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Crosswords and other puzzles

For discussion of all forms of mental gymnastics, especially that baffling final clue

Message board > Crosswords and other puzzles 81 to 100 of 153
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by mjb - Tue 27th Dec 2005, 9:17pm
jmg said: ... questions like "12 DOC" to which the answer is "Days of Christmas" ...
4 CB
4 Calling Birds ?
by jmg - Tue 27th Dec 2005, 6:39pm
jmg said:200 PFPG
Just got this one... pounds for passing go
by Richard - Tue 27th Dec 2005, 6:37pm
10 GB (HOTW)
10 Green Bottles (Hanging On The Wall)
by Richard - Tue 27th Dec 2005, 6:33pm
3 MFASBE
2 WDMAR
3 Minutes For A Soft Boiled Egg?

2 Wrongs Don't Make A Right?
by Richard - Tue 27th Dec 2005, 6:23pm
26 LOTA
26 Letters Of The Alphabet
by jpd - Tue 27th Dec 2005, 5:42pm
9 LOAC
9 Lives of a Cat
by jpd - Tue 27th Dec 2005, 5:39pm
5 OR
3 FIOY
5 Olympic Rings
3 Feet in One Yard
by jmg - Tue 27th Dec 2005, 5:30pm
Okay, so we've got this christmas quiz thing at home. You'll have seen the like before - questions like "12 DOC" to which the answer is "Days of Christmas". Still quite a few left, so here we go...

4 CB
10 GB (HOTW)
2 SDMAS
9 LOAC
36 BKOAP
5 OR
3 MFASBE
18 TITP
30 MSIABUA
12 GMAT
4 PIAPT
2 WDMAR
10 DS
200 PFPG
3 FIOY
3 CIAF
22 TLD
26 LOTA
6 BIAO
7 DS
10 RP

Any ideas?
by Richard - Fri 21st Oct 2005, 3:18pm
And just to keep you going here are a couple for base 8 (containing all of the numbers 1-7)

   **2
  x  *
   ---
   ***

    **
  x *6
   ---
   ***


Both have unique solutions
by Richard - Fri 21st Oct 2005, 1:00pm
Richard said: base 5: (unique solution)
  *
x *
 --
 **
Well, it's unique if you don't count swapping the two numbers being multiplied
by Richard - Fri 21st Oct 2005, 12:53pm
base 5: (unique solution)
  *
x *
 --
 **


base 6: (unique solution)
  **
 x *
 ---
  **


base 7: (2 solutions)
  **
 x *
 ---
 xxx
by Mike - Fri 21st Oct 2005, 9:27am
Dubya said: Are there similar problems in other bases than base 10?
OK: in base 4, the following contains all the digits from 1-3.

  *
+ 1
 --
  *

Or is that not quite what you had in mind?
by Dubya - Thu 20th Oct 2005, 10:12pm
jpd said: OK, you asked for it - prove this is the only solution (or otherwise).
Are there similar problems in other bases than base 10?
by jpd - Thu 20th Oct 2005, 8:04am
Richard said: Are there any other solutions?
OK, you asked for it - prove this is the only solution (or otherwise).
by Richard - Wed 19th Oct 2005, 11:54pm
jpd said:
The following multiplication contains all the digits 1-9. Fill in the blanks:
 186
  39 x
----
7254



Are there any other solutions?
by jpd - Wed 19th Oct 2005, 9:49pm
Richard said: You could also start on any circle which is

1 + 1 / (2n*pi) miles north of the South pole.
Is the right answer.

The following multiplication contains all the digits 1-9. Fill in the blanks:

 ***
  3* x
----
****
by Richard - Wed 19th Oct 2005, 8:23pm
jpd said: Is the right answer.

If we assume the small area around the South pole is flat, our start point would be any point on the circle which is 1 + 1 / (2 * pi) miles North of the South pole.

Where else could I be?
You could also start on any circle which is

1 + 1 / (2n*pi) miles north of the South pole.
by Richard - Wed 19th Oct 2005, 7:48pm
jpd said: Is that inclusive or exclusive?
If it doesn't include 1, then I think 4&13 work.

Peter is told the product, 52
Simon is told the sum, 17

52's factors are;
2&26, 4&13

Simon knows that Peter couldn't tell what the numbers were just from the product.

If Simon was told 28, then one of the possibilities for 28 are 5&23, but the product of 5&23 is 115 (which can be made in no other way) - so if Simon was told 28, then he couldn't know that Peter couldn't tell what the numbers were. Therefore Simon was told the sum 17. Peter now knows what the numbers are.

The possiblities for a sum of 17 are 2&15 (product 30), 3&14 (42), 4&13 (52), 5&12 (60), 6&11(66), 7&10(70), 8&9(72)

product 30 gives 2&15(sum 17), 3&10(13), 5&6(11)
Only a sum of 13 would be eliminated by Peter when told that Simon already knew that Peter couldn't tell what the numbers were (13 can also be 2&11 whose product 22 can only be formed in 1 way), leaving 2 remaining possibilities therefore Peter can't have worked out the right answer. Therefore it's not 2&15 as the answer.

The same argument can be used for 42,60,66,70 and 72, but not for 52.

Therefore Simon will also be able to work out that the numbers must be 4&13
by mjb - Wed 19th Oct 2005, 4:22pm
jpd said: how can you possibly walk at the centre of the Earth?
You obviously haven't read much Jules Verne then.
by jpd - Wed 19th Oct 2005, 4:11pm
Ingers said: I think of 2 integers between 1 and 100.
Is that inclusive or exclusive?

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