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

Latest revision as of 15:37, 8 August 2009

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

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 $\frac {1*2*3}{1*3*5}=\frac {6}{15}= \frac {2}{5}$

Revision as of 15:36, 8 August 2009 (view source) Ihatepie (talk \| contribs) (Created page with '== Problem I1 == Calculate: <math>\frac{123+246+369+4812+51015}{135+2610+3915+41220+51525}</math> == Solution == If you factor the top, you get <math>(123…')		Latest revision as of 15:37, 8 August 2009 (view source) Ihatepie (talk \| contribs) (→‎Solution)
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	If you factor the bottom, you get <math>(135)(1^{3}+2^{3}+3^{3}+4^{3}+5^{3})</math>		If you factor the bottom, you get <math>(135)(1^{3}+2^{3}+3^{3}+4^{3}+5^{3})</math>

−	Dividing out the common factor, you get <~~cmath~~>\frac {123}{135}~~</cmath> <cmath>~~\frac {6}{15}~~</cmath> <cmath>~~\frac {2}{5}</~~cmath~~>	+	Dividing out the common factor, you get <math>\frac {123}{135}=\frac {6}{15}= \frac {2}{5}</math>