Puzzle Corner

[daamon]

The grid has me stumped. enlighten me!

As you know, A=85, B=3612, C=3613. This revolves around Pythagoras Triangles:

3^2 + 4^2 = 5^2
5^2 + 12^2 = 13^2
13^2 + 84^2 = 85^2
85^2 + 3612^2 = 3613^2

Based on this, working out A=85 is straight forward.

I chose the numbers in the first column based on the number in the last column and row above - just to throw people off the scent...

But working out B & C takes a little more thought as there's 2 unknowns. You need to look into how you can identify the 3 sides (x, y and z) of the n-th Pythagoras triangle. A bit of head scratching reveals:

No........x..............y.................z
1..........3..............4.................5
2..........5.............12...............13
n......2n+1......2n(n+1)......2n(n+1)+1

So, you know that 2n+1 = 85 gives n = 42. Plugging this in gives B & C as 3612 and 3613 respectively (not to mention a way of calculating th n-th Pyhtagorean Triangle - whatever use that is!).

[flaninacupboard]

and -that's- why i couldn't figure it

I tried for ages but i don't have fantastic maths skills.

[daamon]

SquirrelDip did it a different way - it purely solves this puzzle rather than calculating any Pythagorean Triangle, but still valid. It was simpler:

Observation suggests 85^2 + B^2 = C^2

But, by observation again, C = B + 1, so:

7225 + B^2 = (B+1)^2

7225 + B^2 = B^2 + 2B + 1

7224 = 2B (subtracting B^2 and 1 from both sides)

B = 3612, hence C = 3613.

Still a bit maths-y though...

[daamon]

HOUSEKEEPING:

New puzzles required. Apply to this thread...

Still outstanding (in order of posting) - as best I can see:

"A Farmer & His Sons" - Much earlier from Devanshu, missed in the original housekeeping. Sorry

"Clever Dead Guy" - flaninacupboard posted, but already knew a similar one so (kindly) edited his post.

Rules / Guidelines

Of course, you don't have to follow these rules, coz I'm not Baldrick - but it'd help.

[tgpo]

edit

[tgpo]

2 6s back to back. Dont you just love the simpsons?

Great thread by the way.

EDIT: Here's mine...

A farmer and his sons are carrying grain sacks to the barn at the end of a day. The farmer is carrying one sack of grain and his sons are carrying two sacks. Who is carrying the heavier sack?

The farmer. He has grain in his.

[daamon]

New Puzzle: "Dr. Jones"

First of all - Please, if you know this one, don't post the answer. I think it's a real challenge. Unless they know it, I think even SquirrelDip, northcat_8 and Capmaster will have fun with this one... (The gauntlet is down gentlemen!).

There are no hidden tricks or "silly" answers, you have all the info you need to solve the problem.

You've all heard of the emininent archealogoist and explorer Dr. Oklahoma Jones? Well, it seems that he's in trouble again...

Dr. Jones finds himself trapped inside a cavern containing 31 bags of gold coins. Each bag has an unknown quantity of coins in, and so many that he would run out of air before he could count them all.

Inscriptions on the wall say that to escape, the one bag of fake coins must be identified by removing it completely from it's spot to reveal the hidden doorway. Removing any other bag will mean that he's trapped until he dies.

Reading on he sees that the fake coins are identical in terms of size, colour, look, feel etc. - the only difference is that a fake coin weighs 0.1g less than a real coin. Though the weight of a real coin isn't given, you do know that it is a whole number of grams...

Fortunately, Dr. Jones has his trusty digital scales with him (accurate to 0.1g funnily enough). Unfortunately, he knows that there's only enough battery life to take one weight measurement. He doesn't have any other batteries or means of generating electricity to power the scales.

How does Dr. Jones escape?

[flaninacupboard]

from each bag he takes a prime number of coins, and weighs them all together. the total weight can then be divided by the total number of coins. from this it should be possible (through some clever mathematics ) to work out which prime multiple of 0.1g is the culprit.

