Common Multiple from a Cubic (Posted on 2016-05-01)

	Submitted by Brian Smith
Rating: 4.0000 (1 votes)
Solution:	(Hide)
My solution is below. Harry provides an alternate method here. Define P(x) = x^3 + a*x^2 + b*x + c Let n=0. Then F(0) = c. Which implies c = 0 (mod m). Let n=-1. Then F(-1) = a-b+c-1. Which implies a = b+1 (mod m). Let n=1. Then F(1) = a+b+c+1. Which implies a = -b-1 (mod m). Combining these results implies 2a = 0 (mod m) and 2b = -2 (mod m). Now let n=2. Then F(2) = 8 + 4a + 2b + c. Combined with the above results yields 6 = 0 (mod m). m must be a factor of 6. The largest factor of 6 is itself. One specific polynomial F(x) = x^3 + 6x^2 + 11x + 6 = (x+1)*(x+2)*(x+3) satisfies the requirements stated in the problem. Therefore the largest m is 6.

	Subject	Author	Date
	Puzzle Thoughts	K Sengupta	2022-12-27 01:25:23
	Difference in approach (spoiler)	Harry	2016-05-02 18:36:32
	possible solution	Ady TZIDON	2016-05-01 14:14:14
	Start	armando	2016-05-01 11:57:48