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General Forum: Love | | PUZZLES PUZZLES AND PUZZLES | |
| *vacuums the room again.....vroom vroom*
Correct RS! :-)
Here is a small calculator puzzle:
Using a simple 8-digit display calculator, what is the max 'Effective Number' you can get?
Some definitions:
Actual Number: The number on the display
N : Number of keys pressed
Effective Number = Actual Number/N
e.g. if you press 99999999, then the effective number would be 99999999/8 = 11111111
Now is this the max or can we get something larger than this?
This is an open puzzle, it just clicked me when I was clicking the calculator buttons! :-)
Some notes:
- use a simple calculator
- exponential results like 9E+04 etc are not valid
- silly results like upside down values are ruled out
Posted by: చంటోడు At: 14, Jul 2004 4:14:02 AM IST
aahaa maLLee ennaaLLaku oka pajil kanabaDimdi ee bOrDulO.......
paarthaa gaaru....meerE choostumDaali komchem ee bOrDuni....
87531 X 9642 = 84,39,73,902
Posted by: Raj Sekhar At: 21, Jun 2004 3:35:04 PM IST
nobody?
Posted by: చంటోడు At: 18, Jun 2004 10:53:44 PM IST
*after vacuuming the room*
Using the digits 1 to 9, create two numbers which when multiplied give you the highest number
For example, 12345678 x 9 = 111111102, get the highest number
Posted by: చంటోడు At: 18, Jun 2004 5:02:01 AM IST
yup u got it :)
Posted by: Ravi At: 1, May 2004 3:05:21 AM IST
ok got it I think!
1 1 2 3 4(1)
5 5 4 6 7(5,4)
8 8 2 6 9(8,2,3,6,7,9)
Posted by: చంటోడు At: 30, Apr 2004 6:04:37 AM IST
Third day:
Orders: 1 1 3
Items : E G H
Posted by: Mr. Venkat venkat At: 30, Apr 2004 6:47:55 AM IST
Posted by: CLUELESS At: 30, Apr 2004 7:13:14 AM IST
both correct but as venkat mentioned ..they need to order all 9 items :) its easy once u get this ..try it :)
Posted by: Ravi At: 30, Apr 2004 5:33:02 AM IST
How about this...
day 1: 1 1 2 2 3(we know 3)
day 2: 2 4 4 5 6(we know 2, 1, 4)
day 3: 5 7 8 8 8(we know 5, 6, 7, 8)
the one not ordered is 9, same logic/confusion as Venkat's for 'eaten all'
(items numbered 1-9)
Posted by: చంటోడు At: 30, Apr 2004 5:13:14 AM IST
I don't think the following is the correct solution, coz it doesn't satisfy: 'they have eaten all nine items on the menu' Even then, following this, we can determine all the 9 items on menu. Here, it is.., (A-I: items on menu, 1,2: no of copies)
First day:
Orders: 2 1 1 1
items : A (B/C/D) (B/C/D) (B/C/D)
Second day:
Orders: 1 2 1 1
Items : B C (E/F) (E/F)
Third day:
Orders: 1 1 2 1
Items : E G H ?
The remaining one is the one never ordered, but you can identify it, cann't you?
Posted by: Venkat At: 30, Apr 2004 4:47:55 AM IST
:( me thinkin all the while its a series :(
ok herez another puzzle
Five people go to a Mesopotamian restaurant. Not being familiar with such food, they do not recognize any of the names for the dishes. Each orders one dish, not necessarily distinct. The waiter brings the dishes and places them in the middle, without saying which is which. At this point, they may be able to deduce some. For example, if two people ordered the same item, and everyone else ordered different dishes, then the item of which two copies arrive must be the one of which two were ordered. They return to the restaurant two more times, following the same drill, though with different orders. After three meals, they have eaten all nine items on the menu, and can tell which is which. What pattern of ordering fulfills this?
Posted by: Ravi At: 30, Apr 2004 3:17:21 AM IST
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