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Difference between revisions of "1976 IMO Problems/Problem 4"
(I SOLVED AN IMO PROBLEM! :D)
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== Solution ==
 
== Solution ==
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{{solution}}
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Since <math>3*3=2*2*2+1</math>, 3's are more efficient than 2's. We try to prove that 3's are more efficient than anything:
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Let there be a positive integer <math>x</math>. If <math>3</math> is more efficient than <math>x</math>, then <math>x^3<3^x</math>. We try to prove that all integers greater than 3 are less efficient than 3:
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When <math>x</math> increases by 1, then the RHS is multiplied by 3. The other side is multiplied by <math>\dfrac{(x+1)^3}{x^3}</math>, and we must prove that this is less than 3 for all <math>x</math> greater than 3.
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<math>\dfrac{(x+1)^3}{x^3}<3\Rightarrow \dfrac{x+1}{x}<\sqrt[3]{3}\Rightarrow 1<(\sqrt[3]{3}-1)x</math>
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<math>\dfrac{1}{\sqrt[3]{3}-1}<x</math>
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Thus we need to prove that <math>\dfrac{1}{\sqrt[3]{3}-1}<4</math>. Simplifying, we get <math>5<4\sqrt[3]{3}\Rightarrow 125<64*3=192</math>, which is true. Working backwards, we see that all <math>x</math> greater than 3 are less efficient than 3, so we try to use the most 3's as possible:
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<math>\dfrac{1976}{3}=658.6666</math>, so the greatest product is <math>\boxed{3^{658}\cdot 2}</math>.
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== See also ==
 
== See also ==
 
{{IMO box|year=1976|num-b=3|num-a=5}}
 
{{IMO box|year=1976|num-b=3|num-a=5}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 09:20, 18 June 2008

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

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
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