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Q. Let $ {{D}_{70}}=\{1,2,5,7,10,14,35,70\} $ . Define $ '+', $ $ ., $ and $ '.' $ by $ a+b=1\,cm(a,b),a.b=\gcd (a,b) $ and $ a'=\frac{70}{a} $ for all $ a,b\in {{D}_{70}} $ .The value of $ (2+7) $ $ (14.10)' $ is:

KEAMKEAM 2005

Solution:

$ (2+7)=1cm\text{ (}2,7)=14 $
and $ (14-10)=gcd\text{ (}14,10)=2 $
$ \therefore $ $ (14.10)=\frac{70}{2}=35 $
$ \therefore $ $ (2+7)(14.10) $
$ =14.35=\gcd (14,35)=7 $