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Use divisions to convert the base ten numeral 107 to base five. How do I solve it?
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To me, 10 = 5*2, hence \(10^0 = 5^0*2^0; 10^1=5^1*2^1;10^2=5^2*2^2\) 107 base 10 is \(7*(10^0) + 0*(10^1)+1*(10^2)\) If we write it under the base 5, then, the number is under the form: \(a*5^0+b*5^1+c*5^2+d*5^3+.....\) now combine them \(a*5^0 + b*5^1 +c*5^2 +d*5^3 +...= (7*2^0)*5^0+(0*2^1)*5^1 + (1*2^2)*5^2\) Now, make comparison we have a = 7 but in base 5, \( 7 = 2*5^0 +1*5^1\) b = 0 c = 4 ------------------------------------------------------ it becomes 412
if you want to use divisions (which is the fastest way) then 107/5=21 R 2 21/5=4 R 1 4/5 = 0 R4 so the answer is 412
wow!!
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