A Fistful of Brain Teasers

Four.

Agreed 4. (You could be luck and need less, buth 4 would ensure certainty however the luck of the pick)

My procedure starts by dividing into groups of 3, then use the balance as follows:

1. One group of 3 vs another 3: if balance then it is in one of the other two groups of 3, otherwise in one of the first two groups.
2. Then one of the two suspect groups of 3 vs one of non-suspect groups.
   If balance then odd one is in the other suspect group of 3, if not balance then it is in the suspect group of 3 picked. At this stage whether heavier or lighter could be evident, but not necessarily.
3. So which 3 it is in known, but not necessarily if heavier/lighter.
   Pick any 1 vs another 1.
4. If balance it is the remaining one (but heavier or not may not be known), otherwise one of the two on the balance. Finally swap 1 of those just tested for the remaining one. If the odd one out was one of the previous pair this will show which and if heavier or lighter. Otherwise if the remaining one was the odd one this would show if heavier or lighter.

Frank, IB, Onlookers,

You can do it in Three. Both identifying the odd ball and whether it's heavier or lighter.

No luck involved.

I thought that might be the case, my method was different than above. I started with 4 and 4.

I'll give it another try. (if only I could find a piece of paper and a pen that works)

fatcat posted:

I thought that might be the case, my method was different than above. I started with 4 and 4.

I'll give it another try.(if only I could find a piece of paper and a pen that works)

In the modern world of computers and i-Pads that could be a real brain teaser !!!!

......but it's what I used to solve this problem > 20 years ago

Hmm, the question is potentially misleading: I took it as meaning the minimum number regardless of where the odd one is. If it meant the possible minimum number if you are lucky, the answer is 2 weighings: a single ball each side, one is heavier than the other, one of those against another and if it is the odd one retained for comparison with a new one it would again be low or high, confirming which it is and whether heavier or lighter.

As for the minimum number regardless of where the odd one is:-

If start with 6:6, could require up to 5 depending where it is 

Start 5:5 could require up to 5 depending where it is 

Start 4:4 - if not in one of first groups, then will require 3 more, 4 in total. But if in one of first groups it could require 4 more,  so as far as I can see the answer is 5 to be sure of finding and confirming if heavier or lighter, wherever it is to start with, not 4.

Start 3:3 - possible in total of 4 regardless of where it is, as explained in my earlier post. I can’t see it definitely possible in less.

Start 2:2 - more than 4 in total unless lucky (I think same as starting 4:4).

Start 1:1 - probably 7 In total.

None appear to be capable of being certain to find the answer in 3 - I will await the answer! (Maybe that is penalty of only using head!)

I think I've got it, but struggling to explain it.

Weigh 1, 2, 3, 4,   v   5,6,7,8.

if balanced, weigh 9,10,  v  11, 1.

if 9,10,  v  11, 1. unbalanced weigh 9  v  10

if 9,10,  v  11, 1. balanced weigh 1  v  12.

Weigh 1, 2, 3, 4,   v   5,6,7,8.

if unbalanced, weigh 1, 2, 5, 6  v  3, 9, 10, 11 (if this is balanced weigh 7 v  8, answer will be 7, 8, or 4)

if 1, 2, 5, 6  v  3, 9, 10, 11 is unbalanced, and 5, 6 and 3 three are still in contention weigh 5  v  6. If 1 or 2 are still in contention weigh 1 v  2.

It looks a lot simpler on paper with a few arrows here and there.

Innocent Bystander posted:

Hmm, the question is potentially misleading: I took it as meaning the minimum number regardless of where the odd one is. If it meant the possible minimum number if you are lucky, the answer is 2 weighings: a single ball each side, one is heavier than the other, one of those against another and if it is the odd one retained for comparison with a new one it would again be low or high, confirming which it is and whether heavier or lighter.

