A Riddle

As this riddle was described to me, a 4 year old will solve it in 5-10 minutes, a programmer will solve it in an hour, and you will solve it in.....

 

9313 = 1

7756 = 1

6855 = 3

9881 = 5

8096 = 5

7111 = 0

2172 = 0

6666 = 4

1111 = 0

3213 = 0

7662 = 2

5531 = 0

2222 = 0

7777 = 0

9999 = 4

8809 = 6

0000 = 4

3333 = 0

5555 = 0

8193 = 3

2581 = ???

 

 

 

 

 

Spoiler buffer.

 

 

 

 

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UsesProzac says...

Ugh, this is the essence of my trouble with numbers. They swim around the screen for me when I try to figure them out. It brings me right back to math classes and the damned frustration I'd feel. I consider myself intelligent, even highly in some areas, but when it comes to numbers, nothing makes me feel like an idiot faster.

Can someone explain what this all means?

deathcow says...

I tried for a few minutes and ran out of time and then used google to solve it within the projected 4 yr old completion time.

lucky760 says...

Son of a...

I knew it had to be something to do with the presentation rather than the mathematics, but I could not spend the requisite 1 hour to solve it, so I had to google the answer and now I'm pissed it wasn't more obvious to me.

Well done.

dystopianfuturetoday says...

If it's something a child is more likely to see, then it can't be anything too complicated. Think simple. FTR, I gave up on this one fearing it would take me an hour, but I wish I hadn't when I looked up the answer. >> ^UsesProzac:

Ugh, this is the essence of my trouble with numbers. They swim around the screen for me when I try to figure them out. It brings me right back to math classes and the damned frustration I'd feel. I consider myself intelligent, even highly in some areas, but when it comes to numbers, nothing makes me feel like an idiot faster.
Can someone explain what this all means?

probie says...

I'm with UsesProzac; numbers are not my forte. Had I not Googled the answer, I can easily tell you that I'd be at:
24 days and still counting...

Issykitty says...

Yep, I stared at it trying to figure it out for a good 20 min and I'm certain I wouldn't have come up with the answer on my own, either.

dystopianfuturetoday says...

The riddle from the movie Labyrinth is a classic:

Before you are two doors. Each door has a guard. One door leads to fortune, the other leads to misery. One of the guards always lies. One of the guards always tells the truth. You may ask one question and one question alone to both of the guards before you decide which door to take.

What question will reveal the correct door?

>> ^eric3579:

about 20-30 min total time. That was fun. More please.

Boise_Lib says...

This is one I've seen used in job interviews:

You're in a boat on a lake. You pick up the anchor that's in the boat and throw it into the lake.
Will the water in the lake; rise a bit, fall a bit, or stay the same level?

alien_concept says...

>> ^Boise_Lib:

This is one I've seen used in job interviews:
You're in a boat on a lake. You pick up the anchor that's in the boat and throw it into the lake.
Will the water in the lake; rise a bit, fall a bit, or stay the same level?


Oh hmm, is it the same because whether in the boat to make it heavier or in the water, it's exactly the same?

Boise_Lib says...

>> ^alien_concept:

>> ^Boise_Lib:
This is one I've seen used in job interviews:
You're in a boat on a lake. You pick up the anchor that's in the boat and throw it into the lake.
Will the water in the lake; rise a bit, fall a bit, or stay the same level?

Oh hmm, is it the same because whether in the boat to make it heavier or in the water, it's exactly the same?


That was my first thought, but I came awake about 3 AM that night with the real answer.

I will reveal the answer when I receive $100,000 in unmarked bills.
(or, in two hours--whichever comes first)

NetRunner says...

I got impatient and looked it up before I solved it, but I was zeroing in on it being a weird coding thing where 8 = 2, and 6 = 1, and 9 = 1, but I probably would have never realized why that was getting me the right answer until someone pointed out what those three numbers had in common...

oritteropo says...

AAAAAAARGH!!!!

I reckon I could've stared at it for a week without twigging, but actually gave up after about 20 mins. Now to try it on a preschool age child to see if it really works as well for them as advertised.

p.s. I am a programmer too, although haven't done much recently, and have studied math at university level Having looked up how a programmer would solve this, it looks about what I'd have done too, given another 20 minutes and more coffee. Giving up and googling after 10 minutes is what a real programmer would do though

oritteropo says...

I tried it on a 4 year old (quite numerate) child, but the outcome was not as described. Instead she guessed 100, and gave up after 5-10 minutes... having fallen into the same trap that the grown-ups were supposed to fall into.

I expect an 8 year old to solve it the same way as a programmer would, in about the same time (5 mins with google, an hour without).

longde says...

The riddle involved two skills, IMO (without googling, so correct me if wrong):

1) addition (which I would guess a 4 year old could do)
2) recognition of sets and set mapping (which I am not entirely sure a 4 year old would think to do)

messenger says...

I started off thinking the right way, but guessed a different feature, then for some reason talked myself out of pursuing it. Once I saw Netrunner's semi-spoiler comment, I realized I'd seen this one before.

messenger says...

If I understand the intention of the puzzle correctly, I'm assuming the lake is a closed system (a large pool, really), that all energies are transferred instantly and evenly, and that all things mentioned have mass.

After you pick up the anchor and start lifting, the boat will sink some amount (not a rate) proportional to the speed you're raising the anchor, and the lake will rise a proportional amount to the amount the boat sank. When you stop lifting the anchor, the boat will go back to the start height, as will the lake. The moment you release the anchor, the lake boat will rise above its start height, and the lake will drop below its start height. As the anchor enters the water again, the lake will rise back to the start height, so the final result is the water will stay the same level.

Or I could say that with any anchor that I could lift and an average-sized lake, the difference in the height of the lake distributed over the entire surface would probably be less than a molecule of water, so even without figuring out the forces, I could claim the difference would be insignificant, so "stay the same" again. But I guess I can't just assume the "lake" is not a wee tarn, where an anchor could make a measurable difference.>> ^Boise_Lib:

This is one I've seen used in job interviews:
You're in a boat on a lake. You pick up the anchor that's in the boat and throw it into the lake.
Will the water in the lake; rise a bit, fall a bit, or stay the same level?

messenger says...

@Boise_Lib

My first answer is definitely wrong because it assumes that the increased displacement of the boat due to the anchor's mass is equal to the displacement by the anchor's volume when it enters the water. This is demonstrably false, as a 5kg lead anchor would cause a boat to displace as much water as a 5kg iridium anchor in the same boat, but the lead anchor would displace about twice as much water as the iridium anchor when thrown in the water.

Let's say my anchor is iridium, and is massive enough it almost sinks the boat. The boat is now displacing much more water than the anchor will when I throw it overboard, so the answer is the water will fall. In my thought experiment, I now decrease the density of my imaginary anchor and increase its size such that the boat is always just barely afloat. When I toss the anchor overboard, the water will fall less and less, but as long as the boat is floating (not much of a boat if it doesn't float) and the anchor alone is denser than water (not much of an anchor if it's not denser than water, I believe the anchor will always cause the boat to displace more water in the boat than outside the boat. I'm sure I could use limits to prove this, but I'm just going to assume I'm right without doing the calculus. So my new answer is: the water will always fall.

Thanks for the mental exercise.

(How did I do?)

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