[daamon]

Nope.

Anyway, too vague - "through some clever mathematics". What "clever mathematics"?

Refer to my PM.

[SquirrelDip]

"Dr. Jones"

Here's how far I've got... daamon - am I on the right track?

Take on coin out of bag 1, two out of bag 2, three out of three and so on to bag 31

Total number of coins = 1 + 2 + 3 + ... + 31 = 496

Let the number of fake coins be F (which is also the bag number)
Let the number of good coins be G
Let the weight of a good coin be W
The weight of a fake coin is therefore (W - 0.1)

Weigh the 496 coins and let their weight be T

Now,

F + G = 496

T = (G * W) + F * (W - 0.1)

Combie the two and you get:

T = (496 * W) - (0.1 * F)

Rearrange for F:

F = ([496 * W] - T) / 0.1

This is where I'm stuck... I think I need another piece of information. Like, if W is an even multiple of 0.1g then you step the last equation starting at W = (T / 496), rounded to the nearest 10th, and increasing in 0.1 increments. There will be only one solution for F (between 1 and 31 inclusively) such that the last equation solves for an integer.


Am I on the right track?

[daamon]

I think I need another piece of information

Yes, you are very much on the right track.

Having re-read the puzzle, and based on your workings, I have a humble apology to make (god, I hate having to do this!) - Although the weight of a genuine coin isn't known, you do know it's a whole number of grams. I've amended the puzzle accordingly... I could've sworn I put that in!!! SquirrelDip - My humblest apology - especially if you've been scratching your head for ages trying to spot what you've missed!

Credit to northcat_8 and Tommyknocker who have both PM'd me (in that order) the right approach - even with the missing info - and assumed that the coins were either 1g or a whole number of grams.

[SquirrelDip]

@daamon: Scratched for a little while - even though I identified that something was missing fairly quick...

So then the problem solves to:

Set W = (T / 496) rounded to the nearest whole.

Then F = ([496 * W] - T) / 0.1

[daamon]

@ SquirrelDip - You're so close. Your approach is correct and your conculsions (so far) are correct. Drop the math and apply logic - think about the problem, don't try to solve it with maths (a bit like the "Who owns the zebra?" problem earlier...)

Take a step back and look at it from a different angle with the facts you've concluded thus far, and the info available (new and original )...

[SquirrelDip]

Without knowing the actual total weight, nor the actual weight of the coin, I have already concluded the puzzle, the only unknown left is the measured weight:

1. Take the total measured weight.

2. Divide by 496 and round up to the nearest whole, this is the weight of a single good coin.

3. Multiply that weight by 496 and subtract the measured weight.

4. Multiply the diference from above by 10. The answer is the bag number of the fake coins.

[daamon]

@ SquirrelDip - You're there.

So, in full, here's the solution:

Take 1 coin from Bag 1, 2 from Bag 2, 3 from Bag 3 ... 31 from Bag 31. If all the coins were genuine, the weight would be an exact mulitiple of 496 (1+2+3+...+31). This is the key, as it later allows you to identify the bag of fake coins.

As there's a bag of fake coins, the actual weight (when using the scales only the once) will be short of any multiple of 496. The amount it's short is X (number of coins) times 0.1g. So, just divide what your short by, by 0.1 (or times by 10) and that will give you your bag number.

e.g. The fake coins are in bag 23. That means that, when weighed, the actual weight will be 23 x 0.1g less than what it should be.

So, if the actual weight was (say) 3.4697Kg (3,469.7g) the nearest multiple of 496 to this is 3,472 (7 x 496). 3,472 - 3,469.7 = 2.3. Hence 2.3 / 0.1 or 2.3 x 10 (same thing) = 23. So there's 23 fake coins, so it's bag 23. You also find out how much a real coin weighs too (7g - coz of the 7 x 496).

Well done t northcat_8, Tommyknocker and SquirrelDip (also for your perseverence - know I now why your avatar is a Jack Russell!!! )

Special mention to flaninacupboard who had the start of the right answer...

