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Crosswords and other puzzles

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

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by Mike - Mon 24th Jul 2006, 10:15am
Neil T said: From Saturday's Times:

Too high to catch? (10)
[me at] frontstops?
by Neil T - Sun 23rd Jul 2006, 4:53pm
From Saturday's Times:

Too high to catch? (10)
by RTT - Wed 22nd Feb 2006, 11:30pm
Funniest clue I've seen in ages (no prizes for guessing the publication):

OBN candidate gets to be repellant penetrating a dog (10)?
by Get your PBs down - Tue 31st Jan 2006, 1:26pm
To all crossworders:

DO THE TIMES TODAY.

I have just reduced my PB to 4'04".
by Neil T - Tue 31st Jan 2006, 1:21pm
This may be too late, but...

10 RP is 10 Rillington Place, the scene of several grisly murders a la Fred West. (Also a film I think.) GET IN!!
by Simon - Tue 17th Jan 2006, 8:50pm
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.
I was in the green that is bordered by the Palace of Westminster, Millbank, and the Thames this morning, and the by-laws posted there named 20 sites which are managed by the Royal Parks.
by Mike - Tue 17th Jan 2006, 5:29pm
The Power said: 1, 18 next?
Yup. I'm ashamed to say I didn't get it at the time.
by The Power - Tue 17th Jan 2006, 3:53pm
Mike said: A nice one from Radio 1 this morning:

What is the next number in this sequence?

16, 8, 11, 14, 9, 12, 5, 20
1, 18 next?
by Mike - Tue 17th Jan 2006, 2:50pm
A nice one from Radio 1 this morning:

What is the next number in this sequence?

16, 8, 11, 14, 9, 12, 5, 20
by Andy - Tue 10th Jan 2006, 11:02am
we had this argument in the boat house, whether, a billion was, 1,000,000,000 or 1,000,000,000,000
And also whether an octillion is 10^48 or 10^27, I'm not using any of that inferior american arithmetic
by jmg - Tue 10th Jan 2006, 10:53am
Andy said: 1,000,000,000,000
Very British of you!
by Andy - Mon 9th Jan 2006, 6:01pm
Mike said: 8
3
5

And then I get stuck.
Ok, you've got it, there's no 10th or 11th term
by Mike - Mon 9th Jan 2006, 5:50pm
Andy said: What comes next in this sequence?
8
3
5

And then I get stuck.
by Andy - Mon 9th Jan 2006, 4:58pm
What comes next in this sequence?
101
1,000,000,000,000
10^48
100
1
4
by not even bothered - Thu 29th Dec 2005, 4:27pm
Tom C said: speed in a built up area
18 TITP

Eighteen Turns in the Park (its a sculpture in the Sepentine Gallery?)
by Tom C - Thu 29th Dec 2005, 2:05pm
jmg said: One of the ones we got before I posted, 180 MSITD was 'maximum score in three darts', so I'm thinking this could be maximum score in something?
speed in a built up area
by jmg - Thu 29th Dec 2005, 9:07am
jmg said: 30 MSIABUA
One of the ones we got before I posted, 180 MSITD was 'maximum score in three darts', so I'm thinking this could be maximum score in something?
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)
by Mike - Wed 19th Oct 2005, 1:58pm
jpd said: I walk one mile South, one mile East and one mile North and end up where I started. Where am I?
I hope you're expecting more than one possible answer to this...
by jpd - Wed 19th Oct 2005, 1:45pm
mjb said: £1.20, £1.25, £1.50, £3.16
Is the right answer. (Out of interest, how long did it take you?)

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)
by mjb - Wed 19th Oct 2005, 1:25pm
jpd said: Obviously I meant a magical flying pen.
£1.20, £1.25, £1.50, £3.16
by Tom C - Wed 19th Oct 2005, 1:02pm
jpd said: This one's a bit harder :-)

I buy four items in a supermarker. Both the sum and product of their values is £7.11. What are their individual values?
As Fitz said, "two packs of sausages constitutes one item", so i don't think this has a solution.
by jpd - Wed 19th Oct 2005, 9:28am
jpd said: ...supermarker...
Obviously I meant a shop in which you can buy all manner of groceries, not a magical flying pen.
by jpd - Wed 19th Oct 2005, 9:22am
Slacker said: BBAB?
This one's a bit harder :-)

I buy four items in a supermarker. Both the sum and product of their values is £7.11. What are their individual values?
by mjb - Wed 19th Oct 2005, 8:59am
jpd said: Answer this multi-choice exam:
BBAB seems to work.
by Slacker - Wed 19th Oct 2005, 8:57am
BBAB?
by jpd - Wed 19th Oct 2005, 7:59am
Bored Lilie said: If so, I make it 7..? I need to do some real work, and/or get a life :)
Is the right answer.

