# Message Board

## Crosswords and other puzzles

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

Message board > Crosswords and other puzzles | 16 to 115 of 153 |

by I've never bothered to call in, but... - Wed 14th Mar 2007, 8:22pm | ||

Simon said: Bear in mind that at one stage during the broadcast this would have earnt you £30k: That's if you managed to get through. I heard somewhere that at busy times, the chance of a caller getting through could be as high as 10,000:1. | ||

by even better.. - Wed 14th Mar 2007, 8:08pm | ||

Really stretching the numbers to fit the answer said: then write down further figures with the leading digits removed: 0p, 5p, 47p, 6p, 0p (total 58p) What I really don't get is the justification of this bit..? surely '16' is '16' not '16+6'? I can kind of accept the rest of it.. then again write down further figures with another leading digit removed: 0p, 0p, 7p, 0p, 0p (total 7p) the 'official' answer (according to the Telegraph) seems even weirder: " Two pounds is 200p plus 2p (two p) [no further justiication given] and 1p (at the beginning of 'pounds') makes 203p25p: 25p plus 5p and 1p (the 'p' in the question) = 31p £1.47 = 147p 16p: 16p plus 6p and 1p = 23p Fifty pence: 50p plus 50p (fifty p, a shortening of pence), 1p (reference to pence) and 1p (from the 'p') = 102p 203+31+147+23+102 = 506 | ||

by Simon - Wed 14th Mar 2007, 8:00pm | ||

Really stretching the numbers to fit the answer said: A convoluted way of looking at it Wll done Richard - that's about it. ITV have now published the official result. Bear in mind that at one stage during the broadcast this would have earnt you £30k:ITV explained that the answer could be reached by breaking up the figures in the list to find all the references to pence. Thus: -Two pounds is 200p plus 2p (two p) and 1p (p at the beginning of 'pounds') which makes 203p -25p: 25p plus 5p and 1p (the p symbol) = 31p £1.47 = 147p -6p: 16p plus 6p and 1p (p again) = 23p -Fifty pence: 50p plus 50p (fifty p, a shortening of pence), 1p (reference to pence) and 1p (p only) = 102p -Adding 203p, 31p, 147p, 23p and 102p gives a total of 506p | ||

by Really stretching the numbers to fit the answer - Wed 14th Mar 2007, 7:23pm | ||

Simon said: As this is crosswords and other puzzles, here's one from ITVPlay, currently subject to inevstigation by Ofcom: A convoluted way of looking at itA graphic asked viewers to "Add the pence" from "Two pounds, 25p, £1.47, 16p, Fifty pence". More than three hours later, the host announced the answer was 506 and that no-one had won. No method was given. ITV has denied the problem was particularly complex but has not commented further. £2 25p £1.47 16p 50p write them in pence: 200p, 25p, 147p, 16p, 50p (total 438p) then write down further figures with the leading digits removed: 0p, 5p, 47p, 6p, 0p (total 58p) then again write down further figures with another leading digit removed: 0p, 0p, 7p, 0p, 0p (total 7p) The sum of these three subtotals = 503p Now look back at the question. Add the pence: Two pounds, 25 p, £1.47, 16p. 50penceThere are 3 p or pence listed. Add this to the toal that we already had. 506p - simple. | ||

by Does this make me a genius? - Mon 12th Mar 2007, 9:26pm | ||

The Times Online said: ..a puzzle intended for an early-hours viewing audience of drunks and poor sleepers... Good to see the Times has so much respect for its readership! ... A reader of The Times said [about the show] ... As for how you get to 506.. Add two pounds worth of old money (2 * 20 shillings/pounds * 12 pence/shilling) to the other numbers in pence = 718 Subtract the 75p you spent calling in your answer = 643 This leaves you £1.37, which is taken into account by the 40.77 minutes you spent complaining to ITVPlay, or the 19.57 minutes complaining to OFCOM = 506p. Obvious. | ||

by Simon - Mon 12th Mar 2007, 4:09pm | ||

As this is crosswords and other puzzles, here's one from ITVPlay, currently subject to inevstigation by Ofcom:A graphic asked viewers to "Add the pence" from "Two pounds, 25p, £1.47, 16p, Fifty pence". More than three hours later, the host announced the answer was 506 and that no-one had won. No method was given. ITV has denied the problem was particularly complex but has not commented further. | ||

by Highest common factor - Fri 9th Mar 2007, 9:46am | ||

Nice. I can't come up with anything to follow that though. | ||

by Neil T - Fri 9th Mar 2007, 8:55am | ||

Fabulous Ming's hot coach of term: number one for two thousand and seven? (7, 6, 6) I'll take the stony silence as a request for hints, so here you go:1) No specialist knowledge is required; anyone at the club (or not at the club) could solve this. 2) Unlike the clue, the answer has nothing to do with rowing. 3) If I'm honest, 'Fabulous' wasn't a great choice of word. It enhances the surface reading but is a bit dubious in the cryptic reading. Better would have been " Staggering Ming's...". | ||

