Brain Teasers
 

I think it's about time we had a puzzle thread! ;)

I'll kick off with two easy alphametics:  the rules of this type of puzzle are (1) each letter stands for a different digit, (2) each particular letter stands for the same digit wherever it occurs, and (3) there are no leading zeroes.

1) A^B=BA ("^" means "raised to the power of")
Only one solution.

2)

Code:

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HALF
+HALF
=====
WHOLE

---

Three solutions.

Perhaps one clarification is needed:

In an alphametic the letters stand for the digits of a number; thus, for example, "BA" means (in algebraic terms) 10*B+A, not B*A as it does in standard algebraic notation.

Come on, people, these two puzzles are easy!  Perhaps I should start laying some difficult ones on you? ;)

Re: Brain Teasers
 

*** SPOILER - I'm posting what I think is the solution ***














5^2 = 25 (i.e. A=5, B=2)

The answer to #1 is 2. 2^2=2x2 which is 4=4. (oops never mind...I thought the notation BA was BxA)....well have to rethink this one...oops looks like Edwin got it. ;)

HALF
+HALF
=====
WHOLE

OK, let's work this out:


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W = 1 because of the overflow.

(F+F) is always divisible by 2., which means E is also a multiple of 2.

Say there is an overflow: (F+F) >=10

then (2L+1) MOD 10 has to be L.
Are there any L's where this is possible?

L=0: 1 = 0 NO
L=1: 3 = 1 NO
L=2: 5 = 2
L=3: 7 = 3
L=4: 9 = 4
L=5: 1 = 5
L=6: 3 = 6
L=7: 5 = 7
L=8: 7 = 8
L=9: 9 = 9

So L has to be 9 if there is an overflow.

No overflow:

2L MOD 10 = L

With an overflow, only 0 is possible.

This means: since (H+H+carry) has to overflow, H has to be 9.
If H=9, then L=0 (no number used twice)

This means: F+F is not allowed to overflow, which means F is anyone of 0,1,2,3,4 - but 0 is already used, so is 1, so any of 2,3,4

E=2F

2A has to overflow, which means A is any one of 5,6,7,8 (9 already used).

O = 2A.

Since 0 for a letter is already taken, O cannot be 5.

No let's assume some numbers.

W=1
L=0
H=9

choose F=2 => E=4,
choose A=6 => O=2 - taken
choose A=7 => O=4 - taken
choose A=8 => O=6 works

So, first solution:

9802 + 9802 = 19604

second solution:

choose F=3 => E=6
choose A=7 => O=4

9703 + 9703 = 19406

third solution

choose F=3 => E=6
choose A=8 => O= 6 - 6 taken

choose F=4, E=8
choose A=6 => O=2

9604 + 9604 = 19208


There are no other solutions for F=4.

Excellent answers — I like the thorough way mklose has shown his working!

Now let's see some more puzzles... ;)

Hey Robert, what about the one I gave you over PM: for you first puzzle, *prove* that there's only one solution... It's actually a pretty easy one...

Now, the brain-teasers that Robert provided are "reasonable problems"--they respond to reason, in that you can follow a set of logical steps to find the answer (as Mklose showed). I like the brain-teasers that are "unreasonable"--there is no particular set of logical steps that can find the answer (these are also known as "rompecabezas"). I came across this interesting classification in David Perkins' excellent The Eureka Effect, from where the following puzzles are drawn:[list=1][*]One day at the office, Alice says to Betty, "I heard this great joke from Cathy." And she begins to tell Betty the joke. But Betty says, "Oh, I already known that joke." Alice says, "Oh, Cathy already told you." "No," says Betty. "In fact, I never heard it or read it before." Explain how this could be.[*]Someone brings an old coin to a museum director and offers it for same. The coin is stamped "540 BC". Instead of considering the purchase, the museam director calls the police. Why?[/list=1]I'll give you two more tomorrow if these get solved.

The second one is very easy :-)

the first though - I have no clue.

#2 is so easy. I won't even bother stating it. ;)

#1 -- could it be that....
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Betty made up the joke, and told it to Cathy (or some intermediary)? Perhaps this is not the right answer, because then technically she would have "heard" it as she spoke it -- but just not have heard it from someone else. :confused:

oOo--Correct. Alright, since you both think #2 is too easy (I know a lot of people who don't work it out, BTW), I'll give you another harder one today:

3: There's a man with a mask at home. There's a man coming home. What's going on here? (This is thin information, so perhaps you would like to hear some questions answered. Is the man with the mask a thief? No. Does the man coming home live htere? No. Is the man with the mask going to hurt the man coming home? No.)

Can we assume there is one home and not two in this scenario? If so, perhaps the man at home is giving himself a facial, and the man coming home is a door-to-door encyclopedia salesman??? (If the mask were on the other man, I would say it was Halloween.)

In case there are people who haven't worked out the coin one yet:
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No date-stamp of "540 BC" can possibly be genuine, because "BC" didn't exist until Dionysius Exiguus established the AD epoch in 523 AD... :D

(edit: forgot this link:

The Calendar FAQ)

Well, you don't have to know about 523 AD, you only have to know what 'BC' stands for to see that it's a logical impossibility...

We wouldn't say the salesman is "coming home" in this case--we'd say he's "going to the other person's home". So no, that's not it.

Just kidding of course about the encyclopedia salesman. ;)

But seriously, in any scenario, wouldn't he be "going to the other person's home" by definition since he does not live there, according to the puzzle?