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Robert Abbott's 1963 game What's That on My
Head? has been republished. You can see more about his game at
www.logicmazes.com.
He's provided the following small puzzle that could occur from real play.
You are playing "What's That on My Head?" with three other players who have these cards on their heads: Ann: 2 4 8
Bob: 4 5 8
Charlie: 1 5 5
Ann picks this question card: "Are there two or more players who have the same total?" She answers, "Yes." (Since Ann cannot see what is on her own head, her answer does not include her own cards.) Bob picks this question card: "Of the four even numbers, how many different even numbers do you see?" He answers that he sees all of the even numbers. At this point Ann says she knows what she has, and indeed she is correct. From previous games you know Ann to be a good logician, and you know she doesn't guess unless she is sure of what she has. Therefore, what is on your own head? This puzzle was solved by Al Stanger, Al Badger, Alexandre Muñiz, Brendand Owen, Carl N. Hoff, Chris Lusby Taylor, Deric L Curtis, Daniel Blum, Dick Saunders Jr, Glenn C. Rhoads, Lance Nathan, Naresh Bansal, Scott Purdy, Stas Soumarokov, Steve Hopkins, Stephen Kloder, Stephen Miller, Steve Walter. |
AROUND GOEDEL'S THEOREM by
Karlis Podnieks introduce axiomatic set theory, logic, and the Axioms of
Zermelo-Fraenkel which provide a foundation for Set Theory.
John Conway recently posted the following problem to the math-fun list:
"Last night I sat behind two wizards on a bus, and overheard this conversation: Wizard A: I have a positive integral number of children, whose ages are positive integers the product of which is my own age, while their sum is the number of this bus. Wizard B: How interesting! Perhaps if you told me your own age and the number of your children, I could work out their individual ages? Wizard A: No, that isn't so. Wizard B: Aha! At last I know how old your are! What was the number of the bus?" |
The rec.puzzles archive contains many classic logic problems, along with solutions.
Linda Aitken has made a new number-based crossword at http://www.frogston.demon.co.uk . I love these things.
Retrograde Analysis is a site devoted to math problems in which deducing the past is object of the problem. There are many beautiful problems here, all from this uniquely logical field of chess problems. Very nice.
This page is under construction. I plan to dig out some logic problems I made a few years ago and represent them. If *you* have a logic problem you'd like to post here, please write me.
Here are some deduction problems, with solutions, from Denis
Borris:
Jack, John and CWS's
====================
Canadian Wild Strawberries (CWS) are tiny but tasty.
A and B each have a jar containing 400 CWS; they decide
to have a CWS eating race; A wins, swallowing his last
CWS when B still has 23 left. Took A 13.2 seconds; burp!
Next, B takes on C, each with a jar containing 261 CWS;
B wins, C left with 117 CWS (C has a bad toothache).
Jack: well, John, A took on C next
John: ya ya I'm sure he did
Jack: each had a jar containing N CWS's
John: oh boy
Jack: wanna try figure out what N is?
John: not really
Jack: here's a hint: in the 200 to 500 range, and they
both swallowed at same speed as
in their 1st race
John: oh ya? (comes back with a printout)
Jack: A beat C by an integral amount
John: ya; I figured as much; need another clue
Jack: the sum of digits of the number of CWS that C had left
when A finished is equal to this
number here
...and John knew.
What is the value of N?
SOLUTION
Result of A:B and B:C races means A:C = 400:208.
So, for all to be integral, jars must contain a
multiple of 25, since 25:13 is lowest.
There are 13 such possibilities in the 200-500 range:
A ate C ate A-C SUMDIGITS
200 104 96
15
225 117 108
9
250 130 120
3*
275 143 132
6
300 156 144
9
325 169 156
12
350 182 168
15
375 195 180
9
400 208 192
12
425 221 204
6
450 234 216
9
475 247 228
12
500 260 240
6
Only 3 is unique as sum of digits of the differences.
So N = 250.
JACK'S KIDS
===========
John: so Jack, how many kids you got now?
Jack: less than 5
John: well...how many?
Jack: figure it out; the product of their ages equals twice the
sum.
John: that's nice; but not enough info.
Jack: my wife gave birth a year ago.
John: still not enough.
Jack: the age of my oldest is the same as the number on this door.
John: hmmm...that helps...still not enough.
Jack: the two in the middle...
John: stop there! I know how many kids you have, and also their
ages...
Can you figure it out also??
SOLUTION:
Age possibilities: 3 6
*
4 4 *
1 3 8 **
1 4 5
2 2 4 *
1 1 4 6 **
1 2 2 5
1 2 3 3 **
2 2 2 2 *
* eliminated due to youngest(s) being 1 (wife gave birth a
year ago).
** eliminated: if no. on the door was equal to oldest, then John
would know.
So, it's 1-4-5 or 1-2-2-5; 1-2-2-5 is it, due to "two in the middle"...
RAIN CONTAINERS:
ONE CONTAINER
=============
OK: my 2 puzzles this week are on rain water containers.
I'll start with the easier one, so you can warm up!
