Difference between revisions of "1976 IMO Problems/Problem 4"
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== Solution == | == Solution == | ||
− | {{ | + | 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
Solution
Since , 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 . If is more efficient than , then . We try to prove that all integers greater than 3 are less efficient than 3:
When increases by 1, then the RHS is multiplied by 3. The other side is multiplied by , and we must prove that this is less than 3 for all greater than 3.
Thus we need to prove that . Simplifying, we get , which is true. Working backwards, we see that all greater than 3 are less efficient than 3, so we try to use the most 3's as possible:
, so the greatest product is .
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 |