Can you solve this?

+ - =2
- = 1
- + =0

Nope

= 1
= 1
= 0

coast2coastam wrote: = 1
= 1
= 0

correctomundo

Is there going to be math quizzes everyday or only on certain days...

Ask for a fren

once we reach 100 members, captcha will be a math problem

Here's a brain-teaser for you guys.

A pot contains 75 white beans and 150 black ones. Next to the pot is a large pile of black beans.

A miscer with too much time on his hands removes the beans from the pot according to the following autistic rule: He removes two beans from the pot at random. If at least one of the beans is black, he places it on the bean-pile and drops the other bean, no matter which color, back in the pot. If both beans are white, on the other hand, he discards both of them and removes one black bean from the pile and drops it in the pot.

At each turn of this procedure, the pot has one less bean in it. Eventually, just one bean is left in the pot. What color is it, and why?

NEETvestor wrote:Here's a brain-teaser for you guys.

A pot contains 75 white beans and 150 black ones. Next to the pot is a large pile of black beans.

A miscer with too much time on his hands removes the beans from the pot according to the following autistic rule: He removes two beans from the pot at random. If at least one of the beans is black, he places it on the bean-pile and drops the other bean, no matter which color, back in the pot. If both beans are white, on the other hand, he discards both of them and removes one black bean from the pile and drops it in the pot.

At each turn of this procedure, the pot has one less bean in it. Eventually, just one bean is left in the pot. What color is it, and why?

the last bean in the pot is probably to be a black bean.

NEETvestor wrote:Here's a brain-teaser for you guys.

A pot contains 75 white beans and 150 black ones. Next to the pot is a large pile of black beans.

A miscer with too much time on his hands removes the beans from the pot according to the following autistic rule: He removes two beans from the pot at random. If at least one of the beans is black, he places it on the bean-pile and drops the other bean, no matter which color, back in the pot. If both beans are white, on the other hand, he discards both of them and removes one black bean from the pile and drops it in the pot.

At each turn of this procedure, the pot has one less bean in it. Eventually, just one bean is left in the pot. What color is it, and why?

Suppose you get to the point where you reach 2 black beans and 2 white beans (drawing at least 1 black bean decreases the number of black beans, drawing 2 white beans increases the number of black beans while decreasing the number of white beans; cbf writing out the induction). Drawing 2 white beans would result in a pot of only black beans, with no chance of the population of white beans ever recovering. Drawing any quantity of black beans would decrease the quantity of one bean while maintaining that at least one black bean stays in the pot.

Eh, that was hard to think about, but I guess the solution is fairly simple in the end:
next step                          =   2 white beans drawn          or  (there is more than 1 black bean and 2 black beans are drawn) or  1 black and 1 white are drawn
black_bean_quantity[i+1] = (black_bean_quantity[i] + 1) V (black_bean_quantity[i] > 1 ^ black_bean_quantity[i] - 1)  V black_bean_quantity[i]


The black beans therefore must always have a minimum quantity of 1, but this minimum does not hold for the white beans.


E: idk how tf to rep, but that was fun, thanks brah

The answer is white. There's an odd number of white beans, and you can only remove them two at a time, so there has to be one white bean left over.


@Legbroke I like the way you approached it but it's really more of a trick question than a true logic puzzle