Doing work and getting a life are not allowed. Answer this multi-choice exam:

1. The first question with A as the correct answer is:

    A. 2
    B. 3
    C. 4 

2. Which answer appears most often:

    A. C
    B. B
    C. A 

3. The answer to Question 1 is:

    A. B
    B. A
    C. C 

4. The answer which appears least is:

    A. A
    B. C
    C. B 
by Richard - Wed 19th Oct 2005, 1:20am
Bored Lilie said: If so, I make it 7..? I need to do some real work, and/or get a life :)
If all of the relations have to be there, then I get 7 as well.

Man 1 and Woman 2 are married, and have a son, Man 3 who marries Woman 4. Man 3 and Woman 4 have 3 children - 2 girls, 5 and 6 and a boy, 7.

Man 1 is a grandfather to 5,6&7 (1 grandfather)
Woman 2 is grandmother to 5,6&7 (1 grandmother)

Man 1 is the father of 3. Man 3 is father of 5,6&7 (2 fathers)
Woman 2 is the mother of 3, Woman 4 is mother of 5,6&7 (2 mothers)
3 is the child of 1&2, 5,6&7 are the children of 3&4 (4 children)
5,6&7 are grandchildren of 1&2 (3 grandchildren)
5 is the brother of 6&7 (1 brother)
6&7 are the sisters of 5 (2 sisters)
3 is the son of 1&2, 5 is the son of 3&4 (2 sons)
6&7 are the daughers of 3&4 (2 daughters)
4's mother-in-law is 2 (1 mother-in-law)
4's father-in-law is 1 (1 father-in-law)
4 is the daughter-in-law of 1&2 (1 daughter-in-law).

That's all of the requirements met. Can it be done in a better way?

If you actually had 23 people, all related closely, then presumably, you'd have more than the required number in each category. There must be a maximum number of closely related people.....
by Bored Lilie - Tue 18th Oct 2005, 9:37pm
Bryn said: If not then I guess you only need two women and two men who can cover everything between them.
If so, I make it 7..? I need to do some real work, and/or get a life :)
by Bryn - Tue 18th Oct 2005, 5:07pm
A certain family party consisted of 1 grandfather, 1 grandmother, 2 fathers, 2 mothers, 4 children, 3 grandchildren, 1 brother, 2 sisters, 2 sons, 2 daughters, 1 father-in-law, 1 mother-in-law, and 1 daughter-in-law. A total of 23 people, you might think, but, no. What is the minimum number of people here?
Does the person to whom that relationship corresponds have to be there as well? For example, does the grandfather's grandchild have to be there? If not then I guess you only need two women and two men who can cover everything between them.
by jpd - Tue 18th Oct 2005, 4:53pm
mjb said: AFAICT, it's the Coffee-drinking German in the Green House
Is the right answer.

A certain family party consisted of 1 grandfather, 1 grandmother, 2 fathers, 2 mothers, 4 children, 3 grandchildren, 1 brother, 2 sisters, 2 sons, 2 daughters, 1 father-in-law, 1 mother-in-law, and 1 daughter-in-law. A total of 23 people, you might think, but, no. What is the minimum number of people here?
by mjb - Tue 18th Oct 2005, 4:46pm
jpd said: There are 5 houses ... but who owns the fish?
AFAICT, it's the Coffee-drinking German in the Green House
by jpd - Tue 18th Oct 2005, 4:01pm
There are 5 houses in 5 different colours. In each house lives a person of a different nationality. The 5 owners drink a certain type of beverage, smoke a certain brand of cigar, and keep a certain pet. Using the clues below can you determine who owns the fish?

