Brain Teaser No 3

SciDoctor (correctly) said:

"The initial premise is ambiguous ; those that have said '4' loads have based this on the statement in the premise that we were dealing with 'loads' as a minimum and no part 'loads'."

In reponse to an early reply to this specific brain teaser, I said:

"The number of loads can include fractions/decimals eg 7.642 loads (this isn't the right answer) but if somebody said that this would effectively mean 8 FULL loads, I wouldn't quibble (8 full loads isn't the answer either, BTW)

We are also looking for the MINIMUM number of loads required.........."

Unfortunately I am not able to modify the original question after it's been posted (other than the usual 45 minutes or so)

This means that it worth reading through subsequent posts for up-dates - which i appreciate is a bit of a bind.

Apologies

Cheers

Don

A bit hurried so I might have gotten this wrong:

the difference I see with the previous case is that you need a fueltrack back.
So let's imagine we're leaving a fuel cache trail of 'one gallon each 10 miles' to use up on the way back.

Then fuel usage to transport fuel loads over some distance is increased with 1 gallon/10 miles.

last 1 load will be transported at (1+1=2)gallon/10m so 500/2 km
then 2 loads at (3+1=)4gallon/10km, so the 500/4=125km before that
then 3 loads at (5+1=)6g/10km, so 500/6= 83.3km
and so on.

Then we need a to add while x<800 do (x=x+500/2i and increment i):
14 loads should get you there and back again.

Crossing the desert

Although Alexander has correctly solved this problem, I feel it only appropriate that I set out the solution in my own words….

One "load" of fuel will take the truck 500 miles in a single trip. A "trip" will be the journey between two stopping points, or refuelling points, in either direction.

Two loads of fuel will take the truck (500 + 500/3) miles. This is done in 4 trips by creating a cache 500/3 miles from the start. The truck begins with a full load, goes to the cache, leaves 1/3 load and returns to the start. It picks up the second full load, travels to the cache where it picks up the 1/3 load in the cache. It now has a full load, sufficient to travel another 500 miles.

Three loads will take the truck (500 + 500/3 + 500/5) miles. The first cache is set up 500/5 miles from the start. The truck begins with a full load, goes to the cache, leaves 3/5 load and returns to the start. It picks up a second full load and repeats the trips to the cache and back thereby increasing the cache to 5/6 load. The truck picks up the third load and travels to the cache where it arrives with 4/5 load, making the total at the cache 2 full loads, sufficient to carry the truck (500 + 500/3) miles as described above.

Three loads takes the truck (500 + 500/3 + 500/5) miles or (1 and 8/15)*500 miles. We need the truck to go 800 miles or (1 and 9/15)*500 miles, so we need to set up a cache 1/15 mile from the start. Do you need to read that bit again?

Six trips between the start and the first cache will build a cache of (13/15 + 13/15 + 13/15) = 39/15 loads. The seventh trip starts with 7/15 load so that on reaching the cache the fuel in the truck is 6/15 load, which when added to the cache make 3 full loads, sufficient to carry the truck (500 + 500/3 + 500/5) miles as described above.

The distance from start to finish is (500 + 500/3 + 500/5 + 500/15) miles = 800 miles.
The amount of fuel required is 3 and 7/15 loads.

In general, more and more full loads will take the truck further, but in progressively shorter steps:

d = m (1 + 1/3 + 1/5 + 1/7 + 1/9 ……..) where d = distance travelled and m = miles per load and each term within the brackets = a load of fuel.

The number of trips involved t = (1 + 3 + 5 + 7 + 9 ……)

Hope this is clear, and helps all those who tried, to sleep well.........

Cheers

Don

Acad Tsunami and the Tibetan Walk

One morning, exactly at sunrise, a Buddhist monk called Acad began to climb a tall mountain. The narrow path, no more that a foot or so wide, twisted around the mountain to a splendid temple at the summit.

Acad climbed the path at varying speeds, as he contemplated his way to the temple, stopping many times to rest, eat the dried fruit he carried with him and meditate. He reached the temple just before sunset. After two or three days of fasting and further meditation he began his journey back along the same path, starting at sunrise and again walking at varying speeds with many stops along the way. His average speed on the way down was obviously faster than on the way up. He arrived back at his monastery well before sunset.

Prove that there is a spot along the path that Acad will occupy on both trips at precisely the same time of day.

