This week, we will relax a little bit with a quite easy problem AQQ273:
Good luck and try to be among the first ones!
This week, we will relax a little bit with a quite easy problem AQQ273:
Good luck and try to be among the first ones!
1/2.
Indeed: a/a = 2a, implies a is not 0 (we can't divide by 0), leaving: a = 2a^2, or 1=2a.
This quizz has a lot of initial challenges.
The visual cues above the question itself hints to a division of dice which leads to a double six: statistics here?
Then the question itself is laced with brackets, and a minus sign, so maybe I need to do some subtraction as well ;-).
But removing the expressions in the brackets, the question is: which number divided by itself equals 2x itself, maybe there is more than 1 answer.
So the math is:
- for which x:
I was inclined to also nominate -1/2, but the sign turns out wrong.
Under modular arithmetic (residue modulo) there is a unique solution to 1 = 2x (mod y) -- Gauss' notation
that is: x = (y+1)/2
So, since there is an infinite number of odd numbers, there is an infinite number of solutions, one for each (mod y), y being a odd integer.
y=3, x = (3+1)/2 = 2. Indeed 2x = 4 = 1 (mod 3) since 4 residue modulo 3 get a rest of 1.
y=5, x = (5+1)/2 = 3. Indeed 2x = 6 = 1 (mod 5),
etc. They all satisfy 2x = 1 (mod y)
You can write this as x/x = 2x and multiplying both sides by x arrive at x = 2x^2
Rearranging, we get 2x^2 – x = 0 then factoring we get x(2x-1) = 0
For the equation to be zero, either factor must be zero
So, either x = 0 or 2x-1=0 => x=0.5, so the number is either 0 or 0.5
Anything divided by zero is undefined, so strictly-speaking the only solution is 0.5.
Bingo vanderghast , you get it, congratulations (too easy for you...)!
Wow, I learn something... I was not aware about the modular arithmetic method...