THE 2015 AUSTRALIAN MATHEMATICAL OLYMPIAD SOLUTIONS
... (m, m + 1) for m = 1, 3, 5, . . . , 2013. Furthermore, we have p(2015) = a(2015) − b(2015) = −b(2015) < 0 and p(x) → +∞ as x → +∞. So the intermediate value theorem asserts that there is also at least one root of p(x) on the interval (2015, ∞). So we have found 1008 disjoint intervals on which p(x) ...
... (m, m + 1) for m = 1, 3, 5, . . . , 2013. Furthermore, we have p(2015) = a(2015) − b(2015) = −b(2015) < 0 and p(x) → +∞ as x → +∞. So the intermediate value theorem asserts that there is also at least one root of p(x) on the interval (2015, ∞). So we have found 1008 disjoint intervals on which p(x) ...
Cayley 15
... C4. Ten balls, each coloured green, red or blue, are placed in a bag. Ten more balls, each coloured green, red or blue, are placed in a second bag. In one of the bags there are at least seven blue balls and in the other bag there are at least four red balls. Overall there are half as many green bal ...
... C4. Ten balls, each coloured green, red or blue, are placed in a bag. Ten more balls, each coloured green, red or blue, are placed in a second bag. In one of the bags there are at least seven blue balls and in the other bag there are at least four red balls. Overall there are half as many green bal ...
2007 Grades 9-10 Solutions English
... Solution: Denote by D1, D2, D3 and D4 the digits of the number. Since D1 is the number of zeros and we have a four-digit number, D1 has to be either 1, or 2, or 3. Case 1: D1=3, so the number must contain three zeroes, but the number 3000 does not satisfy the requirements (the number of the digits 3 ...
... Solution: Denote by D1, D2, D3 and D4 the digits of the number. Since D1 is the number of zeros and we have a four-digit number, D1 has to be either 1, or 2, or 3. Case 1: D1=3, so the number must contain three zeroes, but the number 3000 does not satisfy the requirements (the number of the digits 3 ...
Weber problem
In geometry, the Weber problem, named after Alfred Weber, is one of the most famous problems in location theory. It requires finding a point in the plane that minimizes the sum of the transportation costs from this point to n destination points, where different destination points are associated with different costs per unit distance.The Weber problem generalizes the geometric median, which assumes transportation costs per unit distance are the same for all destination points, and the problem of computing the Fermat point, the geometric median of three points. For this reason it is sometimes called the Fermat–Weber problem, although the same name has also been used for the unweighted geometric median problem. The Weber problem is in turn generalized by the attraction–repulsion problem, which allows some of the costs to be negative, so that greater distance from some points is better.