michaelp Wrote:

Sparker Wrote:OH SHI-

Oh right it's fixed now

60 + 20 + 10 ≠ 100

Quote:u have to buy 100 among all the objects with exactly 100 epigs n u have to buy atleast 1 object of each category

Whoops almost forgot something.

Quote:object c costs 10 epigs

Sparker Wrote:60 C's = 60

Almost got it. :P
4 C's = 40
5 B's = 15
90 A's = 45

99 Objects, 100 e-pigs. Soooo close!

OH SHI-
I really suck at this.

Be glad Assassinator hasn't seen this.

0.5a + 3b + 10c = 100
a + b + c = 100

So here wee go:
c = 100-a-b
0.5a + 3b + 10(100-a-b) = 100
0.5a + 3b + 1000 -10a - 10b = 100
-9.5a - 7b = -900
19a + 14b = 1800

Now, a and b must be integers.  The above is an example of a linear Diophantine equation.
...I just happen to have forgotten how to solve them.
If I can be bothered, I'd look back through my books on how... >_>

zinga i don think u can solve this type of problem in that way
lol maybe i would give the answer after a couple of days if no one figures it out by then

(This post was last modified: 24/01/2008 10:45 AM by u_c_taker.)

umm i already did as did a few others at the start of the thread :/
perhaps you didnt explain it correctly?

ill wait 1 more day that is till tommorow
if no one finds it out then i will reveal the answer

ZiNgA BuRgA Wrote:Be glad Assassinator hasn't seen this.

0.5a + 3b + 10c = 100
a + b + c = 100

So here wee go:
c = 100-a-b
0.5a + 3b + 10(100-a-b) = 100
0.5a + 3b + 1000 -10a - 10b = 100
-9.5a - 7b = -900
19a + 14b = 1800

Now, a and b must be integers.  The above is an example of a linear Diophantine equation.
...I just happen to have forgotten how to solve them.
If I can be bothered, I'd look back through my books on how... >_>

a+b+c=100
10a+3b+0.5c=100

c=100-a-b

10a+3b+0.5(100-a-b)=100
9.5a+2.5b+50=100
19a+5b=100

100-19=5b so b is fraction
100-38=5b so b is fraction
100-57=5b so b is fraction
100-76=5b so b is fraction
100-95=5b so b=1!!!

SomethingX Wrote:94xA, 1xB, and 5xC

Was what I meant.