Author Topic: Fun questions :) (Read 113775 times) Tweet Share

/0 · « **Reply #15 on:** April 25, 2009, 02:16:28 pm »

Let the number of digits be $n$ and the number be $m$ .

The number will then be of the form: $a_{n-1} 10^{n-1}+a_{n-2} 10^{n-2}+...+a_{0},\ \ a_{n-1} \neq 0$

So we have $10^{n-1} < m < 10^{n}$

$n-1< \log_{10} m<n$

$n < \log_{10} m +1<n+1$

Since $n$ is an integer, $\log_{10}m+1$ will be somewhere between $n$ and $n+1$ . But if we use the floor function, we have

$n = \lfloor \log_{10} m\rfloor+1$

Looking at our number, we have $n=\lfloor\log_{10} (200^{1988})\rfloor+1=\lfloor\log_{10}(2 \times 10^{2})^{1988}}\rfloor+1$

$n=\lfloor 1988 (\log_{10}(2)+\log_{10} 10^2) \rfloor+1=\lfloor 1988(\log_{10} 2+2)\rfloor+1=3977+\lfloor 1988\log_{10} 2\rfloor$

Since $\log_{10} 2=\frac{\log_2 2}{\log_2 10} = \frac{1}{\log_2 10} \approx 0.3$

$n \approx 3977+\lfloor 1988 \times 0.3\rfloor=596+3977=4573$

Hence, C.

TrueTears · « **Reply #16 on:** April 25, 2009, 02:17:24 pm »

Correct kamil and /0

kamil9876 · « **Reply #17 on:** April 25, 2009, 02:57:38 pm »

Quote from: TrueTears on April 25, 2009, 11:34:43 am

$\therefore 1^3 +2^3 + ...+ 100^3 = (1^3 + ...+7^3) + (8^3 + ... + 14^3) + ... +99^3 + 100^3 = 0 + 0 + ... + 1 + 1 = 2$

Missing a few mods there

btw, am i missing something, because i didn't see the word remainder in the statement of question 1.

TrueTears · « **Reply #18 on:** April 25, 2009, 03:00:52 pm »

Quote from: kamil9876 on April 25, 2009, 02:57:38 pm

Quote from: TrueTears on April 25, 2009, 11:34:43 am

$\therefore 1^3 +2^3 + ...+ 100^3 = (1^3 + ...+7^3) + (8^3 + ... + 14^3) + ... +99^3 + 100^3 = 0 + 0 + ... + 1 + 1 = 2$

Missing a few mods there

btw, am i missing something, because i didn't see the word remainder in the statement of question 1.

soz for everyone who attempted Q1, i forgot to put the word remainder LOL my bad.

TrueTears · « **Reply #19 on:** April 25, 2009, 03:41:58 pm »

7. Alternative method.

expansion of $10^n$ will require $n+1$ digits.

$\implies$ $2^{10} = 1024 \approx 10^3$

$2^{1000} = (2^{10})^{100} \approx (10^{3})^{100} = 10^{300}$

$2^{1988} = \frac{2^{2000}}{2^{12}} = \frac{(2^{1000})^2}{2^{12}} \approx \frac{10^{600}}{4 \times 10^3} = \frac{10^{597}}{4}$

$\therefore (200)^{1988} = 2^{1988} \times 100^{1988} \approx \frac{10^{597}}{4} \times 10^{3976} = \frac{10^{4573}}{4}$

Hence C

Over9000 · « **Reply #20 on:** April 25, 2009, 04:34:36 pm »

10. Find all the real roots to $x^4 - 40x^3 + 206x^2 -40x +1 = 0$
By observation, $(x^{2} - ax + 1)(x^{2} - bx + 1)$ must be quadratic factors of the quartic as coefficients of a and e are 1.
expanding $(x^{2} - ax + 1)(x^{2} - bx + 1)$ yields $x^{4} - ax^{3} - bx^{3} + abx^{2} + 2x^{2} -ax -bx +1$
collecting like terms gives us $x^{4} - (a + b)x^{3} + (2 + ab)x^{2} - (a + b)x + 1$
so $x^{4} - (a + b)x^{3} + (2 + ab)x^{2} - (a + b)x + 1 = x^4 - 40x^3 + 206x^2 -40x +1$
$\therefore$ $a+ b = 40 .....1$
$2 + ab = 206...2$
from 1
$a = 40 - b$
$\therefore$ $2 + (40 - b) \times b = 206$ from subbing 1 into 2
$2 + 40b - b^{2} = 206$
$b^{2} -40b +204 = 0$
$(b-6) (b-34)=0$
b=6 and b=34
$\therefore$ a=34 and a=6
since when a=34 b=6, or when a=6 b=34, we can choose a=34 and b=6, or b=34 and a=6
so $x^4 - 40x^3 + 206x^2 -40x +1 = (x^{2} - 6x + 1)(x^{2} - 34x + 1)$
$\therefore$ $(x^{2} - 6x + 1)(x^{2} - 34x + 1) = 0$
$x^{2} - 6x + 1 = 0$
$(x-3)^{2} - 8 = 0$
$(x-3 -2\sqrt{2})(x-3 + 2\sqrt{2})=0$
so $x=3 \pm 2\sqrt{2}$

and $x^{2} -34x + 1 = 0$
$(x-17)^{2} - 288 = 0$
$(x-17 - 12\sqrt{2})(x-17 + 12\sqrt{2})$
so $x=17 \pm 12\sqrt{2}$