[daamon]

New Puzzle: "Bridge Crossing"

This is an extension of the usual "4 people stuck at one end of a bridge" problem, in that there's 5 people.

Also, I'll admit that this is a problem I got emailed ages ago (a link) and have re-discovered it to post here. I feel it's justified as I (eventually) solved it...

There's a family of 5 people that have been on a nature walk: spritely young Jimmy, his older brother porky Peter, dainty Daphne, uncle Rex and grandpa Joe.

On their way back they've gotten lost and found themselves at a small, rickety old wooden bridge that can only support 2 people at a time. To make things worse, it's getting dark. Luckily, they've got an oil lamp with them. Unluckily, the lamp only has enough oil left to burn for 30 seconds.

Each person, due to their age and / or fear, takes the following amount of time to cross the bridge and a pair walks at the pace of the slowest person:

Jimmy - 1 second
Peter - 3 seconds
Daphne - 6 seconds
Rex - 8 seconds
Joe - 12 seconds

Because it's dark and the bridge is so dangerous, they must have the oil lamp with them when crossing (and alight).

In what order do they cross the bridge to ensure everyone gets across before the lamp runs out of oil?

If you've seen this, please refrain from posting. SquirrelDip - usual thing - no programs!

To make it easier to visualise / work on, here's the link I found:

http://www.plastelina.net/games/game3.html

NOTE: I'm trusting people not to get the solution (by registering) and posting it - work it out!

Enjoy...

P.S. It's worth checking out game 2 just for the fun of seeing what happens when you get it wrong!!!

[DVD_Ripper]

"Two equals One"

Note: a^2 means "a squared" or "a multiplied by a".

Here's one to test your algebra and maths...

a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b)
a+b = b
But, a = b from line 1, so...
b+b = b
2b = b
Dividing by b on both sides gives:

2=1 !!!

Given that the algebra is correct (it is), what's the flaw that allows this false result?

Hey, No one told me I would need an engineering degree to reply to posts on this forum

[daamon]

@ DVD_Ripper - And that's only the first one...!!!

There's all sorts - maths, logic, lateral thinking, and some (are you reading this Capmaster?) that are just plain sly...

Worth checking out if you're into puzzles...

[daamon]

Re: "Bridge Puzzle"

flaninacupboard romped in with the right answer (by PM) in just under 4.5 hours, with northcat_8 just over 2 hours behind that (PM also).

As there's been a shortage of puzzles lately, I'll hold off on posting the solution.

Well done guys!

A light-hearted semi puzzle (best done verbally): "A Bus Journey"

Imagine you're driving a bus. You leave the depot and, at the first stop, pick up 7 people. At the next stop you pick up 4 more but drop off 2. At the following stop, you pick up a mum with her 2 babies and drop of a pensioner. Now you're approaching town and it's getting busy, at the next 3 stops you pick up 9, 11 and 8 people and drop off 4, 5 and 3 people respectively. On your way back to the start of the route you pick up 3 more at the next stop and then 2 more at the stop after that, with the bus conductor getting on and 2 teenagers getting off (probably haven't paid).

Question: What's the bus driver's name?

[flaninacupboard]

P.S. It's worth checking out game 2 just for the fun of seeing what happens when you get it wrong!!!

I gave it a go, and i feel like doing that to some vocal religous folks! but that puzzle harder than the bridge crossing one

[daamon]

Yeah, I know the feeling...

Strange, I found the "Monks and Cannibals" one easier than "Bridge Crossing" one. Weird...

[flaninacupboard]

ah, got it this time - just required a fresh look!

[daamon]

See, it was worth sacrificing all those monks...

Got any good brain teasers - I'm bored!

[northcat_8]

P.S. It's worth checking out game 2 just for the fun of seeing what happens when you get it wrong!!!