The Brit lives in a red house.
The Swede keeps dogs as pets.
The Dane drinks tea.
The green house is on the immediate left of the white house.
The green house owner drinks coffee.
The person who smokes Pall Mall rears birds.
The owner of the yellow house smokes Dunhill.
The man living in the house right in the middle drinks milk.
The Norwegian lives in the first house.
The man who smokes Blend lives next door to the one who keeps cats.
The man who keeps horses lives next door to the man who smokes Dunhill.
The owner who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
The man who smokes Blend has a neighbour who drinks water.
by jmg - Mon 28th Feb 2005, 2:08pm
Not all that hard, but made me laugh...

Royal autobiography, not appearing all at once? (3,2,3)
by bored at work - Mon 7th Feb 2005, 4:25pm
The code, math and general knowledge rounds were a bit of a disappointment after the image round though.
by Richard - Mon 7th Feb 2005, 9:08am
OK, got it now.
by Richard - Mon 7th Feb 2005, 9:07am
Andy said: http://www.etienne.nu/imagepuz/

umm, i'm stuck on the third one. any ideas?
Any ideas on the 6th one???
by Evil eye - Sat 5th Feb 2005, 8:37am
It's a magic eye thing. My eyes hurtnow, i'msure that can't be good for you on a computersreen. The answer is 'o.htm'
by Andy - Sat 5th Feb 2005, 2:15am
http://www.etienne.nu/imagepuz/

umm, i'm stuck on the third one. any ideas?
by Mike - Tue 1st Feb 2005, 12:18pm
The Times are offering free day passes to their crossword content in celebration of the 75th birthday of the Times Crossword. Among other things, you can try the crossword editor's "fiendishly difficult" clue and win a crate of wine:

One describing a vampire crossing short stretch of water - he apologised (8)

E-mail answers to xchallenge@thetimes.co.uk by Monday 7th February. Perhaps if somebody here solves it, we can all enter and share the wine...
by gf - Tue 2nd Nov 2004, 8:08pm
Mike said: Perhaps Shirley Bassey would have got it?
If she said she did, I wouldn't argue with someone built like that... ;-)
by Mike - Tue 2nd Nov 2004, 4:37pm
gf said: "Is bird-watching an appropriate example of this?" (10)
Perhaps Shirley Bassey would have got it?
by Mike - Tue 2nd Nov 2004, 4:36pm
gf said: "Is bird-watching an appropriate example of this?" (10)
Very good indeed. I have to admit I looked up the answer today though...
by gf - Mon 1st Nov 2004, 4:37pm
Today's 29ac is, imho, the best clue the Daily Telegraph have carried for a fair while...

"Is bird-watching an appropriate example of this?" (10)

...nice smug glow when you get it, as with all good clues!
by Mike - Thu 8th May 2003, 9:40am
Neil said: I have most of the time to stitch, then I iron (9)
Very clever. Nice and politically correct...
by Neil - Thu 8th May 2003, 8:42am
I've just come across this gem:

I have most of the time to stitch, then I iron (9)
by Martin - Sun 9th Feb 2003, 4:24pm
RTT said: Errrr.....why are these messages in reverse order? Is it a new feature? If so, it's a sensible one!!!
I have just received a complaint about it too, though.

I'm afraid its a 'feature' of the way it is constructed: because this topic does not have threads, each post is behaving like each of the (for example) quotations in the quotes topic.
by RTT - Sun 9th Feb 2003, 3:07pm
Errrr.....why are these messages in reverse order? Is it a new feature? If so, it's a sensible one!!!
by Neil - Sun 9th Feb 2003, 2:30pm
Guy said: Is it 'bring' with the 'g' taken off on the outside of 'to'?
That seems plausible, though I think 'footloose' to mean 'with the last letter removed' is very dubious.
by Guy - Sun 9th Feb 2003, 2:04pm
About to fetch footloose islander (6)

The answer was 'Briton'; can anyone explain why?
Is it 'bring' with the 'g' taken off on the outside of 'to'?
by Neil - Sun 9th Feb 2003, 1:41pm
To start the ball rolling...

From a recent Saturday Times Crossword:
About to fetch footloose islander (6)

The answer was 'Briton'; can anyone explain why?

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