2006 SMT/General Problems/Problem 8

Problem 8

Given two $2$ 's, "plus" can be changed to "times" without changing the result: $2+2=2\cdot2$ . The solution with three numbers is easy too: $1+2+3=1\cdot2\cdot3$ . There are three answers for the five-number case. Which five numbers with this property has the largest sum?

Solution

Assuming all the numbers are positive, lets denote the five numbers as $a_0,a_1,a_2,a_3,a_4$ . Assume WLOG that $a_0 \ge a_1 \ge a_2 \ge a_3 \ge a_4$ . The condition we are looking for is $a_0+a_1+a_2+a_3+a_4 = a_0 \cdot a_1 \cdot a_2\cdot a_3 \cdot a_4$ .

Manipulating the LHS, we see that: $\begin{align*} a_0+a_1+a_2+a_3+a_4 &= a_0 \cdot a_1 \cdot a_2\cdot a_3 \cdot a_4\\ 5a_0 &\ge a_0 \cdot a_1 \cdot a_2\cdot a_3 \cdot a_4\\ 5 &\ge a_1 \cdot a_2\cdot a_3 \cdot a_4 \end{align*}$

We now see that we need to find all unordered 4-tuples of positive integers such that the product of all the elements is less that five. Those tuples are: $(a_1,a_2,a_3,a_4) \in \{(1,1,1,1), (2,1,1,1), (3,1,1,1), (4,1,1,1), (2,2,1,1), (5,1,1,1)\}$

We can test each tuple separately, and solve for $a_0$ by plugging them into the original constraint. This yields the solutions $(a_0,a_1,a_2,a_3,a_4) \in \{(3,3,1,1,1),(2,2,2,1,1), (5,2,1,1,1)\}$

We have three solutions, as said in the problem, therefore the tuple with the greatest sum is clearly $(a_0,a_1,a_2,a_3,a_4) = (5,2,1,1,1)$ . Our solution is $\boxed{(5,2,1,1,1)}$

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2006 SMT/General Problems/Problem 8

Problem 8

Solution