What's going on here? ...

[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]

Posted by Joel on October 25, 2002 at 19:20:37:

In Reply to: I think this one has the correct tags (skip one below please) posted by T.Gracken on October 25, 2002 at 17:46:22:

: : 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.

1: 512-64-8+49=489 489/7=483 R 6 489/10=48 R 9
2: 1024-64-8+27=979 979/7=973 R 6 979/10=97 R 9
3: 1024-64-8+97=1049 1049/7=149 R 6 1049/10=104 R 9

Mr. B's triples are sets of values for x,a and b which do satisfy the stated relation, but in each case you have to use the entire set of 3 numbers, so that is not a general solution for ALL x.

Mr. G's solution is the pair {a,b} that seems as if it should satisfy the stated relation for ALL values of x, and it seems to work if you leave it in terms of x:
(-1/3)x^3 -x^2 -x + 25/3
and divide by (x-1), the remainder is 6
or divide by (x+2), the remainder is 9

But when I try plugging in arbitrary values of x, it doesn't seem to work, unless I'm screwing something up.
For example, let x=5:
-125/3 - 25 - 5 + 25/3 = -190/3
How do you divide that by 4 to get a remainder of 6 or by 7 to get a remainder of 9?

Follow Ups:

you are attempting to look at a specific case... this is general case T.Gracken 21:12:01 10/25/02 (7)
- Re: you are attempting to look at a specific case... this is general case Joel 22:37:50 10/25/02 (6)
  - how about this...? T.Gracken 07:54:19 10/26/02 (1)
    - Re: how about this...? Joel 11:29:59 10/26/02 (0)
  - well, ....... Denis Borris 02:21:39 10/26/02 (3)
    - This, I like. nt Joel 11:31:03 10/26/02 (2)
      - nomenclature question Denis Borris 17:47:51 10/26/02 (1)
        
        Re: nomenclature question T.Gracken 07:34:03 10/27/02 (0)

Post a Followup

Name:
E-Mail:

Subject:

Comments:
: : : 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 <b>given</b> 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. : 1: 512-64-8+49=489 489/7=483 R 6 489/10=48 R 9 : 2: 1024-64-8+27=979 979/7=973 R 6 979/10=97 R 9 : 3: 1024-64-8+97=1049 1049/7=149 R 6 1049/10=104 R 9 : : Mr. B's triples are sets of values for x,a and b which do satisfy the stated relation, but in each case you have to use the entire set of 3 numbers, so that is not a general solution for ALL x. : Mr. G's solution is the pair {a,b} that seems as if it should satisfy the stated relation for ALL values of x, and it seems to work if you leave it in terms of x: : (-1/3)x^3 -x^2 -x + 25/3 : and divide by (x-1), the remainder is 6 : or divide by (x+2), the remainder is 9 : But when I try plugging in arbitrary values of x, it doesn't seem to work, unless I'm screwing something up. : For example, let x=5: : -125/3 - 25 - 5 + 25/3 = -190/3 : How do you divide that by 4 to get a remainder of 6 or by 7 to get a remainder of 9?

Optional Link URL:
Link Title:
Optional Image URL:

[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]