Question
these are the problems to convert to base 10.
1. CDXLIV
2. 432 then has five next to it
3. ET0 then has twelve next to it
4. 1011 then has two next to it
5. 4136 then has seven next to it
1. CDXLIV
2. 432 then has five next to it
3. ET0 then has twelve next to it
4. 1011 then has two next to it
5. 4136 then has seven next to it
Answers
Damon
I do not remember Roman numerals
432 base 5 = 2*5^0 +3*5^1 + 4*5^2 = 2+15+100 =117
A = 10
B = 11
C = 12 in base 12. I can not imagine what E and T are
ET0 base 12 = ?
1011 = 1*2^0 + 1*2^1 + 0*2^2 + 1*2^3
= 1+2+0+8 = 11
4136 base 7 = 6*7^0 + 3*7^1 + 1*7^2 + 4*7^3
= 6+21+49+1372 = 1448
432 base 5 = 2*5^0 +3*5^1 + 4*5^2 = 2+15+100 =117
A = 10
B = 11
C = 12 in base 12. I can not imagine what E and T are
ET0 base 12 = ?
1011 = 1*2^0 + 1*2^1 + 0*2^2 + 1*2^3
= 1+2+0+8 = 11
4136 base 7 = 6*7^0 + 3*7^1 + 1*7^2 + 4*7^3
= 6+21+49+1372 = 1448