1, 2, 4, 8, 16, 32, 64, 128, 256 and 489

sjbabbeyMember

662 replies

sjbabbeyMember

1, 2, 4, 8, 16, 32, 64, 128, 256 and 489

Don AtkinsonMember

Last edited by Don Atkinson

Well done sjb.

Nice progression with a "489 top-up"

Just to confirm this is a good solution, try a few numbers yourselves eg

£67 = £64 + £2 + £1

£569 = £489 + £64 + £16

Both myslef and sjb will be astonished if you can find a number that can't be created !

Don AtkinsonMember

**Neighbours !! ** ( I can't remember if I posted this one before)

We have two families living one either side of us – The Suttons to our left and the Suttcliffs to our right.

The united ages of the four Suttons is 100 years, and the united ages of the Suttcliffs also amounts to the same.

We know that in the case of each family, the sum obtained by adding the squares of the children’s ages to the square of the mother’s age equals the square of the age of the father.

In the case of the Suttons however, Clair is one year older than her brother Blain, whereas Sally Suttcliff is two years older than her brother David.

What are the ages of each of the eight individuals ?

stevedMember

The Suttons are 13,14,34,39 and the Suttcliffs are 8,10,40,42.

Regarding the Suttons:- If the youngest child is "a", the oldest child is "a+1". If the mother is "x", and the father is "y", then y = 100 - (2a+1+x). So, 4 ages have only 2 variables. Similarly for the Suttcliffs.

I could get to a fairly clumsy formula, but could only find the solutions by even clumsier Excel trial and error. Knowing Don, I'm sure there must be a nicer solution.......

Don AtkinsonMember

Hi Steve,

I’m afraid to disappoint, but I haven’t been able to find a neat solution either. I also had to resort to a trial and error technique aided by Excel !

What would we do without it these days........!

Don AtkinsonMember

Oh ! I should have also mentioned that your answer was spot-on - well done !

Don AtkinsonMember

MulberryMember

Hi Don,

at least fourteen, but I will keep looking for more

Christopher_MMember

16

Don AtkinsonMember

Mulberry posted:Hi Don,

at least fourteen, but I will keep looking for more

Good shout ! ie "I will keep looking for more"

Don AtkinsonMember

Christopher_M posted:16

Hi Chris, nice try but.............

...................same advice that I proffered to Mulberry............

Remember, there are triangles within triangles etc eg, the two triangles below the bottom horizontal line together make a third triangle

sjbabbeyMember

19

Innocent BystanderMember

Last edited by Innocent Bystander

21 (counted 3 times, changing orientation, same each time - Ie if there are more I can’t see them!)

Don AtkinsonMember

14, 16, 19 and 21 so far....................

Don AtkinsonMember

Ignore the first picture (top left) - it's just a reminder of the puzzle.

The other five show some of the triangles.

Next post will show the rest.

Don AtkinsonMember

...........and this slide shows the rest !

So now it's just a case of counting the shaded triangle and making sure there are no omissions and no duplications !

Just make sure you count carefully !

Don AtkinsonMember

Start anywhere on the top row.

Move through the grid from one connected box to another.

Finnish on the bottom row.

Your total must add up to 100.

What is your route ?

Innocent BystanderMember

Are you allowed to land on same one more than once?

Don AtkinsonMember

Innocent Bystander posted:Are you allowed to land on same one more than once?

No.

No going round in circles.

No retracing steps.

But you could zig-zag up-down, and left-right

Innocent BystanderMember

19, 17, 30, 7, 27

Don AtkinsonMember

Nicely done IB !

Don AtkinsonMember

Don AtkinsonMember

**Arithmetic Squares !**

In this puzzle, you place the numbers 1 thru 9 in the empty boxes such that the equations/relationships are correct, both vertically and horizontally.

You only use each number once.

The sums are progressed Left to Right and Top to Bottom

You**ignore** the usual precedence rules

I have illustrated the concept with an example in the post above !

Innocent BystanderMember

2 7 4

9 6 3

8 1 5

Don AtkinsonMember

Good answer IB.

And nicely laid out.

Don AtkinsonMember

**Odd Ball?**

I have twelve balls that LOOK identical but one (and only one) is either heavier or lighter than the rest.

Using a pair of scales, I can compare the weight of any combination of balls that I chose. What is the MINIMUM number of weighings that I need to make, in order to identify the odd ball and decide whether it is heavy or light?

*PS 'scales' = those old fashioned ones that look a bit like the Scales of Justice. Foe example, I could put three balls on each side and if they balanced, I would know that the odd ball was not one of those six. And that would be ONE weighing gone *

fatcatMember

Four.

Innocent BystanderMember

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:

- 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.
- 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. - So which 3 it is in known, but not necessarily if heavier/lighter.

Pick any 1 vs another 1. - 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.

Don AtkinsonMember

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)

Don AtkinsonMember

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

Innocent BystanderMember

Last edited by Innocent Bystander

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!)

fatcatMember

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.

Don AtkinsonMember

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 numberif 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 !

Don AtkinsonMember

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.

Don AtkinsonMember

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.

fatcatMember

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.

Innocent BystanderMember

Ok I understand.

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

Don AtkinsonMember

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.

Sign In To Reply

When you block a person, they can no longer invite you to a private message or post to your profile wall. Replies and comments they make will be collapsed/hidden by default. Finally, you'll never receive email notifications about content they create or likes they designate for your content.

×
Note: if you proceed, you will no longer be following .