$\therefore$ $x=3 \pm 2\sqrt{2}$ and $x=17 \pm 12\sqrt{2}$ are real solutions to the polynomial $x^4 - 40x^3 + 206x^2 -40x +1 = 0$

TrueTears · « **Reply #21 on:** April 25, 2009, 04:39:21 pm »

Well done over9000, correct method and answer.

TrueTears · « **Reply #22 on:** April 25, 2009, 04:49:13 pm »

6 new questions added, have fun

/0 · « **Reply #23 on:** April 25, 2009, 05:11:29 pm »

6

Assuming we start in 1988, we want to finish in 1787, so we want to go back in time 201 years.

We time travel in the following increments:
+1, +4, +16, +64, +256 for the 1st, 3rd, 5th, 7th, and 9th switch.
-2, -8, -32, -128, -512 for the 2nd, 4th, 6th, 8th, and 10th switch.

The dial looks like this:
_ _ _ _ _ _ _ _ _ _

We can't go back 201 years using -2, -8, -32, or -128, we must overshoot first, choosing -512.

_ _ _ _ _ _ _ _ _ 1

Now we need to go 311 years into the future. We can do this by adding +256 and +64:
_ _ _ _ _ _ 1 _ 1 1

Now we must go 9 years into the past, which we can do by picking -8, -2, +1, hence:

11 0 1 0 0 1 0 1 1

12

If you want $n$ consecutive numbers to sum to 75, then we can case bash:
Solve the following equation for $x$ and if x is not a natural number then disregard it. Stop when x becomes negative.
$nx+\frac{n\times (n-1)}{2}=75$
We see there are 5 ways to sum 75 using at least 2 positive integers.

OK sorry I'll stop now

TrueTears · « **Reply #24 on:** April 25, 2009, 05:18:02 pm »

Well done Q 6 is correct.
Well done Q 12 is correct.

Over9000 · « **Reply #25 on:** April 25, 2009, 05:30:25 pm »

13. How many times in one day do the hands of a clock form a right angle? (Assume the hour hand does not move, eg, say on a real clock when it is 5:50 pm, the hour hand tends to move towards 6. But assume for this question the hour hand stays exactly on the hour it is.)

12am-12:15,12:45
1am-1:20,1:50
2am-2:25,2:55
3am-3:30, 3:00
4am-4:35,4:05
5am-5:40,5:10
6am-6:45, 6:15
7am-7:50, 7:20
8am-8:55, 8:25
9am-9:00, 9:30
10am-now this is the tricky part, only 10:35 is a valid answer, the other right angle that should be formed at 10:05 is actually 11:05 so only 10:35 is an answer
11am-11:40 (same reasoning as 10am)
12pm-12:15, 12:45
1pm-same as 1 am
2pm-same as 2am
...and so on
so there are a total of 22 solutions for each half a day, therefore it makes right angles with both hands 44 times a day.

TrueTears · « **Reply #26 on:** April 25, 2009, 06:08:16 pm »

Correct over9000

/0 · « **Reply #27 on:** April 25, 2009, 06:41:34 pm »

Don't odd switches take the time traveler forwards in time?

Over9000 · « **Reply #28 on:** April 25, 2009, 06:51:20 pm »

Quote from: /0 on April 25, 2009, 06:41:34 pm

Don't odd switches take the time traveler forwards in time?

yep, sorry misunderstood

Over9000 · « **Reply #29 on:** April 25, 2009, 06:59:06 pm »

14. In a circle of radius 10 cm, a chord is drawn 6 cm from the centre. If a chord half the length of the original chord were drawn, its distance, in cm, from the centre would be?

Let centre of circle on Cartesian plane be (0,0)
So equation of the circle is $x^{2} + y^{2} = 100$
When the chord is 6 cm from the centre, it has a y value of 6 on cartesian plane
so chord intersects circle at (+x,6) (-x, 6)
sub in y=6, into $x^{2} + y^{2} = 100$
$x^{2} + 36 =100$
$x^{2} = 64$
$x = \pm 8$
$\therefore$ intersects circle at (8,6) and (-8,6)
Distance of chord =16
so for a chord that has distance $=\frac{1}{2} \times 16=8$
ntersects circle at (4,y) (-4,y)
sub x = 4 into $x^{2} + y^{2} = 100$
$16 + y^{2} = 100$
$y^{2} = 84$
$y = \pm \sqrt{84}$
therefore the chord with half the length would be $\sqrt{84}$ cm from the centre

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