Though I have to say the cube root was clever! :D
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Though I have to say the cube root was clever! :D
Ok, I have to give Dr. Luz the credit b/c all I really did was replace the cube root with something that would work.
(8 / └√8┘ )*8/8 =4
Where |_x_| = floor(x). See Wolfram Research's site (http://mathworld.wolfram.com/FloorFunction.html)
Below is the question posted by SeventhStar....:)Quote:
Originally posted by JPicasso_00
What was the question?
Quote:
you have four digits that are the same for example 1,1,1 and... 1 (funny huh) try to get 4 using all mathematical operations you know but without using anu more digits. for example 1+1+1+1 = 4
So do this for the other nine digits (0,2,3,4,5,6,7,8,9) Good luck!
Strange...????!!!!Quote:
Originally posted by KevinHall
Ok, I have to give Dr. Luz the credit b/c all I really did was replace the cube root with something that would work.
(8 / └√8┘ )*8/8 =4
Where |_x_| = floor(x). See Wolfram Research's site (http://mathworld.wolfram.com/FloorFunction.html)
Yes it is strange! :)
The reason why I quoted Wolfram research is to validate that floor(x) is a mathematical operation and not just a programming function call.
Thank Kevin alot for your explanation...
May I ask how you know about those "#code"_s in your previous post or where you learnt those ?
Thanks...
I selected the times font and cut-and-paste the symbols from MS Word.
If you are going to complain about the cube root then the square
root has got to go as well. Remeber the 2 is just supressed by
convention.
one I can think of that works using radicals is
radical index 2 ( 2^2^2)
How about
2*2 log base 2 of 2
:confused: that really is strangeQuote:
Originally posted by KevinHall
Ok, I have to give Dr. Luz the credit b/c all I really did was replace the cube root with something that would work.
(8 / └√8┘ )*8/8 =4
Where |_x_| = floor(x). See Wolfram Research's site (http://mathworld.wolfram.com/FloorFunction.html)
Oh the cube root (having the digit 3) wouldn't have counted
nooo the square root doesnt have any other digits... ;)Quote:
Originally posted by souldog
If you are going to complain about the cube root then the square
root has got to go as well. Remeber the 2 is just supressed by
convention.
one I can think of that works using radicals is
radical index 2 ( 2^2^2)
How about
2*2 log base 2 of 2
I like your ways of making the equasions they really show that you all are programmers... You know this was a International Math Olympics problem for students between 13 and 15 years
btw here is how I solved it:
0! + 0! + 0! + 0! =4
1 + 1 + 1 + 1 =4
2 + 2 + 2 - 2 =4
(3x3 + 3)/3 =4
sqrt(4) + sqrt(4) + 4 - 4 =4
(5x5 - 5)/5=4
6 - (6+6)/6=4
77/7 - 7=4
8/((8+8 )/8 )=4
(sqrt(9) x sqrt(9) + sqrt(9))/sqrt(9)=4
:p :p kinda easier huh?
Thanks ALOT...:)Quote:
Originally posted by KevinHall
I selected the times font and cut-and-paste the symbols from MS Word.
IMO... isn't that an extension of USAMO... but wait I thought it was for highschoolers? Well I guess this isn't the type of question they'd ask in the IMO I'm thinking of... must be different contests with the same name.Quote:
Originally posted by SeventhStar
I like your ways of making the equasions they really show that you all are programmers... You know this was a International Math Olympics problem for students between 13 and 15 years
1998 in Plovdiv, Bulgaria there was a Mathematical olympiade for Eastern Europe that had students of all ages form Bulgaria, Greece, Macedonia, Russia, Romania, Estonia... I took part in this and I was in highscool (being 14) but there were different groups: to 10, 10-13, 13-15, 15-18, 18+ years and this was a problem that was given for my groupQuote:
Originally posted by solarflare
IMO... isn't that an extension of USAMO... but wait I thought it was for highschoolers? Well I guess this isn't the type of question they'd ask in the IMO I'm thinking of... must be different contests with the same name.
that was my last math contest... then I moved to programming (informatics as we call it)
I recently was tidying up my stuff and I found it and thought it'd be suitable for here. Here is another from these problems:
There are 3 programs that read a pair of numbers (a,b) from a special diskette each outputs another pair on the same diskette, deleting the first. The first program - (a-b, b), the second - (a+b,b) and the third (b,a) You have a diskette with the numbers (19,87). Is it possible using it to get the numbers
a) (41,14)
b) (18,81)
and if yes, how?
The fact is to see what is the generator of these pairs. I call generator at the lowest (a,b) pair we can create with a!=0 and b!=0.
(19,87) has (1,1) as generator.
You can create the pair (1,1) by doing these operations
(19,87) (III)->(87,19) (I)-> (68,19) (I)->(49,19) (I)->(30,19) (I)->(11,19) (III)-> (19,11) (I)->(8,11) (III)->(11,8) (I)-> (3,8) (III)->(8,3) (I)->(5,3) (I)-> (2,3)->(III)->(3,2) (I)->(1,2) (III)->(2,1) (I)->(1,1)
a)
Then from (1,1) applying (II) 12 times you have (13,1), and (III)->(1,13) (II)->(14,13) and (III) (13,14). Then (II) ->(27,14) (II)->(41,14) Then (41,14) has the same generator as (19,87)
b) The generator of (18,81) is (9,9) then you can not obtain this pair from (19,87) because do not have the same generator.