Brain Teasers are Back!

Looks good Mike, I was struggling with that for a while yesterday, got all the easy bits but struggled with the 3 1 3 area and didn’t think to try splitting it as you did.

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Hi Mike, Hi Ian,

My solution below, which looks the same as Mike’s.

Well done to both of you - despite your loss of sanity !!

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A couple of easy-peasy ones, to help Mike (and Ian) recover their sanity !!

Which of the following can be divided by all the numbers 1 to 10 inclusive (without any remainders or fractions)

A 34 x 23
B 34 x 45
C 67 x 56
D 45 x 56
E 78 x 67

Knave of Hearts “I stole the tarts”

Knave of Clubs " The Knave of Hearts is lying"

Knave of Diamonds " The Knave of Clubs is lying"

Knave of Spades " The Knave of Diamonds is lying"

How many of the four Knaves are telling the truth ?

PS. take great care in reading the question !

A 34 x 23 23 is prime and 34 is not divisible by 3 so no for this one
B 34 x 45 34 is 2x17 and 17 is prime, 2x45 = 90 so not divisible by 4 so no for this one too
C 67 x 56 67 is prime and 56 is not divisible by 10 so no for this one
D 45 x 56 = 9x5 x 7x8 = 3x3x5 x 7x4x2 which is divisible by 1…10 since 6=3x2 and 10 = 5x2.
E 78 x 67 67 is prime and 78 is not divisible by 9 so no.

So only D - lots of other rationales possible for ruling our the failing ones.

Hi Ian,

Nicely explained. And of course, absolutely correct !

Cheers, Don

Anybody into stealing tarts ?

ok, I’ll bite into a tart,

2 are lying. Start at the last statement and assume for a moment the Knave of Spades is telling the truth, hence the Knave of Diamonds is lying, hence the Knave of Clubs is telling the truth, hence the Knave of Hearts is lying.

If on the other hand the Knave of Spades is lying the sequence is reversed , but either way two of the Knaves are telling the truth, you just can’t tell which two.

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No wonder I had a sore head with that one…

I worked the other way, assuming either the knave of hearts stole them, or he didn’t.

That’s also the way I worked it, Dozey.

BTW, nice explanation Ian.

Sorry about the sore head again Mike, but that’s why I wrote … PS. take great care in reading the question !

This one is designed to relieve Mike’s sore head…

Blindfold Coins

Start with 10 coins all Heads-up (but the trick works with any number of coins).

You (the magician) are blindfolded. Your audience shuffles the coins and turn over as few or as many coins as they please. There will now be a mix of Heads and Tails, or possibly all Heads or even all Tails. You are blindfolded so you don’t know !

Your audience now declares how many Tails-up coins there are. Let’s say “Four”

You now volunteer to divide the collection of coins (10 in this case) into two sets, each with the same number of Tails-up showing !!

You now select four coins at random (remember, you are blindfold) and these coins form one of the two sets. The other six coins form the other set.

You (still blindfold) now turn over all four of the coins that you selected to create that first set.

There will be the same number of Tails-up in the set of four as there are in the set of six.

You can now remove the blindfold and bask in the inevitable applause !!!

As I said above, this works, regardless of the number of coins at the start.

PS I should add, the reason you pick four coins at random, is because the audience told you there were four Tails-up coins. ie, you always form that first set using the same number of coins as the audience declares are Tails-up.

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I hope one or two of you have now tried the Blindfold Coins trick above and are satisfied that it always works ?

Of course, the Brain Teaser part is “why does it work ?”

Nifty trick, I shall use that the next time parlour conversation dries up :grinning:

One simple way to see why this always works is that the number of heads-up coins you draw out is always exactly the same is the number of tails-up coins left remaining in the original pile. So when you flip all the coins in that set the number of tails-up in each set must match. This is ‘forced’ by the fact that you always draw out the same number of coins as there are tails-up coin. Neat

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Frogs and Toads

Place three coins of one denomination heads up and three coins of the same denomination Tails up on a line as shown in the diagram. On the left are the Frogs and on the right are the Toads.

Frogs can only move left to right and Toads can only move right to left. A Frog or a Toad can either move one place forward to a vacant position, or it can jump over a single coin into an empty space, providing it is moving in the correct direction.

The task is to move all the Frogs into the positions of the Toads, and the Toads into the positions of the Frogs.

The gap between them at the start is equivalent to one coin.

Powers of nine

The following list of four-digit numbers are each the last four digits of the numbers :-

31^9; 32^9; 33^9; 34^9; 35^9; 36^9; 37^9; 38^9; 39^9

…5077

…8759

…1953

…1875

…8832

…0671

…2848

…8416

…6464

The above numbers are listed in a somewhat random order.

All you need to do is re-arrange them in ascending order, ie so as to match their counterparts 31^9; 32^9 etc

If you have a decent spreadsheet, even Excel will do, you can work out the values of 31^9 etc in a couple of minutes.

You can then match the list of “last four digits” in the table to the values of 31^9 etc.

Easy-peasy !

What is far more interesting (IMHO of course !) is that you will notice (OK, you should notice) something rather neat about these numbers.

In fact, you will now (should now) be able to sort that list into ascending order without having to use Excel !

Let us all know how you get on with these neat numbers !

I sort of got on, in that the last digit of each set of four numbers (…1875) matches the last digit of number raised (31 power 5).

As a wild intuitive guess, the 31^9 ends in 1, the 32^9 ends in 2, etc.

Nine is a very interesting number. If the digits of a 2 digit number add up to 9 it is divisible by nine.