Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1976 IMO Problems/Problem 4

Revision as of 16:27, 20 May 2012 by Mahamaya (talk | contribs) (Slightly differnt solution)

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 $\boxed{3^{658}\cdot 2}$.

Alternate Solution

Note: This solution uses the same strategy as the above solution (that having the largest possible number of three's is good), but approaches the proof in a different manner.

Let $f(S) \triangleq \prod_{x \in S} x$, and $g(S) = \sum_{x\in S} x$, where $S$ is a multiset. We fix $g(S) = 1976$, and aim to maximize $f(S)$. Since $3(x-3) > x$ for $x \geq 5$, we notice that $S$ must only contain the integers $1,2,3$ and $4$. We can replace any occurrences of $4$ in $S$ by replacing it with a couple of $2$'s, without changing $f(S)$ or $g(S)$, so we may assume that $S$ only contains the integers $1,2$ and $3$. We may further assume that $S$ contains at most one $1$, since any two $1$'s can be replaced by a $2$ without changing $g(S)$, but with an increase in $f(S)$. If $S$ contains exactly one $1$, then it must also contain at least one $2$ (since $1976 = 2\;(\mathrm{mod }\;3)$). We can then replace this pair of a $1$ and a $2$ with a $3$, thus keeping $g(S)$ constant, and increasing $f(S)$. Now we may assume that $S$ contains only $2$'s and $3$'s.

Now, as observed in the last solution, any triplet of $2$'s can be replaced by a couple of $3$'s, with $g(S)$ constant, and an increase in $f(S)$. Thus, after repeating this operation, we will be left with at most two $2$'s. Since $g(S) = 1976$, and $1976 = 2\;(\mathrm{mod }\;3)$, we therefore get that $S$ must have exactly one $2$ (since we already showed it consists only of $2$'s and $3$'s). Thus, we get $\boxed{f(S) = 2\cdot 3^{658}}$.

See also

1976 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1976_IMO_Problems/Problem_4&oldid=47073"