1976 IMO Problems/Problem 4

Problem

Determine the greatest number, who is the product of some positive integers, and the sum of these numbers is $1976.$

Solution

Since $3*3=2*2*2+1$ , 3's are more efficient than 2's. We try to prove that 3's are more efficient than anything:

Let there be a positive integer $x$ . If $3$ is more efficient than $x$ , then $x^3<3^x$ . We try to prove that all integers greater than 3 are less efficient than 3:

When $x$ increases by 1, then the RHS is multiplied by 3. The other side is multiplied by $\dfrac{(x+1)^3}{x^3}$ , and we must prove that this is less than 3 for all $x$ greater than 3.

$\dfrac{(x+1)^3}{x^3}<3\Rightarrow \dfrac{x+1}{x}<\sqrt[3]{3}\Rightarrow 1<(\sqrt[3]{3}-1)x$

$\dfrac{1}{\sqrt[3]{3}-1}<x$

Thus we need to prove that $\dfrac{1}{\sqrt[3]{3}-1}<4$ . Simplifying, we get $5<4\sqrt[3]{3}\Rightarrow 125<64*3=192$ , which is true. Working backwards, we see that all $x$ greater than 3 are less efficient than 3, so we try to use the most 3's as possible:

$\dfrac{1976}{3}=658.6666$ , so the greatest product is $\boxed{3^{658}\cdot 2}$ .

1976 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
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1976 IMO Problems/Problem 4

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