As for the minimum number regardless of where the odd one is:-

If start with 6:6, could require up to 5 depending where it is 

Start 5:5 could require up to 5 depending where it is 

Start 4:4 - if not in one of first groups, then will require 3 more, 4 in total. But if in one of first groups it could require 4 more,  so as far as I can see the answer is 5 to be sure of finding and confirming if heavier or lighter, wherever it is to start with, not 4.

Start 3:3 - possible in total of 4 regardless of where it is, as explained in my earlier post. I can’t see it definitely possible in less.

Start 2:2 - more than 4 in total unless lucky (I think same as starting 4:4).

Start 1:1 - probably 7 In total.

 

None appear to be capable of being certain to find the answer in 3 - I will await the answer! (Maybe that is penalty of only using head!)

Hi IB,

The question is not at all misleading. It is quite straightforward.

Your understanding....."I took it as meaning the minimum number regardless of where the odd one is....." is quite correct. That is the task.

The difficulty (ie the puzzle) is to figure out how to do it in three weighings. (doing it in four, is a bit of a doddle !)

Luck doesn't come into it. Regardless of which ball is the odd one, you CAN always find it in three weighings

Of course, you do need to remember which ball is which, but that goes without saying. eg if you put two groups of four onto the scales and they balance, then NONE of them can possibly be the odd ball. If they don't balance, then ONE of those eight MUST be the odd ball. You have now used ONE weighing ! I trust this gives you the confidence that your understanding, highlighted above, is correct ?

I do accept it is a more challenging teaser than one or two of my more recent ones, but then, solving it is a lot more satisfying !

Frank,

I think you are heading in the right direction. And I do appreciate the difficulty in providing a clear explanation, so you might already have it !

Bear in mind, for example, if you know the odd ball has to be ball 1 or ball 2, but you don't know whether it is odd because it is light, or odd because it is heavy, it's no good putting ball 1 on the left scale and ball 2 on the right scale. For sure, one will go down and one will go up, but you won't be any the wiser whether its a heavy one making it go down, or a light one making it go up (if you see what I mean ?)

Don.

When you do initial weighing of 1, 2, 3, 4,   v   5,6,7,8. if 1, 2, 3, 4 goes up, you know 1 and 2 is light, when you weigh 1 v 2, if 1 goes up it is light, if 2 goes up it is light.

fatcat posted:

Don.

When you do initial weighing of 1, 2, 3, 4,   v   5,6,7,8. if 1, 2, 3, 4 goes up, you know 1 and 2 is light, when you weigh 1 v 2, if 1 goes up it is light, if 2 goes up it is light.

Hi Frank,

what you say is correct, and that sort of thinking needs to form a clear element of the overall explanation.

as I say, you might well have the solution, it’s just the explanation that possibly needs to be somewhat more detailed. Eg if on the first weighing, 1,2,3,4 goes down and you subsequently determine that one of those balls is the odd one,  then it has to be heavy. If you subsequently determine the odd ball is 2,3 or 4 you would be able to identify which one it is, with one more weighing.

Some further explanation, but probably more confusing.

Weigh 1, 2, 3, 4,   v   5,6,7,8.

 if balanced, weigh 9,10,  v  11, 1.

 if 9,10,  v  11, 1. unbalanced weigh 9  v  10 (you know if 9 and 10 are heavy or light from previous weighing, whichever ball is still showing heavy or light is the odd ball. If 9 and 10 balance, 11 must be odd and whether heavy or light is known from previous weighing))

 if 9,10,  v  11, 1. balanced.  weigh 1  v  12. (12 is the odd ball and 1 is known not to be odd)

Weigh 1, 2, 3, 4,   v   5,6,7,8.

 if unbalanced, weigh 1, 2, 5, 6  v  3, 9, 10, 11 (if this is balanced answer must be 7, 8, or 4, weigh 7 v  8, you know if 7 and 8 are heavy or light from first weighing, whichever ball is still showing heavy or light is the odd ball, if 7 and 8 balance, 4 must be odd and whether heavy or light is known from first weighing )

 if 1, 2, 5, 6  v  3, 9, 10, 11 is unbalanced, either 5, 6 and 3 will be showing the same results as initial weighing  or 1 and 2 will be showing the same results as initial weighing.  If 5, 6 and 3 are showing the same results, weigh 5  v  6, whichever is showing the same result as initial weighing is odd and heavy or light is known. If 5 and 6 balance 3 is odd and heavy or light is known.