-open-top empty container C is outside
-at 0 hours, a steady heavy rain starts
-rain enters C at rate of R gallons per hour (gph)
-Jack jumps in C as soon as the rain starts
-Jack takes water out at 1/6 R gph for 9 hours
-at this point, rain slows to 1/3 R gph, and Jack triples his speed
-13 hours later, the rain stops
-at this point, Jack reduces his current speed by 8 gph,
and empties C in 10 hours
OK; if Jack handled a total of 778 gallons in this 32hour period,
what is the size of C in gallons, and the rain
speed at start?
SOLUTION:
R = Rain original speed
Easy to see that Jack's 778 gallons are this way:
9(R / 6) + 13(R / 2) + 10(R / 2 - 8) = 778
so: R = 66
Total rained gallons = 9(66) + 13(22) = 880
so:1- C overflowed during start 9hour period (102 gallons)
2- 13hour period begins with a full C
3- Jack emptied C plus 286 gallons (the 13hour rainfall),
in last 2 sessions, or over 23 hours:
C + 13(R / 3) = 13(R / 2) + 10(R
/ 2 - 8)
since R=66, then C = 393 gallons.
3 CONTAINERS
============
OK; now that you got practice (or got your feet wet!):
-3 identical empty open-top containers (C1 C2 C3) are outside
-At 0 hours, a steady rain begins
-rain enters containers at R gallons per hour (gph)
-7 hours later, Jack jumps in C1 and empties it at rate of J gph
-as soon as C1 is empty (split second,since it is still raining)
Jack jumps in C2 and empties it in 42 hours
-then Jack jumps in C3; at that point, rain doubles to 2R gph
-Jack also doubles his speed, to 2J gph
-when Jack finishes C3, it is no longer raining
-Jack goes back to C1 then C2 emptying them at J gph in 44 hours
WELL:
-at what hour did Jack enter C3?
-at what hour did it stop raining?
-at what hour did Jack finally finish?
NOTE: if you feel tired for Jack, you can always pretend hours=minutes!
SOLUTION:
R = Rain speed ; J = Jack speed ; A = hours to empty C1
B = hours of rain in C3 ; C = hours to empty C3 (Note: AR means
A*R)
Jack enters C1 at hour 8
========================
Gallons already in C1 : 7R
Rain while Jack empties: AR
Jack's emptying of C1 : AJ
so: AJ = 7R + AR ;
J / R = (A + 7) / A (1)
Jack empties C2 in 42 hours
===========================
Gallons already in C2 : R(A + 49)
Rain while Jack empties: 42R
Jack's emptying of C2 : 42J
so: 42J = R(A + 49) ; J / R
= (A + 49) / 42 (2)
(1) (2) : (A + 7) / A = (A + 49) / 42
A^2 + 7A - 294 = 0
(A + 21)(A
- 14) = 0 ; A = 14
(3)
(1) (3) : 2J = 3R
(and that's all you need to know about these speeds)
Jack empties C3; both speeds double
===================================
Gallons already in C3 : R(A + 49) = 63R
Rain while Jack empties : 2RB
Jack's emptying of C3 : 3RC
so: 3RC = 2RB + 63R ;
C = (2B + 63) / 3 (4)
Jack empties C1 and C2 in 44 hours, normal speed
================================================
Gallons now in C1 and C2: R(42 + 4B)
Jack's emptying of C1/C2: 44(3R / 2)
so: R(42 + 4B) = 44(3R / 2) ;
B = 6 (5)
(4) (5) C = 25
Hour at which Jack enters C3 : 7 + 14 [A] + 42
= 63
Hour at which rain stopped : 7 + 14 + 42 + 6
[B] = 69
Hour at which Jack is finished: 7 + 14 + 42 + 25 [C] + 44 = 132
JIZZ-KIXX
=========
For the purposes of this puzzle, we now have a new weight:
similar to the pound/ounce, we now have the JIZZ/KIXX;
(took opportunity to give the lonely j/k/x/z some work!).
Jack orders a bag of Ontario potatoes,asking for j JIZZES
and k KIXXES.
The potato man goofs and prepares a bag weighing k JIZZES
and j KIXXES.
The total KIXXES Jack ordered are less than 500.
YOU: good; see ya later
ME : the clerk's error caused Jack's order to be
reduced by P%, P being an integral in
the
range 1 to 99.
YOU: gotta go
ME : ah c'mon, how many KIXXES in a JIZZ?
YOU: oh boy (leave, come back with printout);
hmmm...83 possibilities; gimme a clue
ME: P is equal to this number here
YOU: that sure helps, but need another clue
ME : Jack's order was over 100 KIXXES
...and now you know...
How many KIXXES in a JIZZ ?
SOLUTION:
Among the 83 possibilities <500, we have:
(the possibilities cannot include 1 JIZZ or KIXX;
Jack ordered jizzES and kixxES...more than 1)
TOTAL KIXXES IN
KIZZES IF % REDUCTION
KIXXES A JIZZ KIXXES
JIZZES REVERSED TO TOTAL KIZZES
60 8
7 4
39 35 ( 60*.65=39 )
480 43
11 7
312 35 (480*.65=312)
So if I pointed to 35, you would have these 2 choices.
Then the >100 total KIXXES clue eliminates the smaller one.
So there are 43 KIXXES in a JIZZ.
These are the ONLY possibilities that end up with you KNOWING
after the clues.
All others are either 1 choice (then you'd know as soon as
shown the number) or 3 choices or more with more than 1 in
the >100 range.