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Difference between revisions of "1997 PMWC Problems/Problem I1"

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== Solution ==
 
== Solution ==
−
<math>\frac{29 \cdot 28 + 14}{28} \cdot \frac{27 \cdot 15 + 24}{15} = \frac{839 \cdot 419}{420} = \frac{839(420 - 1)}{420} = 839 - \frac{839}{420} = 837 \frac{1}{420}</math>
+
<math>\frac{29 \cdot 28 + 27}{28} \cdot \frac{27 \cdot 15 + 14}{15} = \frac{839 \cdot 419}{420} = \frac{839(420 - 1)}{420} = 839 - \frac{839}{420} = 837 \frac{1}{420}</math>
  
 
== See also ==
 
== See also ==

Revision as of 07:24, 9 October 2007

Problem

Evaluate $29 \frac{27}{28} \times 27\frac{14}{15}$.

Solution

$\frac{29 \cdot 28 + 27}{28} \cdot \frac{27 \cdot 15 + 14}{15} = \frac{839 \cdot 419}{420} = \frac{839(420 - 1)}{420} = 839 - \frac{839}{420} = 837 \frac{1}{420}$

See also

1997 PMWC (Problems)
Preceded by
First question 		Followed by
Problem I2
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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