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Difference between revisions of "1997 PMWC Problems/Problem I14"

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== Problem ==
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If we make five two-digit numbers using the digits <math>0, 1, 2, \ldots 9</math> exactly once, and the product of the five numbers is maximized, find the greatest number among them.
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== Solution ==
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This is a [[greedy algorithm]] question. Let <math>a_{1}, b_{1}</math> be the digits of the first number, etc. [[Without loss of generality]] let <math>10a_1 + b_1</math> be the greatest number. Then we want to maximize the quantity
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<cmath>(10a_1 + b_1)(10a_2 + b_2) \cdots (10a_5 + b_5)</cmath>
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<cmath>=10^5 a_1a_2a_3a_4a_5 + 10^4(b_1a_2a_3a_4a_5 + \ldots) + \ldots + b_1b_2b_3b_4b_5</cmath>
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The greedy algorithm quickly tells us that the first digits of the numbers should be <math>9,8,7,6,5</math>, so <math>a_1 = 9</math>. Now, look at the coefficient of <math>10^4</math>. The product <math>a_2a_3a_4a_5</math> is less than any of the other terms (which all contain the maximal <math>a_1 = 9</math>), so by the greedy algorithm, we should make <math>b_1</math> as small as possible. Hence <math>b_1 = 0</math>, and our answer is <math>90</math>.
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== See Also ==
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{{PMWC box|year=1997|num-b=I13|num-a=I15}}

Latest revision as of 15:05, 15 May 2012

Problem

If we make five two-digit numbers using the digits $0, 1, 2, \ldots 9$ exactly once, and the product of the five numbers is maximized, find the greatest number among them.

Solution

This is a greedy algorithm question. Let $a_{1}, b_{1}$ be the digits of the first number, etc. Without loss of generality let $10a_1 + b_1$ be the greatest number. Then we want to maximize the quantity

\[(10a_1 + b_1)(10a_2 + b_2) \cdots (10a_5 + b_5)\] \[=10^5 a_1a_2a_3a_4a_5 + 10^4(b_1a_2a_3a_4a_5 + \ldots) + \ldots + b_1b_2b_3b_4b_5\]

The greedy algorithm quickly tells us that the first digits of the numbers should be $9,8,7,6,5$, so $a_1 = 9$. Now, look at the coefficient of $10^4$. The product $a_2a_3a_4a_5$ is less than any of the other terms (which all contain the maximal $a_1 = 9$), so by the greedy algorithm, we should make $b_1$ as small as possible. Hence $b_1 = 0$, and our answer is $90$.

See Also

1997 PMWC (Problems)
Preceded by
Problem I13 		Followed by
Problem I15
I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
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