is there a way to cheat in the cubes? like....using 8 and 1 to represent 19(since it's impossible to have 81, when u see these 2 cubes, u just have to add the two to have a 9 to serve as the 2nd digit and the smaller number as the 1st digit...something like that..) sorry if I can't ask in a clearer way>.<
Sayumi wrote:
Yep, I'll work on something as soon as I have the time and a few good ideas . Here just some little things to keep the mind busy...
I got these ones of the internet, but I changed both of them a bit... the second one was with cement originally, but I like chocolate cake better .
The Cubes
Yuusaku Kudo once said to little Shinichi: "I've got two cubes (with a number on each face) on my desk. I arrange the them every day, so that the front faces show the current day of the month. What numbers are on the faces of the cubes?"
(You can't represent the day "5" with a single cube with a side that says 5 on it. You have to use both cubes all the time. So the 5th day would be "05".)
Five liters
You are baking a huuuge chocolate cake and the recipe calls for five liters of milk. You have a cow giving you all the milk you need. The problem is that you only have a four liters and a seven liters bucket and neither has graduation marks. Find a method to measure five liters. (You'll have to waste a lot of milk - but it's an amazing cow with a neverending supply of milk )
when I read your hint I got it. But I also must say I have seen these kind of cubes only for the full year
Spoiler:
Cube 1 - 0 ,1 , 2, 3, 4, 5; Cube 2 - 0, 1, 2, 7, 8, 9 The Catch is 9 is also 6!
One cube has 0 1 2 3 4 5, the other has 0 1 2 6 7 8. We need to use the 6 for a 9!
5 litres of milk from the amazing cow
Spoiler:
Fill up the 7 litre bucket with two full loads of the 4 litre bucket, leaving 1 litre in the 4 litre bucket. Empty the 7 litre bucket and pour in the left-over 1 litre, topping up with another 4 litres. That makes 5 litres in the 7 litre bucket. Do you know a nice recipe for chocolate cake?
Sayumi wrote:Five liters
You are baking a huuuge chocolate cake and the recipe calls for five liters of milk. You have a cow giving you all the milk you need. The problem is that you only have a four liters and a seven liters bucket and neither has graduation marks. Find a method to measure five liters. (You'll have to waste a lot of milk - but it's an amazing cow with a neverending supply of milk )
I think I've solved it:
Spoiler:
First you have to fill the seven liter bucket, then fill as much as fits from the seven liter bucket into the four liter bucket. After getting rid of the 4 liters in the four liter bucket, you have to fill the remaining 3 liters into the four liter bucket, too. Now have the cow fill the 7 liter bucket again, and fill one liter of that into the 4 liter bucket, so that bucket is full. 6 Liters remain in the 7 liter bucket. Get rid of the 4 liters in the 4 liter bucket, and fill it with 4 liters out of the 7 liter bucket once again. Now 2 liters remain in the 7 liter bucket. Now for the last time you have to waste the 4 liters inside the 4 liter bucket and fill the 2 liter from 7 liter bucket to 2 liter bucket. Then you have to make the cow for the last time put 7 liters inside the 7 liter bucket. If you now fill as much milk as fits inside the 4 liter bucket, that would be 2 liters, 5 liters remain inside the 7 liter bucket.
Sayumi wrote:The Cubes
Yuusaku Kudo once said to little Shinichi: "I've got two cubes (with a number on each face) on my desk. I arrange the them every day, so that the front faces show the current day of the month. What numbers are on the faces of the cubes?"
(You can't represent the day "5" with a single cube with a side that says 5 on it. You have to use both cubes all the time. So the 5th day would be "05".)
Although it's not a complete answer I post what I think about this one:
Spoiler:
The problem about this one is: There are 9 possibilities that contain an 0: 01, 02, 03, ... 08, 09. If you put the 0 on one cube you have to put all the numbers between 1 and 9 onto the other in order to be able to create all the above named examples. Since there's ony place for six numbers on the cube, the only way to solve this is to put the 0 on both of the cubes.
Then there are the numbers 11 and 22. You can't create them unless the numbers 1 and 2 are on both cubes.
Now I have 3 free places on each cube which makes a total of 6 places.
But I still haven't used th following numbers at all: 3, 4, 5, 6, 7, 8 and, 9. That's 7 numbers never used. How am I gonna put 7 numbers in 6 places?
"Vad ska jag annars vara?" - "Det vet jag inte. Det måste du svara på. Men om du släpper allt du tror att du måste, och frågar dig vad du vill... Vad vill du då?"
Sry that it took so long to reply:
sstimson: Your answer for the cubes is right!
googleearth has got the cake one correct.
