Difference between revisions of "1998 PMWC Problems/Problem I1"

(Created page with '== Problem I1 == Calculate: <math>\frac{1*2*3+2*4*6+3*6*9+4*8*12+5*10*15}{1*3*5+2*6*10+3*9*15+4*12*20+5*15*25}</math> == Solution == If you factor the top, you get <math>(1*2*3…')
 
(Solution)
Line 9: Line 9:
 
If you factor the bottom, you get <math>(1*3*5)(1^{3}+2^{3}+3^{3}+4^{3}+5^{3})</math>
 
If you factor the bottom, you get <math>(1*3*5)(1^{3}+2^{3}+3^{3}+4^{3}+5^{3})</math>
  
Dividing out the common factor, you get <cmath>\frac {1*2*3}{1*3*5}</cmath> <cmath>\frac {6}{15}</cmath> <cmath>\frac {2}{5}</cmath>
+
Dividing out the common factor, you get <math>\frac {1*2*3}{1*3*5}=\frac {6}{15}= \frac {2}{5}</math>$

Revision as of 16:36, 8 August 2009

Problem I1

Calculate: $\frac{1*2*3+2*4*6+3*6*9+4*8*12+5*10*15}{1*3*5+2*6*10+3*9*15+4*12*20+5*15*25}$

Solution

If you factor the top, you get $(1*2*3)(1^{3}+2^{3}+3^{3}+4^{3}+5^{3})$


If you factor the bottom, you get $(1*3*5)(1^{3}+2^{3}+3^{3}+4^{3}+5^{3})$

Dividing out the common factor, you get $\frac {1*2*3}{1*3*5}=\frac {6}{15}= \frac {2}{5}$$

Invalid username
Login to AoPS