I gave it a go, and i feel like doing that to some vocal religous folks! but that puzzle harder than the bridge crossing one

[daamon]

@ northcat_8 - Is that at the comment, the cannibals animation or both?

Got any fresh puzzles...?

[northcat_8]

I'm just laughing at flaninacupboard

I would get flan on some of my test and bonus questions...

Sometimes I will give a multiple choice quiz where all 15 answers are "A" ask them questions like "a man rides into town on friday, stays 3 days and leave on friday" how is this possible.

Sometimes a preconceived notion that the issue is complex is enough to blind the subject to the obvious.

[daamon]

@ northcat_8 - I can sense you're one of those evil teachers who sees the kids as merely there for your entertainment... Nice one.

Sometimes a preconceived notion that the issue is complex is enough to blind the subject to the obvious.

Don't I know it!!! "You have 2 coins totalling 30 cents, one of them isn't a 5 cent piece..." That still pains me - I even got a PM from Capmaster recently saying that he was shaving one Monday morning and it just popped into his head and he chuckled about it all over again.

You two would get on well as drinking buddies!!!

P.S. Yes, I am a nosey little f*cker. And I'm not even going to go into the other one...

[northcat_8]

Well I will be honest, I really enjoy making the kids entertaining...they are so dumb I guess that's probably why I like little kids in their terrible 2's...they don't know any better.

I usually give the "All A" quiz to start the year to destroy their preconceived idea of "this can't be right" and to make them have to feel strongly about their answer...if they aren't sure on their answer they'll change it. What I have found is that usually the kids who really get math don't really notice that all the answers are A until they are finished, they may be suprised at the result and they may look over a few problems but they will turn it in and let it ride. The average student will question it about 6-8 questions in and the lower students will change answers so they have a more equal distribution of ABCD. I've given that quiz with simple addition questions like 5+8=, 3+4=.... ....and do you know lower level kids will change answers that they know are right just because they cannot believe that all the answers are "A"

Now in upper level classes I give a true/false quiz that is a little different. The directions are: If the statement is true then give a example that supports the "Truth of the statement." If the statement is false, show that the statement cannot be true.

Then my questions are something like:

1. One plus one divided in two is four.

Of course they all answer false as algebraically the result should be 1 or 3/2 depending on how the terms are grouped. However the question is very vague and the result is very dependent on how the math is applied. For example in algebra that statement would be false, but if the 2 ones were in fact 2 sticks and you broke each one in half then you would have 4 sticks...the question makes no claims that the sticks have to be the same length or any other stipulations just that one plus one divided in two is four...and that is possible as applied by my example.

I use this small example to illustrate how they could unintentionally put stipulations or limitations on a given situation.

Kids today are not very good thinkers. They tend to be more fact regurgitators than anything else. There are some thinkers, don't get me wrong but the majority just follow the outline to get through rather than actually grasping the concept of "WHY".

[daamon]

OK, it's been nearly a week now so the correct answer to the "Bridge Crossing" puzzle (as got by flaninacupboard and northcat_8) is as follows:

1. Jimmy (1s) & Peter (3s) - 3s
2. Jimmy returns (1s) - 1s
3. Jimmy (1) & Daphne (6s) - 6s
4. Jimmy returns (1s) - 1s
5. Rex (8s) & Joe (12s) - 12s
6. Peter returns (3s) - 3s
7. Jimmy (1s) & Peter (3s) are the last 2 to cross - 3s

3 + 1 + 6 + 1 + 12 + 3 + 3 = 29 seconds in total.

Crossings 1 and 3 can be swapped - it doesn't matter if Jimmy takes Peter or Daphne first.

[daamon]

Here's one I saw in a newspaper, reasonably simple: "Jugs Of Water"

You have 1 x 3 litre jug, 1 x 5 litre jug (both empty) and a water supply (e.g. a tap). You can only go to the water supply 3 times (i.e. filling up either of the jugs counts as one visit). Using only the 2 jugs, you must end up with exactly 4 litres of water. How's this done?