

TrueTears · « **Reply #77 on:** April 27, 2009, 09:39:09 pm »

anyone care to solve Q 19?

it's quite lonely being stuck between all those solved questions

correct solution btw kamil.

humph · « **Reply #78 on:** April 27, 2009, 10:19:20 pm »

Quote from: TrueTears on April 23, 2009, 05:52:59 pm

40. Take any whole number q. Calculate $q^2-1$ . Factorize $q^2-1$ to give two factors a and b (not necessarily $q+1$ and $q-1$ ). Put $c=a+b+2q$ . Then you will find that $ab+1$ , $bc+1$ and $ca+1$ are all perfect squares. Prove that this method always gives three perfect squares.

Suppose $ab = q^2 - 1$ , where $a,b,q \in \mathbb{N}$ , and set $c=a+b+2q$ . Then
$ab+1 = q^2 - 1 + 1 = q^2$
$bc+1 = ba + b^2 + 2bq + 1 = q^2 - 1 + b^2 + 2bq + 1 = b^2 + 2bq + q^2 = (b+q)^2$
$ca+1 = a^2 + ab + 2aq + 1 = a^2 + q^2 - 1 + 2bq + 1 = a^2 + 2aq + q^2 = (a+q)^2$
and so this method always gives three perfect squares.

TrueTears · « **Reply #79 on:** April 27, 2009, 10:21:59 pm »

nice solution.

Over9000 · « **Reply #80 on:** April 27, 2009, 10:55:57 pm »

39. I start with 3 marbles, one red, one green and one blue. I can trade any one marble for two others, one each of the other two colours. Is it possible to make a number of such trades and end up with five more blue marbles than red? Show all working. (I don't care how many green marbles I have at the end.)

Let G=green, R=Red, B=Blue

Start off with R,G,B (1B,1R) = same

trade a green for red and blue

R,(B,R),B (2B,2R) = same

trade a red for a green and blue

(G,B),(B,R),B (3B,1R) = +2R

trade a green for red and blue again

(R,B,B)(B,R),B (4B,2R) = +2R

trade a red for a green and blue

(G,B,B,B)(B,R),B (5B,1R) = +4R

trade a green for red and blue again

(R,B,B,B,B)(B,R),B (6B,2R) = +4R

This is the result of always trading in red and green to increase blue more than red, however, even if we try to trade a blue for a green and red
We lose one blue and gain one red, so there is still a net even difference between red and blue

So we see a pattern of an even difference between Red marbles and Blue marbles, 5 being an odd number means a difference of 5 is impossible

TrueTears · « **Reply #81 on:** April 27, 2009, 10:57:14 pm »

splendid.

Could you also determine a general case for this? (vectors)

humph · « **Reply #82 on:** April 27, 2009, 11:15:53 pm »

Quote from: TrueTears on April 23, 2009, 05:52:59 pm

31. Given integers m and n such that $1 \le m < n$ , how many positive solutions must the equation $-1 - x^m + x^n = 0$ have?

Let $f(x) = x^n - x^m - 1$ .
Then $f'(x) = nx^{n-1} - mx^{m-1} = x^{m-1}(nx^{n-m} - m)$ , and so $f'(x) = 0$ when $x = (m/n)^{1/(n-m)}$ (and possibly at $x = 0$ if $m > 1$ ).
Moreover, $f'(x) < 0$ for $0 < x < (m/n)^{1/(n-m)}$ and $f'(x) > 0$ for $x > (m/n)^{1/(n-m)}$ (as we know the only turning point of $f$ on $(0,\infty)$ , and so we only need to check the gradient at any point either side of this point to know its behaviour).
Thus $f$ is monotonically decreasing for $0 < x < (m/n)^{1/(n-m)}$ and monotonically increasing for $x > (m/n)^{1/(n-m)}$ .

Now note that $f(0) = -1$ . As $f$ is monotonically decreasing for $0 < x < (m/n)^{1/(n-m)}$ , we must have that $f\left((m/n)^{1/(n-m)}\right) < 0$ , and there must be no zeroes of $f$ in $\left(0,(m/n)^{1/(n-m)}\right)$ .

On the other hand, $f$ is monotonically increasing for $x > (m/n)^{1/(n-m)}$ .
Furthermore, $f(2) = 2^n - 2^m - 1 = 2^m (2^{n-m}-1)-1 > 0$ as $n > m \geq 1$ implies that $2^{n-m} - 1 \geq 2 - 1 = 1$ and hence $2^m (2^{n-m}-1)-1 \geq 2^m - 1 > 0$ . Thus $f\left((m/n)^{1/(n-m)}\right) < 0 < f(2)$ .
As $f$ is continuous and monotonically increasing, this implies by the Intermediate Value Theorem that $f$ has exactly one zero on the interval $\left[(m/n)^{1/(n-m)},2\right]$ .

Finally, $f(2) > 0$ and $f$ is monotonically increasing for $x > 2$ implies that $f$ has no zeroes on the interval $(2,\infty)$ .

Putting these three cases together, we find that the equation $x^n - x^m - 1 = 0$ must have exactly one positive solution.

TrueTears · « **Reply #83 on:** April 27, 2009, 11:23:17 pm »

Nice! you got it

humph · « **Reply #84 on:** April 27, 2009, 11:42:29 pm »

Quote from: TrueTears on April 23, 2009, 05:52:59 pm

41. Semicircles are drawn on the sides of a rectangle ABCD as shown in the diagram. A circle passing through points ABCD carves out four crescent-shaped regions (coloured yellow and green in the diagram). Prove that the sum of the areas of the four crescents is equal in area to the rectangle ABCD.
(Image removed from quote.)

