Question 27

Let $$a_1, a_2, a_3, \ldots, a_n$$ be all natural numbers and also $$n \in \mathbb{N}$$ such that $$a_1 + a_2 + a_3 + \cdots + a_n = 1000$$. When $$a_1 \times a_2 \times a_3 \times \cdots \times a_n$$ would be maximum, then $$n =$$

Correct Answer: 334

Solution

To maximise a product with a fixed sum, the parts must be 2s and 3s, with as many 3s as possible, since $$3^2 > 2^3$$ for the same total 6 and a part of 4 or more can be split without loss. As $$1000 = 3 \times 332 + 4$$ and the leftover 4 is taken as $$2 + 2$$, the maximum product is $$3^{332} \times 2^2$$. This uses $$332 + 2 = 334$$ numbers.

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