The worlds first riddle!

Woops, almost forgot the best bit.

X / I - X = XI
Cross the slash to make an X and get:
XXI - X = XI

VI + VIII = III
Add a horizontal bar over 'I + VIII' that is connected to the first V creating a radical (square root symbol). You get:
sqrt(I + VIII) = III

Very clever. Now, we crank it up a notch.

XII = VII

VI / XXV = XL

BERLY
60-15-30-6=9 dollars

TENNIS
18-1=17 matches

DAYS
Sunday

AUREN
100/3=33.3 mph

Gotta think about the escalator.

VI / XXV = XL

IVI / XXV = XL (1000/25=40)

Whim

ESCALATOR
The escalator has 100 stairs.
The escalator is going at the same speed as Jim (if Jim weren't on the escalator).
They were running with the escalator.
John finishes in half the time of Jim.

Mark, who is clearly destined to go to the top, writes;

ESCALATOR
The escalator has 100 stairs.
The escalator is going at the same speed as Jim (if Jim weren't on the escalator).
They were running with the escalator.
John finishes in half the time of Jim.

The length is 100 stairs, the boys were running along the escalator which was moving with the same speed as the slow boy. Solution: in the time the fast boy stepped on 75 stairs, the slow one could step on only 25, so, since he stepped on 50, he spent twice as much time on the escalator as the fast one. Therefore, his speed relative to the ground was half that of the fast boy, therefore the escalator's speed was the same as the speed of the slow boy, and he counted exactly half the stairs.


A lady is locked in a dungeon, while the vile monster went to get firewood. When the monster is back in 15 min, he will cook and eat the lady. The lock is an advanced coded one: it has 4 oriented (top/bottom) spikes in a drum, arranged in a square and invisible. The lady can put her two hands into the drum, feel 2 of the spikes (adjacent or diagonal), determine their orientation and change it as she wishes.

If all 4 spikes end up the same direction (up or down), the lock opens and she can escape. If not, the drum rotates quickly for 1 min (so that she does not know which spikes she touched), then stops and she can try again.

Can she escape
How


According to a survey, around 90% of the professionals tested got all questions wrong. However, many pre-schoolers got several correct answers. This conclusively proves the theory that most management consultants have the brains of a four year old.

The following short quiz consists of four questions and tells whether you are qualified to be a "Consultant".

1. How do you put a giraffe into a refrigerator

2. How do you put an elephant into a refrigerator

3. The Lion King is hosting an animal conference, all the animals attend except one. Which animal does not attend

4. There is a river you must cross, but it is inhabited by crocodiles, and there is no bridge. How do you manage it

Whim, very clever.
VI / XXV = XL

IVI / XXV = XL (1000/25=40)


Re: XII = VII - Think ?'thick' line


XXII / VII = II

XXII / VII = II
This isn't some pi approximation is it? Connect the tops of the two Is on the right with a horizontal bar to form a pi symbol? 22/7=pi

DUNGEON LADY
I've seen this before as a rotating table problem. I don't recall the answer, but I'm fairly certain the solution required fewer than 15 steps; so she can escape.

Here's an easy one I made up:
V = X / II * L * XX

Here's a sneaky one I made up:
IV - X = VI - I

I give up on XII = VII

The square roots of N-5, N, N+5 are all rational numbers.

Hint to simplify the problem: N-5, N, N+5 can all be expressed as fractions with a denominator equal to 144.

What are N-5, N, N+5

Two right triangles with integer sides have their area equal to their perimeter.

What are their areas

Two similar triangles with integer sides have two of their sides the same. The third sides differ by 387.

What are the lengths of the sides

XII = VII

add a thick line to the bottom part of XII. The top part will show VII.

Mark: Are these still "add or remove one straight line"?


Whim

Whim: Yes they are.

XXII / VII = II
This isn't some pi approximation is it? Connect the tops of the two Is on the right with a horizontal bar to form a pi symbol? 22/7=pi

Absolutely not. A simple / will do the trick :wink:

DUNGEON LADY
I'm fairly certain the solution required fewer than 15 steps; so she can escape.

Thank goodness!

Whim to the rescue:

XII = VII
add a thick line to the bottom part of XII. The top part will show VII.

Mark,

"….Hint to simplify the problem: N-5, N, N+5 can all be expressed as fractions with a denominator equal to 144."

Talking about ?'144'

Prove that the area of a right triangle with integer sides is not a perfect square.

Let a, b, and c be the lengths of the sides of a right triangle with integer sides. These numbers, a, b, and c, are called "Pythagorean Triples". All Pythagorean Triples can be found using the following formula:
a = d*(m² - n²),
b = 2dmn,
c = d*(m² + n²),
where d is any positive integer, m>n>0, and m and n are relatively prime and of opposite parity (i.e. m+n is odd).

Then the area is ab/2, which is d²(m²-n²)(mn), which is d²(m+n)(m-n)(m)(n)
d²(m+n)(m-n)(m)(n) is a perfect square if and only if (m+n)(m-n)(m)(n) is a perfect square. (because only perfect squares have rational square roots) In other words, dividing by d² doesn't change whether the number is a perfect square.

So to show the area, ab/2 is not a perfect square, I need only show that (m+n)(m-n)(m)(n) is not a perfect square.
GCD(m,n)=GCD(m+kn,n), where k is an integer. (Linear Combination)
Since GCD(m,n)=1, then m is relatively prime to m+n and m-n. Similarly, n is relatively prime to m+n and m-n.
GCD(m+n,m-n) = GCD(2m,m-n) is at most 2, but since m-n is an odd number, GCD(m+n,m-n)=1.