Let $\ell$ be the length of the sides $AD$ and $BC$ , and let $w$ be the length of the sides $AB$ and $DC$ . Then the radius of the black circle is half the length of the diagonal, so that by Pythagoras' theorem we have that $r^2 = (\ell/2)^2 + (w/2)^2$ . Thus the area of the black circle (including the rectangle inside it) is $\pi r^2 = \frac{\pi}{4}(\ell^2 + w^2)$ , and so the area of the shaded black region (i.e. the circle without the rectangle inside it) $\frac{\pi}{4}(\ell^2 + w^2) - \ell w$ .
The semicircle connected to $AD$ has radius $\ell/2$ , and so its area (which includes the black bit as well as the green crescent) is $\pi(\ell/2)^2/2 = \pi \ell^2/8$ . Similarly with the other three semicircles. Adding the area of these four semicircles comes out to be $\frac{\pi}{4}(\ell^2 + w^2)$ .
Thus the area of the four crescents is the area of the four semicircles minus the area of the black region;
$\frac{\pi}{4}(\ell^2 + w^2) - \left(\frac{\pi}{4}(\ell^2 + w^2) - \ell w \right) = \ell w$ ,
which is the area of the rectangle.

TrueTears · « **Reply #85 on:** April 27, 2009, 11:46:27 pm »

Nice explanation absolutely nailed it in the head like a pr0

kamil9876 · « **Reply #86 on:** April 28, 2009, 12:15:45 am »

35.)

$r-2+r-1+r+r+1+r+2=n^3$
$5r=n^3$

and similairly:

$3r=k^2$

divide both equations and rearange:

$5k^2=3n^3$

k and n are both natural. Hence they are made from prime factors. The prime factors on the left must equal prime factors on right, hence both sides of the equation must have the same number of prime factors. The smallest possible number of prime factors on each side is 7 since 2*3=6 plus that extra 3 and 5. hence k 3 prime factors while n has 2. subscripting these bad boys:
$ 5(k_1 k_2 k_3)^2=3(n_1 n_2)^3$
$5 (k_1)^2 (k_2)^2 (k_3)^2 = 3 (n_1)^3 (n_2)^3$

Because the prime numbers must be identical, we need some 5's on the right and some 3's on the left. hence say $k_1=3$ and $n_1=5$ . Now we have:

$5 * 3^2 (k_2)^2 (k_3)^2=3 * 5^3 * (n_2)^3$

So now the only possibility is $n_2=3, k_2=5, k_3=3$

So now we know the smallest possible values for $k$ and $n$ . Subbing any of these into the appropriate equation found at the beggining of my post will do the trick... giving $r=3^3 * 5^2$

kamil9876 · « **Reply #87 on:** April 28, 2009, 12:30:46 am »

37. If the line y = 2x+3 is reflected in the line y = x+1, find what its equation becomes.
(sauce: /0)

If it was y=x that would be easy so let's turn it into such a situation. We will superimpose another x axis called x' at y=1 (shifted the x axis one unit up). Hence line y=2x+3 becomes: y'=2x+2 (no need to prime the x). Hence we are reflecting the line y'=2x+2 in the line y'=x. So the line becomes the inverse:

x=2y'+2
y'=0.5x-1

Hence now we need to change this new reflected line back into our old axis: hence

y=0.5x-1+1
y=0.5x

Do i get the sauce?

humph · « **Reply #88 on:** April 28, 2009, 03:12:50 am »

Quote from: TrueTears on April 23, 2009, 05:52:59 pm

36. Find all integers that leave a remainder of 3 when divided by 5 , a remainder of 5 when divided by 7 , and a remainder of 7 when divided by 11.
(sauce: /0)

This is a classic Chinese Remainder Theorem question (if you ever do an elementary number theory course at uni then a question like this will probably be on the exam).

We have the system of equations
$ \begin{align*} x & \equiv 3 \pmod{5} \\ x & \equiv 5 \pmod{7} \\ x & \equiv 7 \pmod{11} \end{align*} $
So from the first equation, we know that $x = 5k_1 + 3$ for some $k_1 \in \mathbb{Z}$ .
Substituting this value of $x$ into the second equation, we have that
$ \begin{align*} 5k_1 + 3 & \equiv 5 \pmod{7} \\ \Longrightarrow 5k_1 & \equiv 2 \pmod{7} \\ \Longrightarrow k_1 & \equiv 6 \pmod{7} \end{align*} $
by multiplying through by $3$ . Thus $k_1 = 7k_2 + 6$ for some $k_2 \in \mathbb{Z}$ , and so $x = 35k_2 + 33$ .
Substituting this value of $x$ into the third equation, we have that
$ \begin{align*} 35k_2 + 33 & \equiv 7 \pmod{11} \\ \Longrightarrow 35k_2 & \equiv 7 \pmod{11} \\ \Longrightarrow k_2 & \equiv 9 \pmod{11} \end{align*} $
by multiplying through by $6$ . Thus $k_2 = 11k_3 + 9$ for some $k_3 \in \mathbb{Z}$ .
We therefore have that solutions to this system of congruences are of the form $x = 385k_3 + 348$ , where $k_3 \in \mathbb{Z}$ .

TrueTears · « **Reply #89 on:** April 28, 2009, 05:02:06 pm »

nice correct answer for Q 35 kamil

and what type of sauce you want for Q 37?

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