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dimis
Responsible
Supreme Hero
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posted May 05, 2010 08:44 PM |
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Edited by dimis at 20:45, 05 May 2010.
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Q9. Logic Behind Coincidences
Nice one angelito!
Q9. I've thought of a three-digit number such that each of the numbers 543, 142, and 562 coincides with it in exactly one decimal location. Guess what this number is.
(V. Proizvolov)
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The empty set
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ohforfsake
Promising
Legendary Hero
Initiate
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posted May 05, 2010 08:52 PM |
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163?
Coincidies with 562 exactly once at 6, with 543 exactly once at 3, with 142 exactly once at 1. It seems awful simple so I've probably misunderstood something, I'd guess.
Nice solution Angeltio, I understood it now, really good thinking.
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Living time backwards
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dimis
Responsible
Supreme Hero
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posted May 06, 2010 07:39 AM |
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Edited by dimis at 07:40, 06 May 2010.
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Q10. Mirror Numbers
Yes OhforfSake, that's it!
Q10. Two numbers are called mirror numbers if one is obtained from the other by reversing the order of digits - for example, 123 and 321. Find two mirror numbers whose product is 92,565.
(A. Vasin, V. Dubrovsky)
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The empty set
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binabik
Responsible
Legendary Hero
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posted May 06, 2010 08:40 AM |
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Off the top of my head....more trial and error than anything else, but the thought process might lead to something.....
1) The number 92565 ends in a 5. So it's obvious that one number starts with a 5 and the other ends with a five.
2) We can easily find that the number is 3 digits.
3) We know that one number starts with a 5. So the beginning 9 must either be a multiple of 5 plus carry over, or the entire digit must be a carry over. The second case can't be true, therefore we have a multiple of 5 plus carry. Therefore the digit on the other end of the number must be a 1.
Now we have the numbers 5?1 and 1?5
5?1 * 1?5 = 92565
The center digit is equal, so we can easily use trial and error to come up with 561 and 165.
So I guess the thought process did lead somewhere.
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dimis
Responsible
Supreme Hero
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posted May 06, 2010 09:53 AM |
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Q11. Dancing Regularities
Alright! Looks like we have a party! So ...
Q11. At a party, each boy danced with three girls, and each girl danced with three boys. Prove that the number of boys at the party was equal to the number of girls.
(V. Proizvolov)
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The empty set
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binabik
Responsible
Legendary Hero
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posted May 06, 2010 10:06 AM |
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As stated it can't be proven because it's not true. And I'm not even sure what you intended by the question that might make it true.
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dimis
Responsible
Supreme Hero
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posted May 06, 2010 10:10 AM |
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Binabik, it is true and it can be proven; just get some sleep.
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The empty set
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Binabik
Responsible
Legendary Hero
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posted May 06, 2010 10:21 AM |
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Lol, I even drew it out and I guess I can't count to 3
OK, I'll think on it, but you're right, I probably need sleep. (But that's always true these days since I haven't slept properly in 3 1/2 years now.)
But I still need a clarification. Do the sets of 3 need to be unique sets? For example:
if A dances with a,b,c
can B also dance with a,b,c?
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dimis
Responsible
Supreme Hero
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posted May 06, 2010 10:33 AM |
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Edited by dimis at 10:33, 06 May 2010.
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Yes, if A dances with a,b,c, then B can also dance with a,b,c.
But my hint is, think simpler than that.
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Badasti
Hired Hero
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posted May 06, 2010 12:41 PM |
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Quote:
About your task, solved by Angelito, I can see it becomes 13 if every ... means +, but what if it's just a random operation (though of course the same all the way through)?
Well in the email there was a spreadsheet attached with the answer being the pw to open it. The puzzle was just given in a table format with pictures. 13 unlocked the spreadsheet. I just had to improvise with the '+' and '$' etc.
What you ask is too technical for me though. I got the solution to that puzzle in under a minute but I also dropped out of school for drugs and alchohol after grade 7. Anything beyond the basics is beyond my knowledge 
And to your reply, glad to see a fellow noobie. That comic is legendary 
http://www.thenoobcomic.com/index.php for anyone who's wondering. Will be most enjoyable if you have played an MMORPG like WoW and raided.
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Badasti
Hired Hero
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posted May 06, 2010 01:17 PM |
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Quote:
Alright! Looks like we have a party! So ...
Q11. At a party, each boy danced with three girls, and each girl danced with three boys. Prove that the number of boys at the party was equal to the number of girls.
(V. Proizvolov)
Unless there was transvestites 
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Badasti
Hired Hero
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posted May 06, 2010 01:19 PM |
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Here's another one for you guys.