Holmes and ayw posted correct answers for both!
Good job everyone!
@S.H.: No you can't cheat like that, but there is a trick to it.
@googleearth: You're close to the answer for the cubes, there is a trick how to solve the problem...
@ayw
ayw wrote:
5 litres of milk from the amazing cow
Spoiler:
Do you know a nice recipe for chocolate cake?
Spoiler:
I do-quite a few! I even got a recipe for one that you bake in a mircowave and that actually tastes good . It haven't got one for which you'd need 5liters of milk for though...
Here is are new ones:
Continue the sequences:
1. Z, O, T, T, F, F, S...
2. M, T, W, T, F...
The 100 Coins
There are 10 sets of 10 coins. You know how much the coins should weigh. You know all the coins in one set of ten are exactly one tenth of a gram off, making the entire set of ten coins one gram off. You also know that all the other coins weight the correct amount. You are allowed to use an extremely accurate digital weighing machine only once.
How do you determine which set of 10 coins is faulty?
Feel free to post answers for the older ones as well of course!
Last edited by Sayumi on March 8th, 2009, 11:18 am, edited 1 time in total.
"It is one of those instances where the reasoner can produce an effect which seems remarkable to his neighbor, because the latter has missed the one little point which is the basis of the deduction."
Sherlock Holmes
Continue the sequence
First letters...
1. zero, one, two, three......= z, o , t ,t f, f,s,s,e,n,t......
2. M T W T F S S <- Monday,Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday
100 coins
Is placing the set of 10 coins one set by one set allowed?
Sample..Place set A, then look at the mass, if there is no problem, place set B(without removing set A),then look at the mass, if there is no problem....continue like that...
Sayumi wrote:
Sry that it took so long to reply:
sstimson: Your answer for the cubes is right!
googleearth has got the cake one correct.
Holmes and ayw posted correct answers for both!
Good job everyone!
@S.H.: No you can't cheat like that, but there is a trick to it.
@googleearth: You're close to the answer for the cubes, there is a trick how to solve the problem...
@ayw
ayw wrote:
5 litres of milk from the amazing cow
Spoiler:
Do you know a nice recipe for chocolate cake?
Spoiler:
I do-quite a few! I even got a recipe for one that you bake in a mircowave and that actually tastes good . It haven't got one for which you'd need 5liters of milk for though...
Here is are new ones:
Continue the sequences:
1. Z, O, T, T, F, F, S...
2. M, T, W, T, F...
The 100 Coins
There are 10 sets of 10 coins. You know how much the coins should weigh. You know all the coins in one set of ten are exactly one tenth of a gram off, making the entire set of ten coins one gram off. You also know that all the other coins weight the correct amount. You are allowed to use an extremely accurate digital weighing machine only once.
How do you determine which set of 10 coins is faulty?
Feel free to post answers for the older ones as well of course!
Spoiler:
Continue the Sequences:
1. Z, O, T, T, F, F, S... S, E, N, T, E, T, T, F... (Zero, One, Two, Three, Four, Five, Six...)
2. M, T, W, T, F... S, S (Monday, Tuesday, Wednesday, Friday, Saturday, Sunday)
The 100 Coins
There are 10 sets of 10 coins. You know how much the coins should weigh. You know all the coins in one set of ten are exactly one tenth of a gram off, making the entire set of ten coins one gram off. You also know that all the other coins weight the correct amount. You are allowed to use an extremely accurate digital weighing machine only once.
How do you determine which set of 10 coins is faulty?
Feel free to post answers for the older ones as well of course!
Ok this is a way to do it with a few measures , but as of yet I not sure how to pick out one bad 'penny' out of ten with but a single measure. Can you tell tell the difference just by picking up a group?
Spoiler:
Weight 5 groups if right bad in the other 5 groups
Weight 2 groups if ok then bad is in the other 3
if not ok then weight one group and that answer will show the bad group
for the group of three with the bad one repeat the above step
wow... S.H., ccppfan, Holmes-all of you got the sequences correct!
now for the coins:
S.H. wrote:
Is placing the set of 10 coins one set by one set allowed?
Sample..Place set A, then look at the mass, if there is no problem, place set B(without removing set A),then look at the mass, if there is no problem....continue like that...
No, your not allowed to put them on one by one. You have to put all the coins on at the same time. Only one measurement.
sstimson wrote:
Ok this is a way to do it with a few measures , but as of yet I not sure how to pick out one bad 'penny' out of ten with but a single measure. Can you tell tell the difference just by picking up a group?
