A Fistful of Brain Teasers

Posted by: Don Atkinson on 13 November 2017

A Fistful of Brain Teasers

For those who are either non-British, or under the age of 65………. The UK used to have a brilliant system of currency referred to as “Pounds, Shillings and Pence”. Simplified to £ ״ s ״ d. No! Don’t ask me why the “Pence” symbol is a “d”, just learn it and remember it !

A £ comprised 20 Shillings and a Shilling comprised 12 Pence. Thus a £ comprised 240 Pence. I reckon that both Microsoft and Apple would have difficulty with these numbers in their spreadsheets, more so if we included Guineas, Crowns, Half-Crowns and Florins. However, I digress..............

The purpose of the explanation is to assist with the first two or three teasers that follow. So just to ensure a reasonable comprehension has been grasped…. ….. if each of three children has £3 − 7s − 9d, then collectively they have £10 − 3s − 3d   Got the idea ? Good ! Just try 5 children, two each with £4 − 15s − 8d and three each with £3 − 3s −  4d. How much do they have between them ? (this isn’t the first brain teaser, just the basic introduction with some “homework”, the Teasers follow)

Posted on: 10 February 2018 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 !

Posted on: 11 February 2018 by Don Atkinson

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 ?

Posted on: 12 February 2018 by steved

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.......

Posted on: 13 February 2018 by Don Atkinson

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........!

Posted on: 13 February 2018 by Don Atkinson

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

Posted on: 13 February 2018 by Don Atkinson

Count the Triangles JPEG

How many triangles can be determined in the above diagram ?

Posted on: 13 February 2018 by Mulberry

Hi Don,

at least fourteen, but I will keep looking for more 

Posted on: 13 February 2018 by Christopher_M

16

Posted on: 13 February 2018 by Don Atkinson
Mulberry posted:

Hi Don,

at least fourteen, but I will keep looking for more 

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

 

Posted on: 13 February 2018 by Don Atkinson
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

Posted on: 13 February 2018 by sjbabbey

19

Posted on: 13 February 2018 by Innocent Bystander

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

Posted on: 14 February 2018 by Don Atkinson

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

Posted on: 14 February 2018 by Don Atkinson

Slide1

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.

Posted on: 14 February 2018 by Don Atkinson

Slide2

...........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 !

Posted on: 15 February 2018 by Don Atkinson

One HundredJPEG

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 ?

Posted on: 15 February 2018 by Innocent Bystander

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

Posted on: 15 February 2018 by Don Atkinson
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

Posted on: 15 February 2018 by Innocent Bystander

19, 17, 30, 7, 27

Posted on: 15 February 2018 by Don Atkinson

Nicely done IB !

Posted on: 15 February 2018 by Don Atkinson

Arithmetic Squares Example JPEG

Posted on: 15 February 2018 by Don Atkinson

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 !

Posted on: 15 February 2018 by Innocent Bystander

2 7 4

9 6 3

8 1 5

Posted on: 16 February 2018 by Don Atkinson

Good answer IB.

And nicely laid out.

Posted on: 16 February 2018 by Don Atkinson

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