In a race from point X to point Y and back, Jack averages 30 miles per hour to point Y and 10 miles per hour back to point X. Sandy averages 20 miles per hour in both directions. If Jack and Sandy begin the race at the same time, who will finish first?
x Jack
x Sandy
x They tie
x Neither
x Impossible to tell
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ihor
Supreme Hero
Accidental Hero
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posted May 06, 2010 01:47 PM |
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Edited by ihor at 13:50, 06 May 2010.
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Q11.
Assume every boy danced exactly 3 times and one pair include 1 boy and 1 girl.
D - number of all dances. Since every boy danced exactly 3 times, then D = 3x.
Similarly D = 3y.
 x = y, number of boys and girls are equal.
@Badasti
S - distance from X to Y, miles.
Jack's time in hours:
t1 = S/30 + S/10 = S * 4/30.
Sandy's time:
t2 = S/20 + S/20 = S/10 = S * 3/30
Since t1 > t2, Sandy will finish first.
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ohforfsake
Promising
Legendary Hero
Initiate
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posted May 06, 2010 01:53 PM |
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I'd say it's a matter of finding the highest total speed. To do that, I'd find the times, add those and take distance and divide with time.
Jack: time travelled = [distance / velocity]_one_way + [distance / velocity]_back_again = X / 30 + X / 10 = (X+3X)/30
Sandy: time travelled = X / 20 + X / 20 = (X+X)/20
Total distance is X+X, so their speed:
Sandy_Speed_Total = 20*(X+X)/(X+X) = 20
Jack_Speed_Total = 30*(X+X)/(X+3X) = 30/2 = 15
Conclusion, Sandy wins the race.
Bonus:
When Jack reaches point X, he does that to time X / 30, at this time, Sandy have travelled the distance = 20 * X / 30 = 2/3 of the way to distance X.
When Sandy reaches point X, she does that to time X / 20, at this time, Jack have been travelling the other direction in X / 20 - X / 30 = X/60 time units and have reached the distance X - 10*X/60 = 5X/6, or in other words, he have moved 1/6th of the way back to start.
Sandy and Jack is at the same distance again (like at starting point) at time: X/20 + (X/6)/(20-10) = X/15 and both are the distance: X-20*X/60 = 2X/3, onr in other words have moved 1/3th of the way back to start.
When Sandy is in Goal at time X/10, Jack is at the distance
X/6 + 10*X/20 = .67X which means Jack is only 1/3X distance away from finishing the race.
Funny how the difference between Jack and Sandy is identical, but with reversed persons in between distance X and returning again.
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Living time backwards
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Badasti
Hired Hero
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posted May 06, 2010 02:15 PM |
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Edited by Badasti at 14:17, 06 May 2010.
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Sorry boys, but you're both wrong... Will post the solution when I get to work tomorrow if we havn't found an answer =)
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Ecoris
Promising
Supreme Hero
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posted May 06, 2010 04:54 PM |
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The time it takes Jack to get back from Y to X is equal to the time it takes Sally to get from X to Y and back again. Clearly Sally finishes first.
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dimis
Responsible
Supreme Hero
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posted May 06, 2010 11:10 PM |
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Edited by dimis at 23:12, 06 May 2010.
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Q12. In their own Worlds
Alright! Let's move around!
Q12. Winnie-the-Pooh and Piglet went to visit each other. They started at the same time and walked along the same road. But since Winnie-the-Pooh was absorbed in composing a new "hum" and Piglet was trying to count up all the birds overhead, they didn't notice one another when they met. A minute after the meeting Winnie-the-Pooh was at Piglet's, and four minutes after the meeting Piglet was at Winnie-the-Pooh's. How long had each of them walked ?
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The empty set
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted May 06, 2010 11:44 PM |
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Edited by Corribus at 03:36, 07 May 2010.
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I'm wondering as a matter of clarification: were they walking at constant velocities?
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I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
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dimis
Responsible
Supreme Hero
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posted May 07, 2010 12:19 AM |
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Edited by dimis at 04:30, 07 May 2010.
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If something is not given, we should make the most logical assumption that yields a solution; so here I guess it is correct to assume that they travel with constant velocity.
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The empty set
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Badasti
Hired Hero
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posted May 07, 2010 12:09 PM |
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Man I'm so stupid... Somehow I misscalculated yesterday and it showed Jack had an edge on shorter distances, yet sally won on longer distances.
After sitting and converting it down even to meters in a 100m sprint it will scale down exactly the same.
So yeah you're right, Sally wins every time.
I'm going to walk away in shame now :<
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