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

Message board > Crosswords and other puzzles 71 to 120 of 153
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by Richard - Wed 28th Dec 2005, 11:52pm
Richard said: The Royal Parks Website suggests that there are only 8 Royal Parks.
The quote from the site is;

Millions of Londoners and tourists visit the eight Royal Parks for free each year.
by Richard - Wed 28th Dec 2005, 11:51pm
gf said: Depending on which one you believe from a selection of four sources I've found so far, there are 8, 9, 17 or... 10 Royal Parks.
The Royal Parks Website suggests that there are only 8 Royal Parks. I'd say that would be a fairly reliable source
by gf - Wed 28th Dec 2005, 12:03pm
Depending on which one you believe from a selection of four sources I've found so far, there are 8, 9, 17 or... 10 Royal Parks.
by gf - Wed 28th Dec 2005, 10:43am
Is it possible that this quiz has been set by somebody non-sporty who thinks there are 18 Teams In The Premiership?
by gf - Wed 28th Dec 2005, 10:33am
mjb said: 3 Coins In A Fountain
12 Good Men And True
by mjb - Tue 27th Dec 2005, 11:13pm
3 CIAF
3 Coins In A Fountain
by Richard - Tue 27th Dec 2005, 10:44pm
4 PIAPT
4 Players In A Polo? Team
by mjb - Tue 27th Dec 2005, 10:28pm
10 DS
10 Downing Street
by Richard - Tue 27th Dec 2005, 10:16pm
22 TLD
22 - Two Little Ducks
by mjb - Tue 27th Dec 2005, 9:26pm
jmg said:

2 SDMAS
36 BKOAP
6 BIAO
7 DS
2 Swallows Don't Make A Summer
36 Black Keys On A Piano
6 Balls In An Over
7 Deadly Sins

So it's just these ones left :

18 TITP
30 MSIABUA
12 GMAT
4 PIAPT
10 DS
3 CIAF
22 TLD
10 RP
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?
by Ingers - Wed 19th Oct 2005, 4:05pm
Ok, this is an old chestnut, but is the single one of these sorts of problems that took me the longest time (and led to at least 1 man-day's loss of productivity at Deloitte).

I think of 2 integers between 1 and 100. I tell Peter the Product and Simon the Sum.

1.) Peter: "I don't know what the numbers are"
2.) Simon: "I knew you wouldn't"
3.) Peter: "Now you've siad that, I do know"
4.) Simon: "Now you've said that I also know"

What are the numbers? No computer programmes allowed
by jpd - Wed 19th Oct 2005, 3:56pm
mjb said: OK, so I rushed the previous one a bit.

How aboot the centre of the earth, or is East undefined there as well. In which case, go out far enough such that your 1 mile East will take you back to where you started the Eastward leg.
East is definitely undefined along the whole of the straight line that runs through the North and the South pole. Also, how can you possibly walk at the centre of the Earth?

You are walking on the Earth's surface. And before you ask, no other form of transport is used.
by Mike - Wed 19th Oct 2005, 3:53pm
mjb said: How aboot the centre of the earth
Have you suddenly turned Canadian?
by mjb - Wed 19th Oct 2005, 3:51pm
jpd said: Where else could I be?
OK, so I rushed the previous one a bit.

How aboot the centre of the earth, or is East undefined there as well. In which case, go out far enough such that your 1 mile East will take you back to where you started the Eastward leg.
by jpd - Wed 19th Oct 2005, 3:41pm
Rich said: At a position one mile north of the place where walking east around the south pole brings you in a full circle 1 mile in circumference, back to the same point.
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?
by Rich - Wed 19th Oct 2005, 3:38pm
jpd said: Also wrong.
Aw, come on! You start "on the equator", you finish "on the equator"!!

At a position one mile north of the place where walking east around the south pole brings you in a full circle 1 mile in circumference, back to the same point.
by jpd - Wed 19th Oct 2005, 3:36pm
Rich, using semantics said: On the equator?
Also wrong.
by Rich, using semantics - Wed 19th Oct 2005, 3:35pm
jpd said: Good. Where else could you be?
On the equator?
by jpd - Wed 19th Oct 2005, 3:34pm
mjb said: Also, depending on which north you take, you could be exactly 1 mile north of the South Pole.
Is also the wrong answer (East is undefined at the South pole).
by jpd - Wed 19th Oct 2005, 3:33pm
mjb said: Anywhere exactly 2 miles north of the South Pole.
Is the wrong answer.
by mjb - Wed 19th Oct 2005, 3:31pm
jpd said: Good. Where else could you be?
Anywhere exactly 2 miles north of the South Pole.

