Equations

[Renevatia]

Forgot how many numbers I was using, but I challenge accepted my dad a long time ago, yeah cool childhood, I know.
Use numbers 1-9 and make an equation with any symbol you like in math or even combine them, try keeping it algebra, and make the answer = 1,2,3,4....as far as it goes. Use all 1-9. Hard to explain, goes a little like this.

List as many as you like, or just list a seemingly hard/impossible number to get. Try not listing pointless numbers like 1233312312312445516351351234234.




1=9-8+7-6+5-4-3+2-1
2=(9-8+7-6+5-4-3+2)1
100=(9+1)(3+7)(6-5)+8-(4x2)

[iironiic]

3=(9-8+7-6+5-4-3+2)+1

xd

[TheSaxRunner05]

3=9-8+7-6+5-4-3+2+1
4=(9+8+7-6-5-4-3-2)1
5=9+8+7-6-5-4-3-2+1
6=(9+8-7-6+5-4-3+1)*2
Edit://
7=9+8-7-6-5+4+3+2-1
8=(9+8-7-6-5+4+3+2)1
9=9+8+7-6-5-4+3-2-1

Edit2://
1=1^2^3^4^5^6^7^8^9 =p

[reuben_tate]

10 = 9 + 1 + dₓ(8) + dₓ(7) + dₓ(6) + dₓ(5) + dₓ(4) + dₓ(3) + dₓ(2)

[TheSaxRunner05]

168,781,726= Sq Root(9)^8 * Sq Root(7)^6 * Sq Root(5)^4 * Sq Root(3)^2 +1

Cant believe that made a whole number lol

[itsmeraph]

Originally Posted by TheSaxRunner05

168,781,726= Sq Root(9)^8 * Sq Root(7)^6 * Sq Root(5)^4 * Sq Root(3)^2 +1

Cant believe that made a whole number lol

They are all raised to even powers, so there aren't really any ugly decimals caused by the square roots: that equation is basically 9^4 + 7^3 + 5^2 + 3 + 1

[TheSaxRunner05]

Originally Posted by reuben_tate

10 = 9 + 1 + dₓ(8) + dₓ(7) + dₓ(6) + dₓ(5) + dₓ(4) + dₓ(3) + dₓ(2)

Lol

Edit://
0=(-1)9+8+7-6+5-4-3+2

[reuben_tate]

11 = 6+5+9-8+1-2+7-4-3

Originally Posted by TheSaxRunner05

Lol

Maybe that's why the OP said Algebra only. If not, I could just spawn as many " + 1"s into the equation by adding "+ dₓ(x)." lmfao

[Renevatia]

1337= 1327+4+9-5+8-6
9001= 9251-((6-9+8)^3)x sqrt(4)
2012= [563-(2+8)(7-1)]4

[TheSaxRunner05]

11= 9 + (Sin(8+5)^2 + Cos(7+6)^2) + (Sin(4+1)^2 + Cos(3+2)^2)

Does squaring count as a digit, or can we square as much as we want?

[Renevatia]

Originally Posted by TheSaxRunner05

11= 9 + (Sin(8+5)^2 + Cos(7+6)^2) + (Sin(4+1)^2 + Cos(3+2)^2)

Does squaring count as a digit, or can we square as much as we want?

well yes, you can argue that if by that logic sqrt should be counted ^(1/2) but theres a symbol (without numbers shown)for sqrt but no symbol for squares.

[TheSaxRunner05]

Gah, I was trying to get 1337 legit but I got ...

1335 = (9*8*7*6/4)*(3/2)+1

[Renevatia]

Originally Posted by TheSaxRunner05

Gah, I was trying to get 1337 legit but I got ...

1335 = (9*8*7*6/4)*(3/2)+1

1337= (5^3)(2+9)-[(4*6)+8+7-1] if you meant not using any double digits from the 1-9 ;o


1000000= (2^((sqrt(9))!!/3!))-485768*1 , someone try to do this legit too lol.

[TheSaxRunner05]

Got it, I found out working backwards is so much easier

1337 = ((1*(4+5+6)-8)*3*9+2)*7

Edit:// Yea work backwards

1,000,000 = 5^6*1(8*7+9+4-3-2)

Edit2:// 5^6 is gold for these multiples of 1,000,000 ....

