[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 490

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joo:

We have to assume that the width of the ladder is equal to 0, i.e. they are just lines drawn on the sides of the cylinders.

From the first glance at the problem: the length of the line is shorter at the smaller diameter of the cylinder. This means that the hobbit will reach the top faster.

This is exactly the wrong view. With a width of 0 and the same angle of inclination to the horizon, the 'ladders' have the same length.
 
PapaYozh:
This is exactly the wrong view. With a width of 0 and the same angle of inclination to the horizon, the 'ladders' have the same length.
don't tell me, either around the house or around the city.
 
sergeev:
You can either run around the house or around town.
Not "around", but up ten metres.
 
PapaYozh:
Not "around", but up 10 metres.
make it simple - turn the cylinders around and draw a line at the same angle on both reamers.
 

Yeah, right. Cylinder reaming takes all the guesswork out of it. How you 'mark it up' afterwards - with a diameter of 2.5 or 10 - doesn't matter. It doesn't affect the answer.

Of course, the width of the ladder is zero.

 
Well, what about the cards?
 

There's already an answer on the five-card forum: count any 10 pieces and turn them over. That's one deck. The rest are left untouched. That's the second one.

 

 
sergeev:

Probably should have put it in Humour. It's a simple task.

 
Honestly, I haven't figured out how to solve it yet. Can the riddled numbers themselves be any two-digit numbers?
1...483484485486487488489490491492493494495496497...628
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