Really Hard Math Problem?

[ikljo]

I have a hard math problem for everyone to try and figure out. It is very tricky, but after seeing #M|A#Wolf's thread, it appears as if we have a few math gods here. So maybe you might be able to figure it out. So here it goes...

The problem:

I've made 100 scripts which I've sold for exactly $100. I've sold python scripts for $5.00 each, javascripts for $2.00, and cube scripts for 10 cents.
How many of each type of script did I make?

Additional Details: I made a total of 100 scripts, no more, no less.


Anyone who solves this problem is a math god!

This problem can be solved

[#M|A#Wolf]

Not enough info.

[GDM]

70 cubescript, 19 javascript, 11 python scripts.

Wolf - the information you were missing was p >= 0, j >= 0, c >= 0

[Oracle]

11 Python Scripts, 19 Javascripts, 70 cubescripts.

Edit: Curses, GDM ;)
(HyPE helping HyPE xiters)

How to do it:

Say X=Python, Y=Java, Z=Cube

X + Y + Z = 100
5X + 2Y + 0.1Z = 100

X, Y, Z must be whole numbers
Z has to be a multiple of 10, since the other two will give only whole numbers (5x, 2y)

So I started from top to bottom, plugging in values of Z - can't be 100, 90, 80

Plug in 70
X + Y = 30
5X + 2Y = 93

Multiply the first equation by -2

3X = 33
X = 11, plugging in X in either one gives Y = 19

Hence 11, 19, 70

[ikljo]

Set up your equations:
P = # of python scripts
J = # of javascripts
C = # of Cubescripts

P + J + C = 100
5P + 2J + C/10 = 100

multiply the 2nd equation by 10 and subtract the two equations from each other

10(5P + 2J + C/10 = 100)
=> 50P +20J + 10C/10 = 1000
Subtract: P + J + C = 100
==> 49P + 19J = 900

Now, there are lots of ways to do that, but you need both P and J to be integers. Notice that 900 is divisible by 10. Divide the whole equation by 10
$4.9 P + $1.9 J = $90

Now, you know that I need to make exactly $90 selling just python scripts and javascripts. Each time I sell a script, I have to give a dime in change. To get back to an exact dollar figure, I will need to sell some multiple of 10 scripts. So, make the substitution

P + J = 10 n

Now, solve for P and substitute
49 n - 3 J = 90

Remembering that n and J are both integers, n must be a multiple of 3 (divide that equation by 3 and see). So try, n=3, 6, 9, etc.
For n=3, J = 19
For n=6, J = 68
For n=9, J = 117 (which is impossible.)

If J = 19, then P = 11 and C = 70.
If J = 68, then P = -8 which is impossible.

So, the only possible answer is 11 python scripts, 19 javascripts and 70 cubescripts.

[Takkunen]

wuts 1+1=?

[Orynge]

...

Window.

[Fate]

1. Ben and Shaun each put $20 into a bag.
2. Shaun then "buys" the bag with the money in it for $30 from Ben.
3. ????
4. Profit?

[MykeGregory]

1. Ben and Shaun each put $20 into a bag.
2. Shaun then "buys" the bag with the money in it for $30 from Ben.
3. ????
4. Profit?

Profit? Yes! Both parties receive $10 profit. BOOM!
http://forum.cubers.net/thread-2305.html

[PolarHedgehog]

Huehuehue

[#M|A#Wolf]

EINS.

[Fate]

1. Ben and Shaun each put $20 into a bag.
2. Shaun then "buys" the bag with the money in it for $30 from Ben.
3. ????
4. Profit?

Profit? Yes! Both parties receive $10 profit. BOOM!
http://forum.cubers.net/thread-2305.html

(╯°□°）╯︵ ┻━┻

[MykeGregory]

(╯°□°）╯︵ ┻━┻

^^ Indeed.

[Aekom]

Huehuehue

That's x-3/x-1 when finding common factors.

[#M|A#Wolf]

Huehuehue

That's x-3/x-1 when finding common factors.

The answer's 1.

[Oracle]

^Nope, Aekom is indeed correct.

