Brain Teaser No 1

Don,

I think I have a viable scheme if you allow a measurement to be off scale - is that allowed?

-Dan

Dan,

The object is to identify the 180s, 250s and 135s and if knowing that "something" weighs MORE THAN 30 kg helps, then this is allowed.

We will assume that the scales will not be damaged by large weights, say up to 100kg, but the only information available above 30kg is that it is more than 30kg.

BTW, from the information you have given, I am certain you have a valid scheme.

Cheers

Don

quote:

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BTW, from the information you have given, I am certain you have a valid scheme.

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Ah, very good -- I'll wait until a few others have had a chance to see the puzzle before posting.
cheers,
Dan

Dan

I'll wait until ........

What a jolly decent chap you are. Remember, the weekend over here leaves very few people at home surfing the net, so it could be Monday before the bright sparks return to their keyboards......

Cheers

Don

One

Don,

I assume you are allowed to open the crates.

If so, I agree that the answer is 1.


Steve D

135s only have one pair of speaker sockets. So I'll take those and you can weigh the remainder as many times as you want...

Paul

Paul,

One is still the answer.

Matthew

Dan

"if you allow a measurement to be off scale" we do.

"I'll wait until a few others have had a chance" Now's your chance.


Mathew "one" Nows your chance to explain

Steved "I assume you are allowed to open the crates" you most certainly are.

Paul R "135s only have one pair of speaker sockets. So I'll take those" not before you describe the system of weighing you won't.

Can I suggest a time, say 22.00 hrs GMT at which you ALL post your solutions, including me and we'll vote on who's is the best?

I find that if I pre-type my entries in Word then I can 'paste' them into the forum pretty quickly at a predetermined time.

Ok, I agree is incredably nerdy, geeky, pathetic, even sad..........so 22.00 hrs it is, unless someone shouts otherwise....

Cheers

Don

Given crates A, B, and C, weigh 3 amps for crate A and 1 amp from crate B.
Possible totals are:

3*6 + 7 = 25 kg => C = 135's, B = 250's , A = 180's
3*6 + 8 = 26 kg => B = 135's, C = 250's , A = 180's

3*7 + 6 = 27 kg => C = 135's, A = 250's , B = 180's
3*7 + 8 = 29 kg => B = 135's, A = 250's , C = 180's

3*8 + 6 = 30 kg => A = 135's, C = 250's , B = 180's
3*8 + 7 > 30 kg => A = 135's, B = 250's , C = 180's

Not very elegent, but it gets the job done.
-Dan

Dan,

I was just about to post my solution when I noticed you had beaten me to it! Mine was identical to yours.

Steve D

Same method. Can't see there is any other way.

Matthew

A number of forum memebers have been frequenting a riddles post for a number of days now and have been keeping track of scores.

Each forum member has scored 100 points, with a 100 points being scored on each day the post has been running. Coincidentally a total of 100 riddles have also been solved.

Scoring works like this. Each member gets one point for the first riddle they solve each day, two points for the second and so on.

The first day saw Don weighing with a mighty 9 riddles and Paul managed a healthy 7. d marsh finsihed the game with 6 riddles on the final day, however he had forgotten what he had scored on the previous days but after a moment (he knew his total was 100) he managed to figure out what he had scored on day one.

What had he scored?

MATTHEW

He scored 6 points (3 riddles) on the first day.

STEVE D

steved hit the nail on the head. Anyone want to explain why...

Matthew

Apologies, but I fell asleep last night and missed the deadline for posting replies. FWIW mine is like others ie

Power Weights

We will take 3 amps from one box and one amp from either of the other two boxes, making four amps to be weighed, altogether.

The possible combined weight of these four amps will be as follows and will enable us to determine which box is which

3 at 6 kg + 1 at 7 kg = 25 kg
3 at 6 kg + 1 at 8 kg = 26 kg
3 at 7 kg + 1 at 6 kg = 27 kg
3 at 7 kg + 1 at 8 kg = 29 kg
3 at 8 kg + 1 at 6 kg = 30 kg
3 at 8 kg + 1 at 7 kg = >30 kg

Cheers

Don

Matthew,
Spent a good hour last night on your riddle teaser and got nowhere.
Obviously I'm missing something here.
cheers,
Dan

The riddle problem has much unnecessary (dis)information.

You're only interested in the first day, you know what everybody else scored, you know the total scored, the rest follows...

Paul

Paul R very helpfully outlined the start of solving Mathew's riddle but I think there a few delicate steps between the rest follows...and the answer!

So you might like to try the 'stocking filler' that follows whilst musing on the various combinations of riddled scores

Cheers

Don

A mere 'stocking filler' to while away the time whilst musing on Mathew's riddled scores....

Choose any 3-digit number and write it down twice to form a 6-digit number eg 267267. Divide this 6-digit number by three prime numbers, thereby reducing the 6-digit number to the original 3-digit number.

What are the three prime numbers ?

Cheers

Don

7, 11, 13?

That's quite profound!

Paul

I get the same answer.

I wouldn't say profound though!

Matthew

I'm going to tell your employers about the lot of you.

malcolm

I wouldn't say profound though!

I'm hurt !

Cheers

Don

You are asking us to solve the following equation.

(1000m + m) / x = m

which means x = 1001

That x exists and is an integer implies that it can be divided by prime numbers, that they are in sequence is somewhat profound I guess.

Matthew