Now it is clear that every factor is relatively prime to every other factor.
So either each of the four factors is a perfect square, or (m+n)(m-n)(m)(n) is not a perfect square.

Now I'm stumped again. I would like to proceed this way: If m-n, m, and m+n are all perfect squares, then n is a multiple of 24. Since n is a multiple of 24 and a perfect square, it must be a multiple of 144.



"Two similar triangles with integer sides have two of their sides the same. The third sides differ by 387.

What are the lengths of the sides "


If I had to give a rough ?'off the top of my head' reply. I might say;
512, 320, 200 and 320, 200, 125

XI / VIII=V


A row of freshly recruited soldiers is standing before a sergeant who gives a command to turn right. Some soldiers turn right, some left, and then they try to correct their mistake: if a soldier sees the face of his neighbour, he assumes that he turned the wrong way and turns around.

Will the situation stabilize

How

How long will it take



Tomorrow - The Top Ten Tablature Taboo Tableau Tangled Tabular Tacautac Tacit Toughest Tricky Testing Tactical Tantalizing Type To Take Talion Tally That Tangible Tangled Tapis.

A.K.A. The Math Riddle Challenge 2004. (Part I)

"XI / VIII=V"

XL / VIII = V


"XXII / VII = II"

X X II / V/ II = II


"V = X / II * L * XX "

V =/= X / II * L * XX
Un-equal-sign

"IV - X = VI - I"
. . . . . . . . _
IV - X = VI - I
There was no zero in roman numerals so perhaps also no negative numbers.

Try wrote:
If I had to give a rough ?'off the top of my head' reply. I might say;
512, 320, 200 and 320, 200, 125 Just popped into your head, eh?

Whim wrote:
V =/= X / II * L * XX
Un-equal-sign 'not equal' not allowed

"IV - X = VI - I"
. . . . . . . . _
IV - X = VI - I
There was no zero in roman numerals so perhaps also no negative numbers. I'm not sure what you're indicating here with the dots and bar.

SOLDIERS
Yes, the situation will stabilize.
If N is the number of soldiers, I believe it will stabilize within N-1 steps after the initial turn. The final configuration will consist of the leftmost M (0<=M<=N) facing left and the rightmost N-M facing right.
One way to convince yourself that it has to stabilize is to consider what happens at the ends. Once an end soldier is facing away from the line, he will not turn again. After that, the soldier next to him will never turn again once he is facing the end soldier. It tends to stabilize from the ends toward the middle. Worst case seems to be a soldier on the end facing in, and the rest of the soldiers facing him. This stabilizes from one end to the other.

Animals:

1. Open the refridgerator door, put the giraffe in.
2. Take the giraffe out, put the elephant in.
3. The elephant doesn't attend because it is in the 'fridge.
4. Swim the river. The crocodiles are at the meeting.

Whim cracks old roman riddle.

"XI / VIII=V"

XL / VIII = V


"XXII / VII = II"

X X II / V/ II = II


Mark cracks me up.

His ?'Soldiers' solution was spot on. (Damn)

As soon as the end soldiers turn outside, they will never turn again, thus induction shows that the situation will stabilize with the left-most k soldiers looking left and the right-most l soldiers looking right. It will take at most N-1 (N being the the number of soldiers) turns to stabilize, the maximum being achieved in the N-1 cases when some left-most soldiers are looking right and the rest are looking left. The average number of turns divided by N seems to be converging to about 0.6 from below.

Since whenever two soldiers look at each other's faces they turn around, the number of soldiers looking left and right is constant, so the number of soldiers who will be finally looking left (right) is the same as the number of soldiers who were looking left (right) initially.

If we encode a left-looking soldier as 1 and a right-looking soldier as 0, this is a parallel sorting routine, where at each step the neighboring unordered digits are transposed.


Merlin cracked the ?'Animals' riddle.

Which was;

The following short quiz consists of four questions and tells whether you are qualified to be a "Consultant".

1. How do you put a giraffe into a refrigerator

2. How do you put an elephant into a refrigerator

3. The Lion King is hosting an animal conference, all the animals attend except one. Which animal does not attend

4. There is a river you must cross, but it is inhabited by crocodiles, and there is no bridge. How do you manage it

Answer:


1. Open the refridgerator door, put the giraffe in.
2. Take the giraffe out, put the elephant in.
3. The elephant doesn't attend because it is in the 'fridge.
4. Swim the river. The crocodiles are at the meeting.

I did not know you were a ?'Management Consultant' :wink:



Free entry - Free questions - Free results

The Math Riddle Challenge 2004. (Part I) For the undisputed World Math Riddle Title.

Do not attempt if you are on medication. Seek professional advice.



1. 5 marks
How many non-congruent strictly isosceles triangles can be constructed using only 3 cm, 7 cm or 10 cm lengths for the edges?


2. 5 marks
Find the range of values of c such that the two lines x-y=2 and cx+y=3 intersect in the first quadrant.


3. 5 marks
A cubic polynomial P(x) is such that P(1)=1, P(2)=2, P(3)=3 and P(4)=5. Find P(6).


4. 5 marks
A rectangle contains three circles, all tangent to the rectangle and each other. If the width of the rectangle is 4, calculate its length.


5. 5 marks
A small class has four students. The teacher collects a test and immediately hands it out again so that the students can correct the test answers themselves. How many ways can the test be handed out so that no student receives his own paper?

MRC 2004

1. Taking strictly isosceles to not include equilateral, the answer is 4 (otherwise 7).

2. -1 < C < 1.5

3. 16

4. There doesn't appear to be enough information.

5. 9