Cheers

Don

Note

For all practical purposes, you can assume that sunrise and sunset are at the same time on each day

Cheers

Don

Acad prefers levitating when sleeping to arduous mountain walking.

Acad

That is absolutely, mind-bogglingly, brilliant.

And all due to the power of the Levitation ENlightenment Society.........

Cheers

Don

quote:

Originally posted by Don Atkinson:
Acad Tsunami and the Tibetan Walk

One morning, exactly at sunrise, a Buddhist monk called Acad began to climb a tall mountain. The narrow path, no more that a foot or so wide, twisted around the mountain to a splendid temple at the summit.

Acad climbed the path at varying speeds, as he contemplated his way to the temple, stopping many times to rest, eat the dried fruit he carried with him and meditate. He reached the temple just before sunset. After two or three days of fasting and further meditation he began his journey back along the same path, starting at sunrise and again walking at varying speeds with many stops along the way. His average speed on the way down was obviously faster than on the way up. He arrived back at his monastery well before sunset.

Prove that there is a spot along the path that Acad will occupy on both trips at precisely the same time of day.

Cheers

Don

The way I see it...

Imagine an Acad-groupie shadowing the journey up the mountain on the same day as Acad is coming back down. He mirrors Acad's upwards journey exactly in space and time. At some point during the day he and Acad must meet and it is at this point that Acad is at the same point and time as on his upwards journey.

Nice puzzle.

Ian

The summit of the mountain is possitioned so that sunset is also sunrise. (possiton of the sun in relation to the planets axial spin and no connection to the time of a day which is rotation of planet around the sun)

So at the summit the sun is viewed as in constant sunset/sunrise.

Therefore as he arrived just before sunset in real time and leaves after sunrise it is possible for him to be in exactly the same spot at exactly the same time of day at the summit of the mountain.

The important point is the time of day he arrived and left the summit in relation to the planets movement around the sun and not the actual visual appearance of the sun in the sky from the summit (which never changes)

quote:

Originally posted by Don Atkinson:
Acad


[QUOTE] And all due to the power of the Levitation ENlightenment Society.........

Nah, trick photography actually.

quote:

The summit of the mountain is possitioned so that sunset is also sunrise. (possiton of the sun in relation to the planets axial spin and no connection to the time of a day which is rotation of planet around the sun)

So at the summit the sun is viewed as in constant sunset/sunrise.

Therefore as he arrived just before sunset in real time and leaves after sunrise it is possible for him to be in exactly the same spot at exactly the same time of day at the summit of the mountain.

The important point is the time of day he arrived and left the summit in relation to the planets movement around the sun and not the actual visual appearance of the sun in the sky from the summit (which never changes)

SciDoctor,

Looking up, and trying to visualise this is making me dizzy........such that even Acad is begining to make sense by comparison.......

Cheers

Don

quote:

Nice puzzle.

Even nicer solution Ian.

Cheers

Don

Counterfeit Coins

Andy C (the detective from Nottingham) has found what he believes to be the evidence that will convict the Counterfeit Cousins - that notorious gang from Leicester led by the Master of the LP12 Voodoo cult - Peter Sxxxx.

Andy has been presented with 11 stacks of £1 coins, each stack containing 10 coins. He has been told that one stack, and one stack only, conprises only conterfeit coins. He also knows that each counertfeit coin weighs exactly 1 gram more than a genuine £1.

There is a pointer scale available that reads in grams, up to 1Kg, accurate to better than one gram.

What's the minimum number of weighings needed to determine which is the stack of counterfeit coins?

Cheers

Don

I am assuming that 50 coins weigh less than 1Kg/1000g so an average coins weighs less than 20g (infact they weigh 9.5g so I am ok)

Split 11 stacks into 5:5:1 a:b:c

Put a on scales weigh remove then b on scales weigh remove.

If a=b then c is the stack of counterfit coins.

Absolute minimum relying on pure luck is two weighs

But realisticly the odds are against us so we should find.

a>b or b>a

Then we take the higher weighed stacks and now split 2:2:1 d:e:f

Weigh d and then e

If d=e then f is the counterfeit stack

Medium luck gives us 4 weighs

But realisticly d<e or d>e

Take the higher weighed stacks and split 1:1 g:h

Weigh each and the higher weighed is the counertfiet stack.

