Posted by T.Gracken on October 25, 2002 at 17:46:22:
In Reply to: Re: equation: question, Mr G posted by Denis Borris on October 25, 2002 at 13:46:45:
: I follow that; a = -1/3 and b = 8 1/3
: BUT:
: if x,a and b are restricted to integer range 1 to 99, there are 1806 solutions:
: (x,a,b in increasing x order)
: 1: 8,1,49
: 2: 8,2,27
: 3: 8,2,97
: ...(did you think I was gonna type 'em all?!)
: 1805: 98,66,39
: 1806: 99,33,73
: Where, oh where have I screwed up, praytell.
I'm not sure I follow what you are doing. In fact, I have no idea what the triples you have listed represent.
the list of values you have for x, a, and b do not satisfy the given conditions as far as I can tell.
The manner in which I interpretted the problem:
we are to find two values (one for a and one for b), so that the expression ax^3-x^2-x+b has a remainder of 6 when divided by (x-1) [for any value of x (except 1)] AND the expression ax^3-x^2-x+b has a remainder of 9 when divided by (x+2) [for any value of x (except -2)]
The only values (for a and b) that will satisfy these two conditions will be a = -1/3 and b = 25/3 (as you stated above)
...of course I may not be seeing something but i am of the mind that you are trying to solve an equation that has no bearing on the stated conditions. That is, we are asked to discover an expression (a number) that we want specific remainders for after dividing by other given numbers (expressions).
...and also, of course, I may just be interpretting the stated problem in an incorrect (or unassumed) manner.
I'm sure you'll let me know if I have "lost it" yet again.