



 
Hi Babawale, I would start by adding $342_5$ and $132_5$ in base 5. Starting at the units column two plus two is four so I get Now the fives column, four plus three is seven and seven in base 5 is $12_5$ so I write the 2 and carry the 1 Finally in the twentyfives column three plus one plus one is five which in base 5 is $10_5$ and hence Now your problem reduces to \[1024_5  223_{5}.\] For that you should look at our response to an earlier question. Penny  


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