Difference between revisions of "1976 IMO Problems/Problem 4"

Revision as of 10:45, 30 July 2013

Problem

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

Solution

Solution 1

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

Solution 2

We demonstrate heuristically that 3's are the most efficient. As we know that the chosen numbers must be equal, by the AM-GM inequality, we wish to maximize $f(x) = \left(\frac{1976}{x}\right)^{x}$ Simple logarithmic differentiation shows that $\frac{S}{x} = e$ maximizes the given function. As $e$ is approximately 2.71828, we use 2's and 3's only. 3's are optimal, but we must use one 2.

Solution 3

Note: This solution uses the same strategy as Solution 1 (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 $\boxed{f(S) = 2\cdot 3^{658}}$ .

@@ Line 18: / Line 18: @@
 <math>\dfrac{1976}{3}=658.6666</math>, so the greatest product is <math>\boxed{3^{658}\cdot 2}</math>.
-=== Solution 2 ===
+===Solution 2===
+We demonstrate heuristically that 3's are the most efficient.
+As we know that the chosen numbers must be equal, by the AM-GM inequality, we wish to maximize <cmath> f(x) = \left(\frac{1976}{x}\right)^{x} </cmath>
+Simple logarithmic differentiation shows that <math>\frac{S}{x} = e</math> maximizes the given function. As <math>e</math> is approximately 2.71828, we use 2's and 3's only. 3's are optimal, but we must use one 2.
-''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.''
+=== Solution 3 ===
+''Note:  This solution uses the same strategy as Solution 1 (that having the largest possible number of three's is good), but approaches the proof in a different manner.''
 Let <math>f(S) \triangleq \prod_{x \in S} x</math>, and <math>g(S) = \sum_{x\in S} x</math>, where <math>S</math> is a multiset.  We fix <math>g(S) = 1976</math>, and aim to maximize <math>f(S)</math>.  Since <math>3(x-3) > x</math> for <math>x \geq 5</math>,  we notice that <math>S</math> must only contain the  integers <math>1,2,3</math> and <math>4</math>.  We can replace any occurrences of <math>4</math> in <math>S</math> by replacing it with a couple of <math>2</math>'s, without changing <math>f(S)</math> or <math>g(S)</math>, so we may assume that <math>S</math> only contains the integers <math>1,2</math> and <math>3</math>.   We may further assume that <math>S</math> contains at most one <math>1</math>,  since any two <math>1</math>'s can be replaced by a <math>2</math> without changing <math>g(S)</math>, but with an increase in <math>f(S)</math>.  If <math>S</math> contains exactly one <math>1</math>, then it must also contain at least one <math>2</math> (since <math>1976 = 2\;(\mathrm{mod }\;3)</math>).   We can then replace this pair of a <math>1</math> and a <math>2</math> with a <math>3</math>, thus keeping <math>g(S)</math> constant, and increasing <math>f(S)</math>.   Now we may assume that <math>S</math> contains only <math>2</math>'s and <math>3</math>'s.
 Now, as observed in the last solution, any triplet of <math>2</math>'s can be replaced by a couple of <math>3</math>'s, with <math>g(S)</math> constant, and an increase in <math>f(S)</math>.    Thus, after repeating this operation, we will be left with at most two <math>2</math>'s.  Since <math>g(S) = 1976</math>, and <math>1976 = 2\;(\mathrm{mod }\;3)</math>,  we therefore get that <math>S</math> must have exactly one <math>2</math> (since we already showed it consists only of <math>2</math>'s and <math>3</math>'s).   Thus, we get <math>\boxed{f(S) = 2\cdot 3^{658}}</math>.
 == 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

Page

Toolbox

Search