Brain Teasers are Back!


#61

On re-reading this today, i’m dismayed that it reads like a criticism of Ian. It was meant to be an explanation as to why it took me so long to fumble through to a solution and to point out how difficult I have found it can be to create an interesting puzzle that is totally unambiguous (as I have found to my dismay, several times in the past).

Apologies if others were likewise dismayed !


#62

No worries Don, I didn’t read it that way - keep up the good work!


#63

Ok, another easy one !

A heap and its 1/7th become 19. What is the heap ?

Answers in vulgar fractions if possible, rather than decimals :sunglasses:


#64

133/8


#65

That’s pretty vulgar, Eoink !

Nice !


#66

I was going to start the reply with “Up Yours Don”, but silly jokes on vulgar don’t always work on the Internet, so took no chances. :grinning:


#67

You can hopefully print this off and try it out using £’s or tiddly-winks…

The aim is to fill nine of the ten red circles with coins or tiddly-winks or anything else that is handy.

Place a coin on any EMPTY circle on the star. Jump it over any ONE adjacent circle (either empty or filled) to another empty circle in a straight line. “A” jumping over “B” to land on “D” would be an example and D would then be filled and no longer available as a subsequent starting point.

It is possible to fill nine of the ten circles in this way.

Let me know how you get on.


#68

There’s a handful of unsolved/un-explained Teasers in the thread. I’ll re-post them below and provide the solutions/explanations later today or tomorrow, unless others jump in before hand.

Just a “tidy up” exercise.


#69

Ian has got the answer, but I will post an explanation, (unless someone wishes to do so first ?)


#70

As I said before, the first part is relatively easy to resolve, the second part is tough !


#71

I didn’t think this was was at all difficult. !


#72

Largest square has side length 60/17 ch’ih.


#73

Looks like I was right. Ie it wasn’t that difficult - only took you a couple of minutes after I posted !

Well done !


#74

I cheated and used algebra and not geometry :wink: !
The line of the hypotenuse is y= 5-5/12 x and the diagonal line of the square is simply y=x. Therefore these two lines cross when x=5-5/12 x or x=60/17.

If you want to use geometry you can calculate the area of the overall triangle is 30. The square has area a^2 (and we’re searching for a) and the two small triangles have areas of 1/2 base x height and the base and height depend on a. Equating these three areas to the total yields the same answer as the algebra but in a few more lines.


#75

Nicely done Ian.

My solution (above) is a lot less elegant but still gets there.


#76

I don’t expect anybody to go to the trouble of explaining the “Tethered Goat” problem that IanG solved a few days ago, so, in order to put all your minds at rest, and send most of you to sleep as well, here goes…:sunglasses:


#77


#78


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#80