by Neil T - Thu 8th Mar 2007, 12:53am | ||

RTT said: ...(not that I can see a great deal of use for it here). I think the 1st Men might disagree with that.So here's one for them: Fabulous Ming's hot coach of term: number one for two thousand and seven? (7, 6, 6) | ||

by RTT - Wed 7th Mar 2007, 10:21pm | ||

Olly Olly said: Oh that's just silly. Apologies to the 99% of people reading this who are confused. I did say it was not for everyone. In fact, I guess it was really just aimed at you (although clearly there are plenty of others who would be able to get it).Having said that, I like the new chant (not that I can see a great deal of use for it here). | ||

by Olly Olly - Wed 7th Mar 2007, 10:16pm | ||

RTT said: Sorry, no. I was originally going to use the word "masters" instead of "leaders", but decided the latter sounded better in relation to the bumps. Oh that's just silly. Apologies to the 99% of people reading this who are confused. I prefer 'King Ming' as an answer anyway, and may chant it a few times now to reinforce that view. | ||

by RTT - Wed 7th Mar 2007, 10:06pm | ||

Neil T said: Is it 'King Ming'? (As in Ming Campbell, leader of the Lib Dems.) Sorry, no. I was originally going to use the word "masters" instead of "leaders", but decided the latter sounded better in relation to the bumps.If it isn't, it should be. | ||

by Neil T - Wed 7th Mar 2007, 10:03pm | ||

RTT said: A further hint to the previous one: Two years ago the clue would not have worked. Is it 'King Ming'? (As in Ming Campbell, leader of the Lib Dems.)If it isn't, it should be. | ||

by RTT - Wed 7th Mar 2007, 9:39pm | ||

Neil T said: Now how about you post one that some of us can solve, instead of one that most (indeed all) of us can't?! A further hint to the previous one: Two years ago the clue would not have worked. | ||

by Neil T - Wed 7th Mar 2007, 9:33pm | ||

RTT said: 1) Valkyrie That was quick! Now how about you post one that some of us can solve, instead of one that most (indeed all) of us can't?!2) Black Prince (Valkyrie: V (= 5* in Roman numerals) + A + L (= learner = novice) + KYRIE (Kylie with 'R[ight]' for 'L[eft]', i.e. 'having changed sides'). Black Prince: see here for the 'other', non-boat-burning, reference.) | ||

by RTT - Wed 7th Mar 2007, 9:25pm | ||

Neil T said: Two more meta-clues, as noone seems to be able to get Tom's: 1) ValkyrieShe carried heroes to glory, including '5' (a novice) and a stunner who'd changed sides (8) Who had burning desire for Fair Maid of Kent? (5, 6) (Wikipedia might be helpful in understanding the second.) 2) Black Prince | ||

by Neil T - Wed 7th Mar 2007, 9:15pm | ||

Two more meta-clues, as noone seems to be able to get Tom's: She carried heroes to glory, including '5' (a novice) and a stunner who'd changed sides (8) Who had burning desire for Fair Maid of Kent? (5, 6) (Wikipedia might be helpful in understanding the second.) | ||

by RTT - Wed 7th Mar 2007, 7:11pm | ||

Neil T said: 1) The definition is 'A rallying cry', and the answer is formed from two shorter words (total 8 letters) meaning 'leader', e.g. 'king', 'CO', etc. This is the correct interpretation. I'd refer you back to the qualifier stated before the clue - not everyone reading this site will be able to get it. | ||

by gf - Wed 7th Mar 2007, 2:03pm | ||

Bit of a guess said: clutter? Indeed. | ||

by Bit of a guess - Wed 7th Mar 2007, 1:24pm | ||

gf said: How about a "meta-clue" in the meantime... clutter?RTT clue could create chaos. (7) | ||

by gf - Wed 7th Mar 2007, 11:49am | ||

How about a "meta-clue" in the meantime... RTT clue could create chaos. (7) | ||

by Neil T - Wed 7th Mar 2007, 11:25am | ||

gf said: Hmm... within rally co-drivers' pacenotes, there are various four-letter words which might somehow be applicable ("bump", "fast", "easy"...) but I can't find a combination which sits well with the rest of the clue. I'm baffled by this. The only two interpretations I can come up with are:1) The definition is 'A rallying cry', and the answer is formed from two shorter words (total 8 letters) meaning 'leader', e.g. 'king', 'CO', etc. 2) The whole clue is a cryptic definition, possibly punning on 'rally' meaning 'to drive a car in a rally race' or 'to get better' or 'to demonstrate/protest', and with 'the two leaders' possibly referring to the double headship. | ||

by gf - Wed 7th Mar 2007, 11:12am | ||

RTT said: One for the venatoris: Are we over-complicating this one? Could it just be "Hold fast" ???A rallying cry for the two leaders (4,4). | ||

