> Well this is a bizarre 4th degree polynomial equation, \(x^4-2x^2 -400x=9999\) and he applies one approach to solving it, then comes to a dead end and says "Hence ingenuity is called for" and finally finds that his unknown number is 11. I, personally, would not have had a clue how to solve it.
Here is what you can do. If there is an integer solution, we can keep the constant term 9999 on the opposite side and factor the formula.
As in:
x^4 - 2x^ - 400x = 9999
x(x^3 - 2x - 400) = 9999
Ok, so no we have x f(x) = 9999. From this we know that if there is an integer solution for x, x itself must be a factor of 9999: either a prime factor, or a product of factors.
So we just factor 9999:
3^2 11 101 = 9999
9999 has 3 as a degree 2 factor, and then also 11, and 101.
Looking at just the factorization breakdown alone, x could be the product of these combinations of factors: { (3), (11), (101), (3, 3), (3, 11), (3, 101), (11, 101), (3, 3, 11), (3, 3, 101), (3, 11, 101) }.
If we substitute these factors for x, in order from least to greatest, we will soon hit upon the solution.
Intuitively we can guess that x is small, because the opposite factor is cubing it. So for instance, it can't be the case that x is the product of 3, 11, and 101, such that x^3 - 2x - 400 then works out to the remaining factor of 3.
It's obvious that x can't be too small, like 3, because x^3 is only 27; that's not going to leave us a positive factor when we subtract 400 alone, let alone -2x.
Here is what you can do. If there is an integer solution, we can keep the constant term 9999 on the opposite side and factor the formula.
As in:
Ok, so no we have x f(x) = 9999. From this we know that if there is an integer solution for x, x itself must be a factor of 9999: either a prime factor, or a product of factors.So we just factor 9999:
9999 has 3 as a degree 2 factor, and then also 11, and 101.Looking at just the factorization breakdown alone, x could be the product of these combinations of factors: { (3), (11), (101), (3, 3), (3, 11), (3, 101), (11, 101), (3, 3, 11), (3, 3, 101), (3, 11, 101) }.
If we substitute these factors for x, in order from least to greatest, we will soon hit upon the solution.
Intuitively we can guess that x is small, because the opposite factor is cubing it. So for instance, it can't be the case that x is the product of 3, 11, and 101, such that x^3 - 2x - 400 then works out to the remaining factor of 3.
It's obvious that x can't be too small, like 3, because x^3 is only 27; that's not going to leave us a positive factor when we subtract 400 alone, let alone -2x.