Good Riddles

[fazstp]

Quote: Originally Posted by J Christopher 111 To be fair, 107 does not belong in the series. It should be: {1, 7, 11, 27, 77, 111, 127, 177, 777, 1127, … } It is the smallest integer (greater than zero) with n syllables in the English language: "One" has 1 syllable. "Seven" has 2 syllables. "Eleven" has 3 syllables. "Twenty-seven" has 4 syllables. "Seventy-seven" has 5 syllables. "One hundred seven" does not have 6 syllables. "One hundred eleven" has 6 syllables. "One hundred twenty-seven" has 7 syllables. Etc.

Bing bing bing bing bing!

I was counting the and in "one hundred and seven".

[J Christopher]

A man takes bids on a landscaping job. He wanted five rows of McIntosh apple trees with four trees in each row.

Upon having his bid selected, the landscaper goes to the nursery to purchase the trees. He tells the owner of the nursery that he has to plant five rows of McIntosh apple trees, with four trees in each row. The owner starts to fill out the invoice, and the landscaper notices that he has written down twenty trees.

"Oh no, I only need ten trees to do the job," the landscaper says.

"Do you mean you already have ten trees, and just need to buy the other ten?"

"No," replied the landscaper, "I only need ten trees to do the job."

"Are you getting the other ten trees from somewhere else?" the nursery owner asked.

"No," replied the landscaper, "I only need ten trees to do the job."

The owner of the nursery decided he would sell the ten trees, and then, when the landscaper realized that he needed ten more trees, he'd be back to buy them. Arrangements were made to have the ten trees delivered to the job site first thing the following Monday.

The landscaper never did return to the nursery for more trees. He was able to complete the job, planting five rows of McIntosh apple trees with four trees in each row, with only ten trees.

How did he do it?

[Felix_MC]

Haha, they gave this to us geometry last year, at the end of the year when we were doing games and stuff.
First, they never said that the rows should be parallel. So the landscaper planted the trees in the form of the Star of David, as there are 5 lines (or rows) in the Star of David. Then he planted 4 trees on each line since there are 2 intersections on each line and 1 tree to start the line and 1 to end it. If that doesn't make much sense, here's my illustration of it.
Sorry to have given away the answer so soon

[kel101]

but thats 12 trees not 10

[edalzell]

It works with a 5 pointed star, not 6.

[Felix_MC]

oh.. i was sleepy when I posted that..
The thing we had in geometry was with 6 rows of 4 trees I guess.. Cuz I no for sure it was a 6 pointed star
So I guess for this one it would be something like this?

[J Christopher]

Five pointed star is the correct answer, as illustrated in Felix_MC's link.

[EatsWithFingers]

I got told this one a few years ago, and have only just remembered it (and I do have a solution, although it's not short).

You have 12 coins and a set of scales. 11 of the 12 coins are of equal weight.

Using only the scales provided, work out which coin is the odd one out and whether it is heavier or lighter than the other 11.

Now, here's what makes it difficult: you cannot use the scales more than three times.

To make it a bit easier, I'll give you a clue. You start by weighing four coins against another four coins.


Good luck. And no cheating. I'll know.


PS - it's not a lateral thinking problem - there is a logical solution.

[Jay Carr]

The problem seemed easy until I ran into the problem of figuring out if it was heavier or lighter than the other coins...

[ThreeDee]

I don't think this one is even possible:


Connect each of the rectangles to each of the circles. Without any lines overlapping.

[Jay Carr]

Quote: Originally Posted by ThreeDee I don't think this one is even possible: Connect each of the rectangles to each of the circles. Without any lines overlapping.

So long as the lines don't have to be straight, I think I see a way to do it.

[ThreeDee]

They don't have to be straight, they can zigzag anywhere, as long as they don't overlap anything.

I got the puzzle from a friend. I think he was trying to play a joke on me.

[fazstp]

Does this count?

[EatsWithFingers]

Quote: Originally Posted by ThreeDee They don't have to be straight, they can zigzag anywhere, as long as they don't overlap anything. I got the puzzle from a friend. I think he was trying to play a joke on me.

It's a known fact in graph theory that K(3,3) cannot be drawn on a 2D plane (i.e. there is no solution to the problem). See this Wikipedia article for details.

