For those who are morbidly curious in the end if there's other symbol that satisfy diamond: Let # be diamond. Note that (a#b)*b=a#(b#b)=a#1=a, so a#b=a/b. Therefore, # is division.
I have a more simple solution here :) let's say the diamond notation is @. From a@a=1, it can be either (a * 1/a) or (a - a), and we can predict that a@b = a*1/b or a - b However, a@(b@c) = (a@b)*c implies that a@a is (a*1/a) because there is multiplication of c, not addition or subtraction. If we calculate a@(b@c) with a@b = a*1/b, it works. With simple calculation with 2016@(6@x)=100, we can solve the problem. Also, if you just think, the problem-producer(?) will not make the notation too complicated because then he should add more limitations to the problems exponentially.
Let diamond represent any equation and what do you have? More complex algebra or something else by another name? I do like the idea of just plugging in a simple variable, 1 in this case, to at least start somewhere...
Alternatively, he could've factored a 6 out of 2016#6. (#=diamond) Then it turns into 6(336#1)=600/x. Divide by 6 and you get 336=100/x, which simplifies into x=100/336, which is 25/84.
That's wrong. We know diamond is division, so if you had like 2#2, that 2/2 = 1. But you're assuming you can do 2#2 = 2(1#1), which would equal 2, not 1. You made an assumption that was not stated and could not be concluded from the given properties.
Srsly, a 5 minute video on this joke problem? Anyone could see that diamond is a binary operations so there aren't that many choices. The obvious one is division and it satisfies the first premise so it's division. This is a 20 second problem... And also what happened to you Richard ;_;.
I loooove the way Deven does this - making the correct solution so intuitively acceptable (and also replicable in future problems).
For those who are morbidly curious in the end if there's other symbol that satisfy diamond:
Let # be diamond. Note that (a#b)*b=a#(b#b)=a#1=a, so a#b=a/b. Therefore, # is division.
He mentions that it's division in the video...
u changed a lot Richard
(a joke)
I have a more simple solution here :)
let's say the diamond notation is @.
From a@a=1, it can be either (a * 1/a) or (a - a), and we can predict that a@b = a*1/b or a - b
However, a@(b@c) = (a@b)*c implies that a@a is (a*1/a) because there is multiplication of c, not addition or subtraction.
If we calculate a@(b@c) with a@b = a*1/b, it works.
With simple calculation with 2016@(6@x)=100, we can solve the problem.
Also, if you just think, the problem-producer(?) will not make the notation too complicated because then he should add more limitations to the problems exponentially.
Let diamond represent any equation and what do you have? More complex algebra or something else by another name? I do like the idea of just plugging in a simple variable, 1 in this case, to at least start somewhere...
WHERE IS RICHARD RUSCEHYTPT%ROG
Alternatively, he could've factored a 6 out of 2016#6. (#=diamond) Then it turns into 6(336#1)=600/x. Divide by 6 and you get
336=100/x, which simplifies into x=100/336, which is 25/84.
You can't assume that you can factor like that.
That's wrong. We know diamond is division, so if you had like 2#2, that 2/2 = 1. But you're assuming you can do 2#2 = 2(1#1), which would equal 2, not 1.
You made an assumption that was not stated and could not be concluded from the given properties.
this is very misplaced for a #23
Srsly, a 5 minute video on this joke problem? Anyone could see that diamond is a binary operations so there aren't that many choices. The obvious one is division and it satisfies the first premise so it's division. This is a 20 second problem... And also what happened to you Richard ;_;.
Did you not even listen to him when he explained what a binary operation is, or are you just that stupid