# Download 102 Combinatorial problems from the training of USA IMO team by Andreescu T., Feng Z. PDF

By Andreescu T., Feng Z.

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Additional resources for 102 Combinatorial problems from the training of USA IMO team

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9. 48 Discrete Mathematical Adventures The same reasoning used for the dollar-changing theorem leads to the following general result. Three denominations theorem. With coins of three denominations d1 , d2 , and 1, the number of ways to make d1 d2 N units of money is d1 d2 d1 + d2 + gcd(d1 , d2 ) N2 + N +1 2 2 for N = 1, 2, . . We will see ways to answer questions involving four denominations later. As a preview, try your hand at the following problem. Challenge 2. How many ways are there to make change for a dollar if quarters, dimes, nickels, and pennies are available?

4) The three inequalities determine a lattice right triangle with vertices (0, 0), (4D, 0), and (0, 10D). Each lattice point in the triangle corresponds to a way to make change for D dollars. Apply Proposition 1 with a = 4D and b = 10D to see that the number of lattice points is L= (4D + 1)(10D + 1) + gcd(4D, 10D) + 1 . 2 Because gcd(4D, 10D) = 2D, we have L = 20D 2 + 8D + 1. The lattice points in the interior of the triangle correspond to ways to make change for D dollars using at least one coin of each of the three denominations.

As before, we ignore the nickels and focus on the pairs (q, d) satisfying the inequalities 25q + 10d ≤ 100, q ≥ 0, d ≥ 0. 5. Consider the integer pairs in1 the triangle—points (q, d) whose coordinates q and d are both integers. Each integer 1 When we refer to points “in” a polygon, we include the boundary as well as the interior. 5: Integer pairs in a triangle and change for a dollar pair in the shaded triangle corresponds to a way to make change for a dollar. For instance, (q, d) = (2, 3) corresponds to two quarters, three dimes, and four nickels.