math brain teaser.... are you brain enough?

[ChefWide]

using the numbers:

21 ,19, 8, 10 and 6

add them, subtract them, square them, square root, cube, cube root, mulitply, divide.... you name it... but the answer MUST be '9' and you must use all the numbers every time... 5k in k to the one with the most correct variations.

 

[hanselthecaretaker]

(21+19)/8=5 +10=15 -6= "9"

 

[ChefWide]

(21+19)/8=5 +10=15 -6= "9"

lead dog. and yes, The Black Keys RULE.

 

[hanselthecaretaker]

lead dog. and yes, The Black Keys RULE.

I've only heard that one song off a promotional cd; I'm gonna check out their album though since I've read they have a lot of meaty blues type guitar on it. Good stuff.

 

[1_more_rep]

1. (21+19)/8=5 +10=15 -6= 9

2. 21-19 = 2(10-6)=8 + 8^0 = 9


i win with 2 !

 

[jackangel]

went down to tin pan alley

see what was going on

things was too hot down there

couldn't stay very long

heyyyyyyyyyyyyyyyyyyyyyyyy

alley's the roughest place...i've ever...beeeeeeeeeeeen

 

[hamstershaver]

(21+19)/8=5 +10=15 -6= "9"

not taking anything away from hansel but how is this answer correct when hes adding 2 numbers in the equation that you didnt have listed
5 and 15

 

[UA_Iron]

1. (21+19)/8=5 +10=15 -6= 9

2. 21-19 = 2(10-6)=8 + 8^0 = 9


i win with 2 !

the exponent of zero was not in the problem either....

21 ,19, 8, 10 and 6

6! = 720; 720/8 = 90; 90/10 = 9; 9 + 21 = 30; 30/10 = 3; 3+6 = 9; 9 + 21 = 30; 30 - 19 = 11; 11 - 10 = 1; 1 + 8 = 9

condensed: [(6!/8)*(1/10) + 21](1/10) + 6 + 21 - 19 - 10 + 8 = 9


10^6/(21+19) = 25000; 25000/10 = 2500; 2500 - (10*10*21) = 400; 400/8 = 50; 50-10 = 40; 40 - 21 = 19; 19 - 10 = 9

condensed: [(10^6)/(21+19) * (1/10)] - (10*10*21)*(1/8) - 10 - 21 - 19 = 9

there are infinite combinations fyi....

 

[Beachbum1546]

not taking anything away from hansel but how is this answer correct when hes adding 2 numbers in the equation that you didnt have listed
5 and 15

those are just the equalled amounts. they are not computated in the problem.

 

[hamstershaver]

those are just the equalled amounts. they are not computated in the problem.

yes they are computed in his equation

 

[UA_Iron]

those are just the equalled amounts. they are not computated in the problem.

the biproducts of the operation, they are not limited to the numbers listed, only the operatations on them are.

 

[UA_Iron]

did you want to see who could do it in the least amount of operations?

 

[jackangel]

i think he wants people to keep churning away like goodmonkeys, results irrelevant.

 

[hanselthecaretaker]

the biproducts of the operation, they are not limited to the numbers listed, only the operatations on them are.

Yes, thanks for clarifying.

 

[Beachbum1546]

I tried doing it with only using the number once, unlike what UA_Iron did. I could go on for an eternity if I used the numbers over and over again.

 

[samoth]

add them, subtract them, square them, square root, cube, cube root, mulitply, divide.... you name it

Well, that took all the fun out of it...

I define my own operator and product space such that any action on those numbers is confinded to the set of R which I define here to be a subset of the integer 9.

QED

 

[redguru]

Well, that took all the fun out of it...

I define my own operator and product space such that any action on those numbers is confinded to the set of R which I define here to be a subset of the integer 9.

QED

How many different operators? I declare an infinite number of operators that when used with the numbers given by Chef is also confined to the set noted by R

R being a subset of the integer 9. Therefore the number of operations using those numbers is without boundary.

 

[hamstershaver]

alert!!!!!!!!!!!!

 

[samoth]

How many different operators? I declare an infinite number of operators that when used with the numbers given by Chef is also confined to the set noted by R

R being a subset of the integer 9. Therefore the number of operations using those numbers is without boundary.

Damn you!

 

[redguru]

Damn you!

I was too damn lazy to set up a Euclidian proof table. Almost wanted to make it an elaborate non-answer.

 

[Beachbum1546]

I spent close to half hour trying to make a perfect equation but I couldn't make a good one without using one of the numbers twice. Here's the best I could do with minimizing the use of an extra number.


{[(10^2 - 6^2)/8] + 21-19} [(100^-.5)(10)]=9


only used extra 10.

 

[samoth]

I spent close to half hour trying to make a perfect equation but I couldn't make a good one without using one of the numbers twice. Here's the best I could do with minimizing the use of an extra number.


{[(10^2 - 6^2)/8] + 21-19} [(100^-.5)(10)]=9


only used extra 10.

The way he worded the problem, you can use any algebra or ring you want, define whatever sets you see fit, and use any operators necessary.

You're working entirely too hard. Math is about being as lazy as possible. Usually, knowing a solution exists is good enough. Anything past that is called "applied math".

 

[Devastation]

10 - ((21-8)/(19-6)) = 9