"Basic" 11 (corrected)

Let ( a ) 2 = ( b ) 3 = ( c ) 4 = ( d ) 5 = ( e ) 6 = ( f ) 7 = ( g ) 8 = ( h ) 9 = ( i ) 10 = ( j ) 11 = 11 (a)_2 = (b)_3 = (c)_4 = (d)_5 = (e)_6 = (f)_7 = (g)_8 = (h)_9 = (i)_{10} = (j)_{11} = 11

1 11 ( a + b + c + d + e + f + g + h + i + j ) = ? \frac{1}{11}(a+b+c+d+e+f+g+h+i+j) = ?

Note : ( n ) b (n)_b is the representation of the number n n in base b b (e.g. ( 101 ) 2 = 5 (101)_2 = 5 .)


The answer is 112.

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1 solution

Romain Bouchard
Jan 22, 2018

Relevant wiki: Number Base - Converting to Different Bases

We know that :

11 = 8 + 2 + 1 = 2 3 + 2 1 + 2 0 a = 1011 11 = 8 + 2 + 1 = 2^3 + 2^1 + 2^0 \Rightarrow a = 1011

11 = 3 2 + 2 × 3 0 b = 102 11 = 3^2 + 2 \times 3^0 \Rightarrow b = 102

11 = 2 × 4 1 + 3 × 4 0 c = 23 11 = 2 \times 4^1 + 3 \times 4^0 \Rightarrow c = 23

11 = 2 × 5 1 + 5 0 d = 21 11 = 2 \times 5^1 + 5^0 \Rightarrow d = 21

11 = 6 1 + 5 × 6 0 e = 15 11 = 6^1 + 5 \times 6^0 \Rightarrow e = 15

11 = 7 1 + 4 × 7 0 f = 14 11 = 7^1 + 4 \times 7^0 \Rightarrow f = 14

11 = 8 1 + 3 × 8 0 g = 13 11 = 8^1 + 3 \times 8^0 \Rightarrow g = 13

11 = 9 1 + 2 × 9 0 h = 12 11 = 9^1 + 2 \times 9^0 \Rightarrow h = 12

11 = 1 1 1 j = 10 11 = 11^1 \Rightarrow j = 10

1 11 ( a + b + c + d + e + f + g + h ) = 1011 + 102 + 23 + 21 + 15 + 14 + 13 + 12 + 10 + 11 11 = 1232 11 = 112 \Rightarrow \frac{1}{11}(a+b+c+d+e+f+g+h)=\frac{1011+102+23+21+15+14+13+12+10+11}{11} = \frac{1232}{11} = 112

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