Taking into account luck there are actualy three answers 2,4 and 6 weighs.

For those that have active DBL's they would only take 2 weighs to find the counterfeit but mere mortals would take 4 or 6 weighs.

quote:

There is a pointer scale available that reads in grams, up to 1Kg, accurate to better than one gram.

We can assume that if we put all the evidence (ie all the coins) on the scales, they would weigh less than 1 Kg.

Cheers

Don

quote:

Andy has been presented with 11 stacks of £1 coins, each stack containing 10 coins. He has been told that one stack, and one stack only, conprises only conterfeit coins. He also knows that each counertfeit coin weighs exactly 1 gram more than a genuine £1.

Only the "counterfeit" stack contains counterfeit coins. All the other stacks only contain genuine coins.

Cheers

Don

Assuming that the weight of a genuine coin is known - say 10 grams - rather than deduced, I think I can do it with 1 weighing.

Number the piles from 1 to 11.
Pick one coin from pile 1, two coins from pile 2, etc, and none from pile 11.

If all coins were genuine, the total weight should be 550 grams; therefore if the total weight is 550 grams, the counterfeit pile is number 11.

If the total weight is more than 550 grams, the number of excess grams is the same as the pile number of the counterfeit pile.

Steve D

quote:

Originally posted by steved:
Assuming that the weight of a genuine coin is known - say 10 grams - rather than deduced, I think I can do it with 1 weighing.

Number the piles from 1 to 11.
Pick one coin from pile 1, two coins from pile 2, etc, and none from pile 11.

If all coins were genuine, the total weight should be 550 grams; therefore if the total weight is 550 grams, the counterfeit pile is number 11.

If the total weight is more than 550 grams, the number of excess grams is the same as the pile number of the counterfeit pile.

Steve D

I tried this answer but the premise didn't mention the known weight of a single non-counterfiet coin.

So an extra weigh is required , weighing eleven coins (one from each stack) deducting the 1gm from the total weight and dividing by eleven to get the now known actual weight for a single non counterfeit coin.

Then follow with your single method weigh.

Two weighs minimum

It is a neat weigh to do it though when more info is available

quote:

I tried this answer but the premise didn't mention the known weight of a single non-counterfiet coin.

The main thing is, that between you, you got there. Well done.

Yes, this puzzle often crops up with 10 piles of coins and you know the weight of a genuine coin as well as the excess of the counterfeits. But as Steve showed, you can sort the counterfeits from 11 piles if you know the standard weight. However, as SciDoctor noted, you need another weighing session to find the standard weight if it isn't given.

Cheers

Don

Getting Home Early

An office worker regularly arrives at his home station at exactly 5 pm (this is a fairy tale where the trains run EXACTLY to time !!!!)

His wife always meets him and drives him home (I already said it was a fairy tale!)

Today he took an earlier train and arrived at his home station at exactly 4 pm. It was a nice fine day today, so instead of telephoning, he set off to walk home, following the precise route his wife always takes and they met along the way. He got into the car and they drove home arriving 10 minutes earlier than usual.

Assuming the wife drove at a constant speed and that today she left home just in time to arrive at the station at 5 pm, can you figure out how long the office worker had been walking when his wife picked him up?

Cheers

Don

55 mins?

The wife's journey was 10 mins shorter overall so she must have met her husband 5 mins from the station at 4.55.

Ian

50 minutes?
Steve D

Ian, Steve d,

Well, one of you is right, but a little bit more "explanation" wouldn't go amiss......

Cheers

Don

Having thought about it, I have to admit to a "blonde" moment (which is an achievement with my lack of hair!). I think Ian is right, so it's only fair to let him explain.
Steve D

quote:

Originally posted by Don Atkinson:
Ian, Steve d,

Well, one of you is right, but a little bit more "explanation" wouldn't go amiss......

Cheers

Don

Sorry I thought I did - I'll try again.

The wife's journey was 10 mins shorter overall. She travels at constant speed with no time for pickups etc and the to and fro legs of the journey are the same. So she must have saved 5 mins per leg.

As she timed her run to arrive at the station at 5pm that means she must have turned around at 4.55, thereby saving the 2 x 5mins.

She picked he husband up when she turns around so she must have picked him up at 4.55. Since he arrived on the 4pm train he'd been walking for 55 mins at than point.

... and so she drives 11 times faster than he walks



Ian