Peckris 2

The first 1 being overstruck is highly worthy of note, as they tended to reuse the '186' matrix, only punching the final numeral on the die. Obviously there was more than one matrix but only a fraction of the number of dies used.

Also known as 'stop over numeral' (son)?

As with most logic puzzles, the actual thing is the calculation - the rest is just window dressing to make it seem more fun, or to misdirect. To restate : Person 1 takes 1 minute to cross the bridge in either direction; Person 2 : 2 minutes, Person 3 : 5 minutes, Person 4 : 10 minutes. They can only cross in pairs (or singly) and have only 17 minutes to do it. (You also have to factor in that one of those who cross has to bring the torch back each time).

Not allowed! D takes 10 minutes there and 10 minutes back.. Try again.

In my opinion - no. The final date numeral of 1861 is so variable, virtually every die seems to be different, with variable heights, overstrikes, spacing, slants etc. I'd say no-one except a rare specialist collects all these minuitiae. Oh, and I just can't see any evidence of a 6/G?

Four members of a band are walking to a night concert. They decide to take a shortcut, but must cross a bridge. Luckily they have one flashlight. Because of the varying size of their instruments, it takes each member a different amount of time to cross the bridge - it takes the first person one minute, the second person two minutes, the third person five minutes and the fourth person ten minutes. They must cross the bridge in pairs, travelling at the slower speed so if the one minute person went with the ten minute person, it would take a total of ten minutes. Since there is only one flashlight, one person must come back across the bridge, then another pair can cross. They only have 17 minutes to cross the bridge and still get to the concert on time. What order should they cross to get everyone across and get to the concert?

a) is correct - as you say, the other factors (unshaven, untidy, scruffy) are classic misdirections.

That may well be true, but the puzzle doesn't specify which of the two was busier.

A logician with some time to kill in a small town decided to get a haircut. The town had only two barbers, each with his own shop. The logician glanced into one shop and saw that it was extremely untidy. The barber needed a shave, his clothes were unkempt, his hair was badly cut. The other shop was extremely neat. The barber was freshly shaved and spotlessly dressed, his hair neatly trimmed. Why did the logician return to the first shop for his haircut?

It's actually even easier than that! You're right that the second man catches up with the first at 10:15. That's exactly two hours after he set out. The dog is running at 5mph, so in 2 hours he runs 10 miles.

Correct. The question was phrased deliberately misleading (rather than simply "How far did the dog run?") in order to give the impression that the calculation could be very difficult, i.e. trying to add up the ever decreasing distances between the two men as the method .. which would be VERY difficult! Did anyone first think - however fleetingly - that that was how it would have to be done?

An elderly guest at a hotel went out for a walk at 7:15 in the morning, ambling at a sedate two miles per hour. At 8:15 another guest went for a walk along the same path with his dog. He walked at a bit more active three miles per hour. The dog was full of beans and trotted off at five miles per hour. The dog chased the first man along the path. When it reached him, it immediately turned around and ran back to its master. When it reached him, it turned right around and ran back to the first man, and kept this up - to and fro - right up until the second man caught up with the first man. Are you able to calculate how far the dog ran altogether?

Category 2 - close facsimiles - is worrying. The other three are easy to identify as gaming tokens.

Nothing.

He needs to show great resistance to going outside especially as going out is kind of hard-wired into our brains.

Great idea. Just for Elizabeth II so far: each year has its own reverse with often very slight differences (see Freeman) there are two reverses for 1954 - short teeth and long teeth there are 4 varieties for 1956 - two obverses each paired with two reverses (the second obverse next appears as 1967 Type II, the reverses are long and short teeth) there are two reverses for 1957 which you probably already know about - calm and wavy sea (from 1958 the sea is always wavy) there are 2 varieties for 1967, two obverses

That's a very difficult one! The solution will be a pointed oval (like an American football) but I have no idea how to calculate half the circle in that shape!

That's not in the original riddle.

Question - must the dates be as you've shown, or is it just that you must convert 2 rows into a circle, i.e. any old dates?

"You are allowed just one question to one of the robots". Your version allows a question to each robot which is not allowed.

That's another solution. Too easy!

That's more or less it. I'll try to think of a harder one next time!

Anyone?

That's no good. If it was the truthful robot, you'd get the answer 'yes', but you'd also get the answer 'yes' from the lying robot if it WASN'T the safe door, so you'd be no better off.