Also, depending on which north you take, you could be exactly 1 mile north of the South Pole.
by mjb - Wed 19th Oct 2005, 3:28pm
jpd said: How did you deduce that 3.16 was the "most likely" multiple of 0.79?

jpd also said: If we don't spot this it's not too difficult to prove that the other three multiples of 79 (so 79, 158, 237) can't be an item value: they must all have two items with three fives (or 125) as a factor (we can't have four fives as the sum would be too large, and otherwise all three other item values would end in zero of five, which when summed with 79, 158 or 237 can't make a number ending in one).
because of the fives and zeros.
with 3,3,1,0.79 as a start point, multiplying one of 3,3,1 with a non-trivial number of pence would leave a partial pence result after the balancing division operation and a small pence multiplication would leave too large a result after division.
but 5's and zeros are easy to come by in the pence column.
by jpd - Wed 19th Oct 2005, 3:15pm
Richard said: North pole?
Good. Where else could you be?
by jpd - Wed 19th Oct 2005, 3:13pm
mjb said: Factorise 711 into 3,3,79. For addition, most likely multiple of 79 to use is 316, so we have 1,3,0.75,3.16 or similar. Then some guided trial and error to find the three smaller numbers.

so 1,3,0.75
becomes 1,1,2.25
then 1.25,0.8,2.25
and eventually hits 1.25,1.2,1.5 which is the answer.
I wasn't this efficient. How did you deduce that 3.16 was the "most likely" multiple of 0.79?

I dealt entirely in pence, so the sum was 711 and the product 711 million. 711 million factorises easily into six twos, two threes, six fives and 79. Each item must have a value which is a product of the factors, with the four items using all factors.

The item with 79 as a factor could only be 79 (on its own), 158 (with one two), 237 (with one three), 316 (with two twos), 395 (with one five), 474 (with a two and a three) or 632 (with three twos). All other possible combinations resulted in the item being worth more than 711.

For the last three possible values (395, 474, 632) it is trivial to prove that the other factors cannot be combined to form a sum less than or equal to 711 for the last three.

The minimum sum of item values for an item of value 316 is 710. Two of the item values for this minimum sum are 125 (which is 5 * 25) and 144 (which is 2 * 3 * 24); swapping the 5 with the (2 * 3) will obviously increase the sum by one to 711.

If we don't spot this it's not too difficult to prove that the other three multiples of 79 (so 79, 158, 237) can't be an item value: they must all have two items with three fives (or 125) as a factor (we can't have four fives as the sum would be too large, and otherwise all three other item values would end in zero of five, which when summed with 79, 158 or 237 can't make a number ending in one). From here the number of potential item values is small enough to discount them easily for each of these three multiples of 79.

We now know that one item is worth 316. We can easily discount having two items with three fives as factors as this would leave an item ending in an even number not ending in zero, so all three items must have at least one five as a factor and in fact two must have at least two fives as factors (otherwise we'd have to have an item with four fives as a factor which is too large). So our other three items have as factors five, 25 and 25.

Since our known item value ends in six, our desired sum ends in one, and our other items all have five as a factor, we must have either one or three items ending in five. Any item with two and five as a factor will end in zero, so two of these items must have the four twos distributed between them.

I'm sure you could continue like this but getting the answer through intuition at this point is trivial.
by mjb - Wed 19th Oct 2005, 3:12pm
Richard said: North pole?
That seems to be the trivial answer.
by Richard - Wed 19th Oct 2005, 3:09pm
jpd said:
I walk one mile South, one mile East and one mile North and end up where I started. Where am I?

(obviously don't answer if I've posed this one to you before)
North pole?
by mjb - Wed 19th Oct 2005, 2:22pm
mjb said something
also making use of the fact that (not sure it this is generally true, but it works for these numbers)

for a,b and x,y where x=a*f, y=b/f
if a < x < b and a < y < b then (x+y) < (a+b)
by jpd - Wed 19th Oct 2005, 2:17pm
Mike said: I hope you're expecting more than one possible answer to this...
I couldn't say how many answers I'm expecting.
by mjb - Wed 19th Oct 2005, 2:16pm
Mike said: Also out of interest, how did you do it? I'm wondering if my own method was hideously inefficient (even though I tried to make it logical rather than trial and error)
Factorise 711 into 3,3,79. For addition, most likely multiple of 79 to use is 316, so we have 1,3,0.75,3.16 or similar. Then some guided trial and error to find the three smaller numbers.

so 1,3,0.75
becomes 1,1,2.25
then 1.25,0.8,2.25
and eventually hits 1.25,1.2,1.5 which is the answer.
by Mike - Wed 19th Oct 2005, 2:02pm
jpd said: (Out of interest, how long did it take you?)
Also out of interest, how did you do it? I'm wondering if my own method was hideously inefficient (even though I tried to make it logical rather than trial and error)

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