11th form to 7th form

Question

I can't solve this one: Consider an integer expressed in the 11-based form. In the 11-based form, digits 0 to 9 correspond to their decimal values, and A corresponds to 10. For example (15)_11 = (1 * 11 + 5)_dec = (16)_dec and (2A)_11 = (2 * 11 + 10)_dec = (32)_dec. Given (10A)_11 expressed in 11-based form, determine its expression in the 7-based form. In 7-based form there are only 7 different digits, i.e. 0 to 6

Answer 1

$\overline{10A}_{11} = 1 \times 11^2 + 0 \times 11 + 10 \times 11^0 = 131$.

We first find $m$ such that $7^m < 131 < 7^{m+1}$. We find $m = 2$, so we know that in base-$7$, this is an integer with three digits. Now we find an integer $k$ such that $k 7^2 < 131 < (k+1)7^2$, and find $k = 2$, so it's first digit is $2$.

Now we subtract $2*49$ from $131$ to get $33$. Now we find an $n$ such that $7n < 33 < 7(n+1)$, so $n = 4$, and its second digit is $4$. Subtracting $4*7$ from $33$ we are left with $5$, so its last digit is $5$.

In conclusion, $\overline{10A}_{11} = \overline{245}_7$.

This is a formalized version of what goes through our mind when finding the representation of a number in a certain base. First find the number of digits, then find what the first digit is, then the second, then the third, etc. until we're done.

Answer 2

Having first determined, as above, that we need to convert $131_{10}$ to base - $7$; we could proceed as follows: (all calculations below are in base - $10$)

Divide $131$ by$ 7$; the result is $18$, remainder $5$

Divide $18$ by$ 7$; the result is $2$, remainder $4$

Divide $2$ by$ 7$; the result is $0$, remainder $2$

Now read these remainders, from last to first as the digits of the required number, $245_7$