Answer this
You call this a mathematical riddle...?
This is a mathematical riddle.
Let (an) be a sequence defined as follows: a1 is a positive integer and an = int((3/2) × a(n-1)) + 1, where int = rounding down to the next smallest integer.
Is it possible to choose a1 such that a100001 is odd and an is even for n < 100001 ?
This is a mathematical riddle.
Let (an) be a sequence defined as follows: a1 is a positive integer and an = int((3/2) × a(n-1)) + 1, where int = rounding down to the next smallest integer.
Is it possible to choose a1 such that a100001 is odd and an is even for n < 100001 ?
Originally Posted by TinyWeeWee
Well, most people make the mistake of thinking $10 is the game price.
The game costs $10 + .5x
Since x is also the price of the game, 10 + .5x must equal x.
So, 2(10 + .5x) = x(2)
20 + x = 2x
x = 20
