APMO

Question

Let n be an integer such that n>3. Suppose that we choose three numbers from the set 1,2, , n. Using each of these three numbers only once and using addition, multiplication, and parenthesis, let us form all possible combinations. a. Show that if we choose all three numbers greater than n / 2, then the values of these combinations are all distinct. b. Let p be a prime number such that p n. Show that the number of ways of choosing three numbers so that the smallest one is p and the values of the combinations are not all distinct is precisely the number of positive divisors of p-1.

Accepted Answer

In both items, the smallest chosen number is at least 2: in part (a), n / 2 > 1 and in part (b), p is a prime. So let 1 < x < y < z be the chosen numbers. Then all possible combinations are x + y + z, x + y z, x y + z, y + z x, (x + y) z, (z + x) y, (x + y) z, x y z. Since, for 1 < m < n and t > 1, (m-1)(n-1) 1 2 m n > m + n, t n + m - (t m + n) = (t-1)(n-m) > 0 t n + m > t m + n, and (t + m) n - (t + n) m = t(n-m) > 0, x + y + z < z + x y < y + z x < x + y z and (y + z) x < (x + z) y < (x + y) z < x y z. Also, (y + z) x - (y + z x) = (x - 1) y > 0 (y + z) x > y + z x and (x + z) y - (x + y z) = (y - 1) x > 0 (x + z) y > x + y z. Therefore the only numbers that can be equal are x + y z and (y + z) x. In this case, x + y z = (y + z) x (y - x)(z - x) = x(x - 1). Now we can solve the items.…