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Difference between revisions of "1997 PMWC Problems/Problem I13"

(I'll get the right problem and the right solution in a couple minutes...)
 
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== Problem ==
 
== 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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A truck moved from A to B at a speed of 50 km/h and returns from B to A at 70 km/h. It traveled 3 rounds within 18 hours. What is the distance between A and B?
  
 
== Solution ==
 
== 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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The truck moved from A to B back and forth once in 18/3=6 hours. Let the time that it takes to go from A to B be <math>T_1</math>, and let the time it takes to go from B to A be <math>T_2</math>.
  
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<cmath>(10a_1 + b_1)(10a_2 + b_2) \cdots (10a_5 + b_5)</cmath>
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<math>T_1+T_2=6</math>
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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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<math>50T_1=70T_2\Rightarrow T_1=\frac{7}{5}T_2</math>
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<math>\frac{12}{5}T_2=6\Rightarrow T_2=\frac{5}{2}</math>
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Therefore the distance from A to B is <math>\frac{5}{2}*70=\boxed{175}</math>.
  
 
== See also ==
 
== See also ==
 
{{PMWC box|year=1997|num-b=I12|num-a=I14}}
 
{{PMWC box|year=1997|num-b=I12|num-a=I14}}

Revision as of 22:27, 9 January 2008

Problem

A truck moved from A to B at a speed of 50 km/h and returns from B to A at 70 km/h. It traveled 3 rounds within 18 hours. What is the distance between A and B?

Solution

The truck moved from A to B back and forth once in 18/3=6 hours. Let the time that it takes to go from A to B be $T_1$, and let the time it takes to go from B to A be $T_2$.

$T_1+T_2=6$

$50T_1=70T_2\Rightarrow T_1=\frac{7}{5}T_2$

$\frac{12}{5}T_2=6\Rightarrow T_2=\frac{5}{2}$

Therefore the distance from A to B is $\frac{5}{2}*70=\boxed{175}$.

See also

1997 PMWC (Problems)
Preceded by
Problem I12 		Followed by
Problem I14
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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