ok! long time no brain torture. i m going to keep updating this thread often so no more separate threads. first one. the question has definite scientific reasoning (a big clue) so tech freaks can try i m reducing the question length. a person decides to fly around, he is from UK (we know someone is ) he takes off in northern direction drives 500 km, then 500 km eastwards after turning, turns again drives 500 km southwards , finally turns westward and drives in that direction for another ..yeah right 500km. does he reach the same point he started at? of course he does not can anyone dare explain why and how? i wud also like this thread to be borrowed/hijacked if someone wishes so.

The traveler will in fact be short of his original point of departure at the end of the exercise you describe. This is because traveling in the directions north, south, east, and west infer travel along constant lines of latitude and longitude. It is a well known fact that circles of constant latitude have different lengths and so 500 kilometers at one latitude does not insure that the traveler will end up at the same longitude when compared to the same distance traveled at a latitude 500 kilometers further north. hgmjr

nooooo SSS darn it! of course you are on course. right on target sir. will comeback with a revenge question after i find one. till then be afraid be very afraid

Well if the world was flat then you would arrive back in the same place as when you started. However a 'straight-line' on the Earth's surface is a geodisic Get your geometry textbooks out! Dave

no not another one revenge , retribution vengeance wait for two days i didnt name this thread teasers just like that. cant believe you ppl only took 2 min to get the answer. well done everybody!

Can anyone come up with an estimate of how far the traveler will fall short of his destination? hgmjr

wont that depend on the exact place he left off the question actually mentioned leningrad and the person landed in lake lagoda abt 77 km away

I can point you in the direction of the Mathematics (ref. http://en.wikipedia.org/wiki/Geodesic). Bit of homework! Dave

It will indeed depend on the latitude and longitude at the point of departure. You mentioned the UK so maybe we could agree that the individual began his journey in London. We may need to assume that he was traveling by air since there is likely to be a portion of the journey over water. hgmjr

what the hell, why wait ? here is a question that is not as tricky as the trick required to solve it. this isnt as hard as promised but will serve the intended purpose of selling few aspirins (isnt it banned since its a habit forming compound) so, lets say you have a 364 kg block, u are req to break it into six pieces which will measure everything from 1 - 364kg (zero can also be measured if u so desire) almost forgot : the weighing will be done with a balance. (thankfully i recalled or it wud have been game over) the good thing abt this is, it is a question which gets asked in different forms yet answer(trick) remains the same. by trick i do not mean its a question where u need to read between the lines. (also if you are super skilled in answering at least pretend that it was a hard one ,it takes me minutes to write it and u ppl mercilessly answer it in secs) __________________ my idea of balanced diet is a beer in both hands

Okay, maybe its just me and old age, but I fail to see the problem. We have a 364 kg block. You want six pieces. Okay, I just broke it. The sum total of the pieces remains 364 kg. Now what?

sorry didnt explain it properly now we have six peices and now we can measure any weight from 1-364kg using these six pieces and a balance. say we have to measure 92 kg suppose pieces are 45 64 95 3 1 155 we can put 95 in one side and 3 in other this way 92 can be measured if kept in 3 kg pan. similarly 109 kg can be measured by keeping 45+1 in one pan and 155 in other. what i want is the pieces be broken in such a way that any weight measurement between 1-364 is possible with combination of six weights in either side of the pan. did i make it clear? one hint : the combination of six weight is so much unique such that even if u dont consider the "six weights must add to 364" rule you will still get the answer. ironically teasing is the fact though it seems to be a math trick no integration/diff can solve these types of problems, shud we call it a limitation of our knowledge of maths?

ok so i think i have given a difficult conundrum here, so as kind as i am, here is a good hint try solving for 13 kg for 3 pieces you will see a pattern developing based on which you can solve for 364, if still you cant get it i will give another clue after you solve it for 13 kg.

yeah you got it right, very well done indeed, if u observe they are all multiples of three, one idea that i got was that you take the summation previous number double it and add 1 this is so that you have a continuous nmber after addition of previous numbers and subtraction of new one after that other numbers can also be obtained. right now i can see your evil grin but i will come up with another torture a bit later.

this one is very easy and a bit amusing u might enjoy this. take any 3 digit number,any 3 digits wud do (now is there anyone so generous?) now place the same number before the original number so that it forms a 6 digit number (like xyzxyz) ok now get a calculator divide the number by 7 seven isnt that odd(literally&mathematically) will there be a remainder? if there is one please sell your calculator its gone bad. divide the answer now by 11. come on now 11! isnt division wise 11 a notorious number? this shud leave a remainder--------surprise surprise it doesnt. ok now divide the answer by 13. 13! i keep raising the stakes dont i ? this must leave a remainder --well does it ? of course not. so whats the final answer look carefully does it not look familiar? ok so the question why ? care and dare to explain?

anything times 10..01 will be itself repeated, ie xyz*1001 = xyz*1000 + xyz*1. 4 digit number 10001, 5 100001 etc. therefore xyzxyz/1001=xyz. 7,13 and 11 multiply to 1001 for some reason, luck I guess. I find it interesting that 1^3+2^3+3^3+...+(n-1)^3+n^3 = (1+2+3+...+(n-1)+n)^2. Somthing I discovered randomly, I cant remember if I proved it or not, but I guess it shouldnt be too hard.

darn it! seems like maths comes easy to you, wait for another update(tomorrow same time) , i promise to make it difficult for you ^_^v

ok awesome, look forward to it. Hopefully not too hard as I should probably do some study tomorrow...just like today.

ok here is a puzzle which is fairly easy one and will get you confused if u think like me A friend of mine who owns a bicycle shop narrated the following story, A man who looked like a tourist came to his shop one day and bought a bicycle from him for $350 (fairly expensive what say?).the actual cost of bicycle was $300 so the owner was happy with the prospect of making $50 profit. however at the time of paying the tourist offers to pay in traveler's cheques . the owner had no arrangements with the bank to encash the cheque but he remembered his neighbour could. so he exchanged the cheques for the money from him. the traveler gave 4 cheque of $100 each. so the bicycle owner gave $ 50 back to the traveler after encashing it from his neighbour . the traveler happily pedaled away whistling a tune (u'll soon know why) the next morning the neighbour found that the cheques were fraudulent. he demanded a refund which the bicycle owner quietly did. can anyone tell how much loss did the owner suffer from this transaction. the answer is a fairly simple one (but can sometimes really confuse u) post your explanation if possible. (cheque= check neighbour=neighbor )