by gf - Wed 7th Mar 2007, 11:06am | ||

RTT said: One for the venatoris: Hmm... within rally co-drivers' pacenotes, there are various four-letter words which might somehow be applicable ("bump", "fast", "easy"...) but I can't find a combination which sits well with the rest of the clue.A rallying cry for the two leaders (4,4). | ||

by Joff - Wed 7th Mar 2007, 10:44am | ||

well yes, I got that much. didn't know if there was an additional definition or bit that I missed. I suppose 'that's what andy thinks we are' but surely the 'but different' should go next to the 'same woe'? bah. one day, one day I will understand these stupid things.. | ||

by Richard - Wed 7th Mar 2007, 10:29am | ||

MC Chung said: same woe - that's what we are, according to Andy, but different this time! (7) I'm not particularly good at cryptic crosswords, but "same woe" is an anagram of awesome, which seems to fit the bill. | ||

by RTT - Tue 6th Mar 2007, 8:24pm | ||

Neil T said: I guess one of the words is probably 'head'? Nope. | ||

by Neil T - Tue 6th Mar 2007, 9:49am | ||

A rallying cry for the two leaders (4,4). Struggling a bit with this one. I guess one of the words is probably 'head'?A couple of people have asked for an explanation of the (6, 8) clue below, so here it is: 'First and Third decked in gold' = 'gold' with the first and third letters removed ('decked') = OD so: 'First and Third decked in gold and blue remarkably' = OD and BLUE 'remarkably' (anagrammed) = DOUBLE 'crown' = HEAD 'end of Bumps' = S 'with it' = HIP (as in 'trendy') and the whole clue is also the definition. | ||

by RTT - Mon 5th Mar 2007, 12:51am | ||

One for the venatoris: A rallying cry for the two leaders (4,4). | ||

by awesome! - Sun 4th Mar 2007, 11:22pm | ||

but I don't think I got most of that clue..! (which reflects my crossword inability rather than your clue ming!) | ||

by MC Chung - Sun 4th Mar 2007, 10:57pm | ||

same woe - that's what we are, according to Andy, but different this time! (7) | ||

by MC Chung - Sun 4th Mar 2007, 9:50pm | ||

Double Headship. Not bad Neil, I'm impressed. | ||

by Neil T - Sun 4th Mar 2007, 9:02pm | ||

Not from tomorrow's Times Crossword (Mars Bar for the first correct answer): First and Third decked in gold and blue remarkably crown end of Bumps with it (6, 8) | ||

by past it - Mon 22nd Jan 2007, 10:05am | ||

Neil T said: I don't know of anything lower than this. Diving is.. I'd be eligible to dive 'masters'. If I could dive that is. It caters for 'adults of all abilities', so I presume the lower limit is 18. | ||

by Neil T - Fri 28th Jul 2006, 7:59am | ||

RTT said: We were wondering at work today which sport has the lowest age for veteran / master status. Rowing (27) is obviously pretty low, but then this is mainly because it is sub-categorised and thus flexible enough to have something approaching a fair system for any age. Still, there must be something lower. Women's gymnastics maybe (bet that'd be about 16....)? In swimming the qualification age for 'Masters' events is 25. I don't know of anything lower than this. | ||

by RTT - Fri 28th Jul 2006, 7:31am | ||

We were wondering at work today which sport has the lowest age for veteran / master status. Rowing (27) is obviously pretty low, but then this is mainly because it is sub-categorised and thus flexible enough to have something approaching a fair system for any age. Still, there must be something lower. Women's gymnastics maybe (bet that'd be about 16....)? | ||

by Neil T - Tue 25th Jul 2006, 1:55pm | ||

Neil T said: From Saturday's Answer: UltrasonicTimes:Too high to catch? (10) | ||

by Roy Walker - Mon 24th Jul 2006, 10:38am | ||

Mike said: [me at] frontstops? It's good, but it's not the one. | ||

by Mike - Mon 24th Jul 2006, 10:15am | ||

Neil T said: From Saturday's [me at] frontstops?Times:Too high to catch? (10) | ||

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: 1, 18 next?What is the next number in this sequence? 16, 8, 11, 14, 9, 12, 5, 20 | ||

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 Ok, you've got it, there's no 10th or 11th term3 5 And then I get stuck. | ||

by Mike - Mon 9th Jan 2006, 5:50pm | ||

Andy said: What comes next in this sequence? 83 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 TITPEighteen 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 Swallows Don't Make A Summer2 SDMAS 36 BKOAP 6 BIAO 7 DS 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 Calling Birds ?4 CB | ||

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 3 Minutes For A Soft Boiled Egg?2 WDMAR 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 5 Olympic Rings3 FIOY 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) Well, it's unique if you don't count swapping the two numbers being multiplied* x * -- ** | ||

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 Is the right answer.1 + 1 / (2n*pi) miles north of the South pole. 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. You could also start on any circle which isIf 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? 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. 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?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. 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? because of the fives and zeros.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). 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. I wasn't this efficient. How did you deduce that 3.16 was the "most likely" multiple of 0.79?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 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. |

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