1st Brazil 1979 Problem 2 The remainder on dividing the polynomial p(x) by x 2 - (a+b)x + ab (where a and b are unequal) is mx + n. Find the coefficients m, n in terms of a, b. Find m, n for the case p(x) = x 200 divided by x 2 - x - 2 and show that they are integral.
m = (2 200 -1)/3, n = (2 200 +2)/3. Solution p(x) = q(x)(x-a)(x-b) + mx + n. So putting x = a, b we get p(a) = ma + n, p(b) = mb + n. Solving, m = (p(a)-p(b))/(a-b), n = (p(b)a-p(a)b)/(a-b). In the case given a = 2, b = -1, so m = (2 200 -1)/3, n = (2 200 +2)/3. Note that 2 = -1 mod 3, so 2 200 = 1 mod 3 and hence 2 200 - 1 is a multiple of 3, so m is integral. n = m+1, so n is also integral. 1st Brazil 1979 © John Scholes jscholes@kalva.demon.co.uk 23 Oct 2003 Last corrected/updated 23 Oct 03