Maths Olympiad Prep

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Algebra Difficulty 5.4 AIME, harder Prove it United States

Problem:

Several positive integers are given, not necessarily all different. Their sum is 20032003. Suppose that n1n_{1} of the given numbers are equal to 11, n2n_{2} of them are equal to 22, …\ldots, n2003n_{2003} of them are equal to 20032003. Find the largest possible value of
n2+2n3+3n4+⋯+2002n2003 n_{2}+2 n_{3}+3 n_{4}+\cdots+2002 n_{2003}

Solution

Solution:

The sum of all the numbers is n1+2n2+⋯+2003n2003n_{1}+2 n_{2}+\cdots+2003 n_{2003}, while the number of numbers is n1+n2+⋯+n2003n_{1}+n_{2}+\cdots+n_{2003}. Hence, the desired quantity equals
(n1+2n2+⋯+2003n2003)−(n1+n2+⋯+n2003)=(sum of the numbers)−(number of numbers) \begin{gathered} \left(n_{1}+2 n_{2}+\cdots+2003 n_{2003}\right)-\left(n_{1}+n_{2}+\cdots+n_{2003}\right) \\ =(\text{sum of the numbers})-(\text{number of numbers}) \end{gathered}
=2003−(number of numbers), =2003-\text{(number of numbers)},
which is maximized when the number of numbers is minimized. Hence, we should have just one number, equal to 20032003, and then the specified sum is 2003−1=20022003-1=2002.

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