Spoiler:
Weight 5 groups if right bad in the other 5 groups
Weight 2 groups if ok then bad is in the other 3
if not ok then weight one group and that answer will show the bad group
for the group of three with the bad one repeat the above step
four weights
Later
Not a chance to notice just by picking them up, the difference is not big enough.
Well... it would work, but you're only allowed to to weigh them once.
Remember: There are 10 coins in every set, not only 10 coins in total.
"It is one of those instances where the reasoner can produce an effect which seems remarkable to his neighbor, because the latter has missed the one little point which is the basis of the deduction."
Sherlock Holmes
Assuming the all the faulty coins in one set are all equal in mass....
Get 1 coin from set A, 2 from set B, 3 from set C, continue the process...10 from set J
When you got the measurement...you will notice that:
If Set A is the faulty then measurement is 54/100 gram + x
If Set B is the faulty then measurement is 53/100 gram + 2x
If Set C is the faulty then measurement is 52/100 gram + 3x
If Set D is the faulty then measurement is 51/100 gram + 4x
If Set E is the faulty then measurement is 50/100 gram + 5x
If Set F is the faulty then measurement is 49/100 gram + 6x
If Set G is the faulty then measurement is 48/100 gram + 7x
If Set H is the faulty then measurement is 47/100 gram + 8x
If Set I is the faulty then measurement is 46/100 gram + 9x
If Set J is the faulty then measurement is 45/100 gram + 10x
Should know the x since "You know how much the coins should weigh."
Haven't test it though...sleepy already...will test it tomorrow
The idea for the solution is right -well done! .
But I don't really understand your explanation...how did you get to the 54/100g - 45/100g and x - 10x??? What is x?
"It is one of those instances where the reasoner can produce an effect which seems remarkable to his neighbor, because the latter has missed the one little point which is the basis of the deduction."
Sherlock Holmes
The 100 Coins
There are 10 sets of 10 coins. You know how much the coins should weigh. You know all the coins in one set of ten are exactly one tenth of a gram off, making the entire set of ten coins one gram off. You also know that all the other coins weight the correct amount. You are allowed to use an extremely accurate digital weighing machine only once.
How do you determine which set of 10 coins is faulty?
[...]
Here's an idea:
Spoiler:
As far as I understand there's no limit for the extremely accurate digital weighing machine, so it's not like I could only put 10 coins on it. Because my idea is to weight 55 coins.
Out of every set you put a differnent number of coins onto the weighting machine. For instance: You take one coin out of the first set, two out of the second, 3 outta the fourth, .... and 10 out of the tenth set. That exactly 55 coins. With simple mathamatics you will be able to determine how much 55 coins are supposed to weigh, since you know how much one coin weighs. You compare that number with the coins' actual weight, and since you know that one faulty coin's weight is 0.1 gramm off you can easily find out how many of these 55 coins are faulty. The set you put out the same number of coins, that coins turned out to be faulty, is the one with faulty coins.
Sayumi wrote:
[...]
@googleearth: You're close to the answer for the cubes, there is a trick how to solve the problem...
[...]
Can you give me a hint, I don't think I'll be able to get the answer without one.
See you later
googleearth
"Vad ska jag annars vara?" - "Det vet jag inte. Det måste du svara på. Men om du släpper allt du tror att du måste, och frågar dig vad du vill... Vad vill du då?"
as for the hint... hmmm. Look closely at the remaining numbers. You have to write two of them on the same side, without writing two different numbers. Did that make sense? Probably not.
Maybe you could try writing all the remaining numbers on little pieces of papers and just move them back and forwards and twist them around a bit... and: the answer is easy. It's one of these things where you smack your forehead once you get it Sorry if I confused you more than anything else
Here is my mathematical formula to solve the coin one... ;-)
Spoiler:
take one coin from set1, two coins from set2...ten coins from set10 normal weight - measured weight= x/10
depending on whether the faulty coins are lighter or heavier than the normal ones x will be positive or negative. /10 beause then x will give you the number of the set that's faulty.
e.g. normal weight 500g, measured weight 500.5g
500-500.5=x/10
-0.5=x/10
x=-5
->set5 is faulty, (the coins are heavier than the normal ones)
Last edited by Sayumi on March 9th, 2009, 1:36 pm, edited 1 time in total.
"It is one of those instances where the reasoner can produce an effect which seems remarkable to his neighbor, because the latter has missed the one little point which is the basis of the deduction."
Sherlock Holmes
like....using 8 and 1 to represent 19(since it's impossible to have 81, when u see these 2 cubes, u just have to add the two to have a 9 to serve as the 2nd digit and the smaller number as the 1st digit...something like that..) sorry if I can't ask in a clearer way>.<
. Here just some little things to keep the mind busy...
. 
...these two are really good - quite tricky!
