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lurey
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Posted on 06-07-07 9:15
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Beauty of Maths: 1 x 8 + 1 = 9 12 x 8 + 2 = 98 123 x 8 + 3 = 987 1234 x 8 + 4 = 9876 12345 x 8 + 5 = 98765 123456 x 8 + 6 = 987654 1234567 x 8 + 7 = 9876543 12345678 x 8 + 8 = 98765432 123456789 x 8 + 9 = 987654321 1 x 9 + 2 = 11 12 x 9 + 3 = 111 123 x 9 + 4 = 1111 1234 x 9 + 5 = 11111 12345 x 9 + 6 = 111111 123456 x 9 + 7 = 1111111 1234567 x 9 + 8 = 11111111 12345678 x 9 + 9 = 111111111 123456789 x 9 +10= 1111111111 9 x 9 + 7 = 88 98 x 9 + 6 = 888 987 x 9 + 5 = 8888 9876 x 9 + 4 = 88888 98765 x 9 + 3 = 888888 987654 x 9 + 2 = 8888888 9876543 x 9 + 1 = 88888888 98765432 x 9 + 0 = 888888888 and finally take a look at this symmetry 1 x 1 = 1 11 x 11 = 121 111 x 111 = 12321 1111 x 1111 = 1234321 11111 x 11111 = 123454321 111111 x 111111 = 12345654321 1111111 x 1111111 = 1234567654321 11111111 x 11111111 = 123456787654321 111111111 x 111111111=12345678987654321
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TutunKaMun
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Posted on 06-08-07 10:02
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HOW ABOUT THIS ONE 3 GUYS WALKS IN TO THE HOTEL AND ASK FOR A ROOM. RECEPTIONIST SAYS ITS £30 PER NIGHT. 3 OF THEM DECIDES TO SHARE AND PUTS IN £10 EACH. LATER ON, THE RECEPTIONIST FINDS OUT THAT ACTUAL PRICE OF THE ROOM WAS £25, SHE DECIDES TO SEND £5 BACK TO THE GUYS WITH THE BELL BOY. BELL BOY BEING GREEDY DECIDES TO KEEP £2 FOR HIMSELF AND GIVES £1 EACH BACK (£3 IN TOTAL). NOW GUYS HAVE PAID £9 EACH FOR THE ROOM. NOW YOU DO THE SUM £9 PER PERSON TIMES 3 = £27, AND BELL HAS GOT £2,, EQUALS £29, WHERE THE HECK IS £1?
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blue_moon
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Posted on 06-08-07 10:09
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Three guys total pay $27. $25 for room 2 for Bell boy. Nothing wrong.
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TutunKaMun
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Posted on 06-08-07 10:19
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Blue_moon, come on man, everybody knows that, but if you follow as you read, it is bit confusing. I bet you got £29 at the begining.
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gemmi_auj
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Posted on 06-09-07 12:01
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One more ugly maths: In a closed room 30ft x 12ft x 12ft, there is a spider in the middle of one of the square walls exactly 1ft below the ceiling. In the middle of the opposite wall, exactly 1 feet above the floor, lies a fly. What is the shortest distance that the spider will have to crawl to reach fly?
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tregor
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Posted on 06-09-07 8:04
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in an ordinary case, the spider has to crawl at least 42 feet.. if there is a channel between them, then minimun distance by pythagorus theorem = 31.62 feet
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slowPoison
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Posted on 06-09-07 8:27
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by pythagorus theorem my answer is 43.17 feet (for a right-angle triangle of 10 feet height and 42 feet base)
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tregor
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Posted on 06-09-07 9:20
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the base is supposed to be 30 ft
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gemmi_auj
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Posted on 06-10-07 1:42
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Well, the spider has to crawl along the wall; it cannot fly in the air. The floor is 30 ft x 12 ft, and the height of the room is 12 ft.
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Thyangboche
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Posted on 06-10-07 2:52
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gemmi_auj
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Posted on 06-10-07 6:56
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tregor, slowPoison, thyangboche, The straight line joining the two locations will be 10*sqrt(10) ft. But what is required is the distance that the spider will have to crawl along the wall / ceiliing / floor to reach the fly. Yes, 42 ft is the ordinary guess, but that's not the length of the shortest possible route to crawl along.
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tregor
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Posted on 06-10-07 7:17
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if you assume that the spider can drop itself from the top of the ceiling to the floor rather than crawling, the shortest distance will be 31 feet.........
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gemmi_auj
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Posted on 06-10-07 8:04
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It's a problem on ordinary geometry.. no tricks are to be assumed.
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Thyangboche
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Posted on 06-10-07 8:47
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Man what comes 2 my head are:: 31ft, if you dont consider the jump 'or' 42ft if it can't jump. but are u sure my answer is wrong???
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Thyangboche
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Posted on 06-10-07 8:48
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gem.... use separate thread next time coz itz messy!!!
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tregor
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Posted on 06-10-07 11:51
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ya thts what i said earlier 42 and 31
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gemmi_auj
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Posted on 06-10-07 9:58
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Thyangboche, The spider cannot jump. And the correct answer is less than 42 ft.
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Thyangboche
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Posted on 06-10-07 9:59
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buddhu spider!! this spider is good for nothing......it'll crawl 42 feet or 31 ft(if it's a real SPIDERMAN) .....and see....how sweet the fly is......she is eagerly waiting there only meanwhile for her dear Spider....???
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ma_thito_ho
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Posted on 06-11-07 1:38
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The spider has to move: 38.6349082376012411 FT.
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gemmi_auj
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Posted on 06-11-07 6:48
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But how's that??? This is not the correct answer.
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tregor
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Posted on 06-11-07 10:33
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the nearest possible distance is the straight line which is 42 or 31
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