If 1 and 2 are showing the same results, weigh 1  v  2, whichever is showing the same result as initial weighing is odd and heavy or light is known.

Ok I understand.

next time I think Il’ll have to find myself some nominal objects!

what is 111 111 111 x 111 111 111 = ?

yep ! that's it, straight forward one hundred and eleven million, one hundred and eleven thousand etc.

Naming depends whether you use traditional British terminology, or the now more widely and possibly near universal formerly American system.

i would say: Twelve quadrillion, three hundred and forty five trillion, six hundred and seventy eight billion, nine hundred and eighty seven million, six hundred and fifty four thousand, three hundred and twenty one.

But it looks better written out - actually a rather lovely product,  12345678987654321  

Or with commas but doesnt look so neat: 12,345,678,987,654,321

12345678987654321 is quite a neat outcome.

hopefully, others will also have appreciated this.

11 x 11 = 121 and similar products yield similar results.

the given 111 111 111 example provides the largest product comprising single digit components. The symmetry continues beyond eg 1 111 111 111 But the appearance is complicated because the product includes double-digit components.

Anyway, I hope this little diversion was interesting.

0 appears eleven times (that’s a Zero to avoid ambiguity)

2 appears twenty times

3 appears twenty times

4 appears twenty times

5 appears twenty times

6 appears twenty times

7 appears twenty times

8 appears twenty times

9 appears twenty times

There is nothing sinister or obscure about the list in which all the above things occur.

How many times would the number 1 appear ?

21 in the sequence of numbers from 1 to 100

You’re spot-on sjb, as usual.

But , i’m Wondering how many others came up with a slightly smaller number, first time round ?

Well, I simply didn’t twig

Going to the Pub

I live in the house shown in the bottom left of the diagram and pop up to the pub, top right, most evenings. The diagram is orientated North-Up and the Pub is NE of the house.

The possible paths linking the two places are indicated by the lines in the diagram and I always walk either due N, due E or NE. That is, I always go so that every step brings me closer to the pub !

What is the greatest number of different routes I can take ?

345 (I think, and I’m not trying again!) 

IB has ruled himself out of any further attempt at this one. (nice try IB BTW - and not too far wide of the mark !

So, to encourage "the others" ..........

I have entered the number of possible routes available to reach a dozen of the "nodes"

You might be able to detect a pattern developing and....

this might lead to a accurate solution

Total frustration, I have a pile of envelopes with numbers scribbled on. I got 347, then 375. Then, I have been told there is a similar puzzle about getting to see a mistress without being seen en route every day of the year, so without counting again, 365?

notnaim man posted:

Total frustration, I have a pile of envelopes with numbers scribbled on. I got 347, then 375. Then, I have been told there is a similar puzzle about getting to see a mistress without being seen en route every day of the year, so without counting again, 365?

Tantalising, Frustrating, annoying............That's what some of these Brain Teasers are all about Also a bit of fun andgreat satisfaction when you get there !

Take a careful look at the "13" Node. ie there are 13 different routes to that specific Node.

They are quite easy to follow. and en-route you will have passed through Nodes that could have been reached by 5, 3 and another 5 routes respectively. (You might have noticed, there is only 1 route to the NW corner and there is only 1 route to the SE corner).

There is a "pattern" developing, with a Pub at the end.

(but sorry, the number of routes isn't 345, 347, 365 or 375)

More Frustration ! or Great satisfaction ?