A Fistful of Brain Teasers

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 ?

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.......

Arithmetic Squares !

Arithmetic Squares 1st Puzzle JPEG

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

Youignore the usual precedence rules

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

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

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.

 

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!)

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