Of course, if we're talking 3D, then there would be a solution. For example, a square-bottomed pyramid, where two of the bottom corners and the top are one shape, and the other two bottom corners and the middle of the base are the other shape.

EDIT: A rough picture is attatched (remember, it's in 3D though, so there are overlapping lines in the picture)

[fazstp]

Quote: Originally Posted by EatsWithFingers I do have a solution, although it's not short.

I can see why. I've worked through a couple of variations and they're almost harder to explain than to work out in the first place.

[EatsWithFingers]

Quote: Originally Posted by fazstp I've worked through a couple of variations and they're almost harder to explain than to work out in the first place.

Care to try?

[ThreeDee]

Quote: Originally Posted by EatsWithFingers It's a known fact in graph theory that K(3,3) cannot be drawn on a 2D plane (i.e. there is no solution to the problem). See this Wikipedia article for details. Of course, if we're talking 3D, then there would be a solution. For example, a square-bottomed pyramid, where two of the bottom corners and the top are one shape, and the other two bottom corners and the middle of the base are the other shape. EDIT: A rough picture is attatched (remember, it's in 3D though, so there are overlapping lines in the picture)

It was in 2D. So that means I wasted about an hour (and a bunch of paper), trying to figure out the impossible???

[fazstp]

Quote: Originally Posted by EatsWithFingers Care to try?

Ok here goes. Excuse the shorthand. Let me know if you see any flaws here.

Labelling weights A-L

Scenario 1)
ABCD < EFGH
AFG < EBI
AE < BC

A is lighter


Scenario 2)
ABCD > EFGH
AFG > EBI
AE > BC

A is heavier


Scenario 3)
ABCD < EFGH
AFG > EBI
GB < AI

B is lighter


Scenario 4)
ABCD > EFGH
AFG < EBI
GB > AI

B is heavier


Scenario 5)
ABCD < EFGH
AFG = EBI
CH < EB

C is lighter


Scenario 6)
ABCD > EFGH
AFG = EBI
CH > EB

C is heavier


Scenario 7)
ABCD < EFGH
AFG = EBI
CH = EB

D is lighter


Scenario 8)
ABCD > EFGH
AFG = EBI
CH = EB

D is heavier


Scenario 9)
ABCD > EFGH
AFG > EBI
AE < BI

E is lighter


Scenario 10)
ABCD < EFGH
AFG < EBI
AE > BI

E is heavier


Scenario 11)
ABCD > EFGH
AFG < EBI
FB < EI

F is lighter


Scenario 12)
ABCD < EFGH
AFG > EBI
FB > EI

F is heavier


Scenario 13)
ABCD > EFGH
AFG < EBI
FB = EI

G is lighter


Scenario 14)
ABCD < EFGH
AFG > EBI
FB = EI

G is heavier


Scenario 15)
ABCD > EFGH
AFG = EBI
CH < BI

H is lighter


Scenario 16)
ABCD < EFGH
AFG = EBI
CH > BI

H is heavier


Scenario 17)
ABCD = EFGH
ABC > IJK
I < J

I is lighter


Scenario 18)
ABCD = EFGH
ABC < IJK
I > J

I is heavier


Scenario 19)
ABCD = EFGH
ABC > IJK
I > J

J is lighter


Scenario 20)
ABCD = EFGH
ABC < IJK
I < J

J is heavier


Scenario 21)
ABCD = EFGH
ABC > IJK
I = J

K is lighter


Scenario 22)
ABCD = EFGH
ABC < IJK
I = J

K is heavier


Scenario 23)
ABCD = EFGH
ABC = IJK
A > L

L is lighter


Scenario 24)
ABCD = EFGH
ABC = IJK
A < L

L is heavier

[fazstp]

Quote: Originally Posted by ThreeDee It was in 2D. So that means I wasted about an hour (and a bunch of paper), trying to figure out the impossible???

This link from the Wikipedia page goes into this specific problem.