[PolarHedgehog]

Party A received 36% of the votes in the first elections. The voting percentage was 85%. In the second elections, party A received 38% of the votes, and the voting percentage was 80%. How many percent more or less votes did party A get in the second elections than in the first? The number of voters did not change between the elections.

ps 9th grade ftw.

[1Cap]

lol----
Want to find out how many steps are visible in an escalator. To this was done the following: two people began climbing an escalator together, one going one step at a time while the other climbed two . On top the first person counted 21 steps while the other 28. With this data it was possible to answer the question. How many steps are visible on that escalator? (Note: escalator is operating).
[/u]

[GDM]

lol----
Want to find out how many steps are visible in an escalator. To this was done the following: two people began climbing an escalator together, one going one step at a time while the other climbed two . On top the first person counted 21 steps while the other 28. With this data it was possible to answer the question. How many steps are visible on that escalator? (Note: escalator is operating).
[/u]

The person who was skipping every other step took more total steps than the person who wasn't? Were they going up a down escalator?

[Flames]

k here's an easy question .
How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that odd digits occupy even position ?

[Oracle]

I think I did this wrong (I kinda forget the permutations/combinations w/ repeats) but...
(5! x 4!)/(2! x 2! x 2! x 3!) = 60 numbers?

[Flames]

yep

One more question :

Let ABC be a triangle such that angle ACB = Pi/6 [ :D no idea how to insert the image] and let a,b and c denote the lengths of the sides opposite to A,B and C respectively. The value of x for which a= x^2+x+1,b=x^2-1 and c=2x+1 is ..

[Oracle]

The value of x for which a= x^2+x+1=x^2-1 and c=2x+1 is ..

Was there supposed to be a "b=x^2-1"?

[Flames]

The value of x for which a= x^2+x+1=x^2-1 and c=2x+1 is ..

Was there supposed to be a "b=x^2-1"?

Oops fixed now

[Roflcopter]

Let ABC be a triangle such that angle ACB = Pi/6 [ :D no idea how to insert the image] and let a,b and c denote the lengths of the sides opposite to A,B and C respectively. The value of x for which a= x^2+x+1,b=x^2-1 and c=2x+1 is ..

Hint: use the cosine law rearranged into cos C = (a^2+b^2-c^2)/(2ab). We know angle C and we know a, b and c. This will give us a quadratic equation in R[x] with only one viable root (the other is negative).

[GDM]

Let ABC be a triangle such that angle ACB = Pi/6 [ :D no idea how to insert the image] and let a,b and c denote the lengths of the sides opposite to A,B and C respectively. The value of x for which a= x^2+x+1,b=x^2-1 and c=2x+1 is ..

Hint: use the cosine law rearranged into cos C = (a^2+b^2-c^2)/(2ab). We know angle C and we know a, b and c. This will give us a quadratic equation in R[x] with only one viable root (the other is negative).

Yeah. No one is bothering because this isn't a "really hard" or even "hard" math problem. It's an obvious use of a common formula.
Also: plugged into wolfram

[Roflcopter]

What are the last 10 digits of tet(2337, 2339)?

You're given that tet(a,0) = 1 and that tet(a,b+1) = a^tet(a,b) for a and b non-negative integers.
See Wikipedia on tetration.

[Marti]

the answer always is:

[Aekom]

What are the last 10 digits of tet(2337, 2339)?

You're given that tet(a,0) = 1 and that tet(a,b+1) = a^tet(a,b) for a and b non-negative integers.
See Wikipedia on tetration.

I'm not going to pretend I know anything about Tetration, but I thought I'd have a go anyway. As far as I understand, Tetration is iterated powers (powers of a power). I'm assuming that tet(a,0) = 1 is another notation style for (please correct me if I'm just making stuff up here).

I tried to give it a go through WolframAlpha using the information from this thread:

http://community.wolframalpha.com/viewto...20&t=73980

and it came back with "indeterminate": http://www.wolframalpha.com/input/?i=+Po...2C+{2339}]

[ExodusS]

1. Ben and Shaun each put $20 into a bag.
2. Shaun then "buys" the bag with the money in it for $30 from Ben.
3. ????
4. Profit?

Profit? Yes! Both parties receive $10 profit. BOOM!
http://forum.cubers.net/thread-2305.html

SCAM