10,000,000 = 5^6*8*(3^4-1)+(9-7-2)

Edit3:// Gah, I almost got my number of games played, too bad the 65* is cheating

21733 = (65*(7/2)*8-9)(3*4)+1

Edit4://
Nvm, got it!

Games played
21,733 = ((6+7*(7/2)*8*5-9)*3*4+1

[Renevatia]

Originally Posted by TheSaxRunner05

Got it, I found out working backwards is so much easier

1337 = ((1*(4+5+6)-8)*3*9+2)*7

Edit:// Yea work backwards

1,000,000 = 5^6*1(8*7+9+4-3-2)

Edit2:// 5^6 is gold for these multiples of 1,000,000 ....

10,000,000 = 5^6*8*(3^4-1)+(9-7-2)

Edit3:// Gah, I almost got my number of games played, too bad the 65* is cheating

21733 = (65*(7/2)*8-9)(3*4)+1

Edit4://
Nvm, got it!

Games played
21,733 = ((6+7*(7/2)*8*5-9)*3*4+1

lol nice.

Try your total score lolol.
8533539750

[reuben_tate]

100,000,000 = (5^6)*(4^3)*[(9+1)^2]*(8-7)
EDIT 1: 1,000,000,000 = (5^6)*(4^3)*[(9+1)^(2+8-7)]
EDIT 2:10,000,000,000 = (5^6)*(4^3)*(9+1)^[((8!)/(7!))/2]
EDIT 3: 100,000,000,000 = (5^6)*(4^3)*{(9+1)^(7-[sqrt(8)/sqrt(2)])}
EDIT 4: 1,000,000,000,000 = (5^6)*(4^3)*{(9+1)^[((8!)/(7!))-2]}
EDIT 5: 10,000,000,000,000 = (5^6)*(4^3)*(9+1)^((7!)/((8-2)!))
EDIT 6: 100,000,000,000,000 = (5^6)*(4^3)*(8+2)^((9-1)!/(7!))
EDIT 7: 1,000,000,000,000,000 = (5^6)*(4^3)*(9+1)^(sqrt(8)/sqrt(2)+7)

EDIT: Thanks for the 5^6 trick Sax! ^_^
EDIT: I have too much time on my hands right now ._.

[iCeCuBEz v2]

equations are a tool overall used as a method to find a particular answer.....

Here's an example I had for a question on my "calc III" test.... too vague for me to answer I wanna see what you guys come up with....

Find a symmetrical graph going through the points P(1,2,4) and P(3,4,5

how would YOU guys find the "answer" to "that question"

(specifically it was a test written on paper so I'd have to answer with respect to a "cartesian coordinate plane") (x,y)

[qqwref]

What does it mean by "symmetrical"? I mean, I'd consider planes and lines to have symmetry, so any plane going through those two points (e.g. z=(x+7)/2) seems to work.

PS: Here's a fun one:
123456789 = ((7+3)^(8+sqrt(4))-1)/(9^2)-6+5

[iCeCuBEz v2]

yah ok you're confused as well..... I could've made any "graph" I wanted but it was wayyyy to general with its definition of "symmetrical"

It could've had translational symmetry...... symmetry with respect to x-axis..... with respect to y-axis..... rotational symmetry...... across that particular plane in which those points intersect......

I really had no ****ing idea how to answer this question......

If I made a circle on a cartesian coordinate plane.... how would I figure out the radius.....???

I "could" take those two points and create a function that could be written as "symmetrical" across the y-axis.... which would appear as a quadratic function or maybe even a quartic function or something asinine like that..... but was I given enough "information" to conceptualize this problem to a particular formula (or function as what it was defined as in this question.....) this little information could be manipulated into so many "cartesian functions"....... ugh

Wait: I think I got it now..... If I find the distance between those 2 points in the 3 dimensional coordinate plane..... that new found line will create a plane that it is sitting upon.... so I can just use distance formula between those 2 points and then THAT is the radius??? (if i want to graph a circle on a cartesian coordinate plane)

[qqwref]

If you create a circle (or sphere!) so that the two points are exactly opposite each other on the curve, the radius will be half of the distance between those points (and the center will be the midpoint of the two points). This is getting a bit off-topic, though