[Felix_MC]

Quote: Originally Posted by EatsWithFingers I got told this one a few years ago, and have only just remembered it (and I do have a solution, although it's not short). You have 12 coins and a set of scales. 11 of the 12 coins are of equal weight. Using only the scales provided, work out which coin is the odd one out and whether it is heavier or lighter than the other 11. Now, here's what makes it difficult: you cannot use the scales more than three times. To make it a bit easier, I'll give you a clue. You start by weighing four coins against another four coins. Good luck. And no cheating. I'll know. PS - it's not a lateral thinking problem - there is a logical solution.

AHA!! After like an hour, some 7 flash-cards and a mild headache, I think I've finally got it!!
I've put it into C++ code (hence I'm not that good with explaining, and it made more logical sense when I was trying to figure it out)
In the code 'a' through 'l' are the coins 1 through 12.
I designed the code so it works for each of the numbers if either they are heavier or lighter then the rest. Hopefully it works, I haven't really tested it

{
	if (a+b+c+d==e+f+g+h)
	{
		double sum1,sum2;
		sum1 = a + k;
		sum2 = j + i;
		if (sum1=sum2)
		{
			if (a<l)
			{
				cout<<"The 12th coin is heavier than the rest."<<endl;
			}
			if (a>l)
			{
				cout<<"The 12th coin is lighter than the rest."<<endl;
			}
		}
		if (sum1>sum2)
		{
			if (j=i)
			{
				cout<<"The 11th coin is heavier than the rest."<<endl;
			}
			if (j<i)
			{
				cout<<"The 10th coin is lighter than the rest."<<endl;
			}
			if (j>i)
			{
				cout<<"The 9th coin is lighter then the rest."<<endl;
			}
		}
		if (sum1<sum2)
		{
			if (i=j)
			{
				cout<<"The 11th coin is lighter than the rest."<<endl;
			}
			if (i<j)
			{
				cout<<"The 10th coin is heavier than the rest."<<endl;
			}
			if (i>j)
			{
				cout<<"The 9th coin is heavier than the rest."<<endl;
			}
		}
	}
	if (a+b+c+d<e+f+g+h)
	{
		double sum3,sum4;
		sum3 = a + b + e;
		sum4 = c + f + i;
		if (sum3=sum4)
		{
			if (g=h)
			{
				cout<<"The 4th coin is lighter than the rest."<<endl;
			}
			if (g<h)
			{
				cout<<"The 8th coin is heavier than the rest."<<endl;
			}
			if (g>h)
			{
				cout<<"The 7th coin is heavier than the rest."<<endl;
			}
		}
		if (sum3<sum4)
		{
			if (a=b)
			{
				cout<<"The 6th coin is heavier than the rest."<<endl;
			}
			if (a<b)
			{
				cout<<"The 1st coin is lighter than the rest."<<endl;
			}
			if (a>b)
			{
				cout<<"The 2nd coin is lighter than the rest."<<endl;
			}
		}
		if (sum3>sum4)
		{
			if (c=k)
			{
				cout<<"The 5th coin is heavier than the rest."<<endl;
			}
			if (c<k)
			{
				cout<<"The 3rd coin is lighter than the rest."<<endl;
			}
		}
	}
	if (a+b+c+d>e+f+g+h)
	{
		double sum5,sum6;
		sum5 = f + i + c;
		sum6 = g + h + d;
		if (sum5=sum6)
		{
			if (a=b)
			{
				cout<<"The 5th coin is lighter than the rest."<<endl;
			}
			if (a<b)
			{
				cout<<"The 2nd coin is heavier than the rest."<<endl;
			}
			if (a>b)
			{
				cout<<"The 1st coin is heavier than the rest."<<endl;
			}
		}
		if (sum5>sum6)
		{
			if (f=g)
			{
				cout<<"The 3th coin is heavier than the rest."<<endl;
			}
			if (f<g)
			{
				cout<<"The 7th coin is lighter than the rest."<<endl;
			}
			if (f>g)
			{
				cout<<"The 8th coin is lighter than the rest."<<endl;
			}
		}
		if (sum5<sum6)
		{
			if (d=k)
			{
				cout<<"The 6th coin is lighter than the rest."<<endl;
			}
			if (d>k)
			{
				cout<<"The 6th coin is heavier than the rest."<<endl;
			}
		}
	}	
	return 0;
	
}

EDIT: Looks like fazstp beat me to it, and posted the solution before me