# Thread: Numbers in different bases

1. ## Numbers in different bases

Hello Everyone. I need help starting this problem. I do not need anyone to do it all for me. Im just looking for some help getting pointed in the right direction. Thanks.

If we write a number in a base other than base 10 (how we usually write numbers),
it is customary to write the base as a subscript as a clarification. For example,
345(
16) signifies the number 345 in base 16. If we convert this to base 10, we find
that this is equal to 837. In other words 345(
16) = 837(10).

Problem:
Find all three digit sequences XY Z and two digit sequences AB such
that
AB(16) = XY Z(10) and AB(10) = XY Z(8).

2. Hello, Twinsfan85!

Find all three-digit numbers $\displaystyle XYZ$ and two-digit numbers $\displaystyle AB$
. . such that: .$\displaystyle AB_{16} \:=\:XYZ _{10}\;\text{ and } \;AB_{10} \:=\: XYZ_8$
Note that: .$\displaystyle \begin{array}{ccccc} A,B & \in & \{0,1,2,3,\hdots,9\} & A \neq 0 \\ X,Y,Z & \in & \{0,1,2,\hdots,7\} & X \neq 0\end{array}$

We have: .$\displaystyle \begin{array}{ccccccccccc} AB_{16} &=& XYZ_{10} && \Rightarrow && 16A + B &=& 100X + 10Y + Z & {\color{blue}[1]} \\ AB_{10} &=& XYZ_8 && \Rightarrow && 10A + B &=& 64X + 8Y + Z & {\color{blue}[2]} \end{array}$

Subtract $\displaystyle {\color{blue}[1] \text{ - } [2]}$: .$\displaystyle 6A \:=\:36X + 2Y \quad\Rightarrow\quad A \:=\:6X + \frac{Y}{3}$

Since $\displaystyle A$ is a digit, $\displaystyle Y$ must be a multiple of 3: .$\displaystyle Y \,=\,3m\;\;{\color{blue}[3]}$
. . Hence: .$\displaystyle A \:=\:6X + m \;\;{\color{blue}[4]}$

Substitute into $\displaystyle {\color{blue}[2]}$: .$\displaystyle 10(6X + m) + B \:=\:64X + 8(3m) + Z \quad\Rightarrow\quad B \:=\:4X + 14m + Z\;\;{\color{blue}[5]}$

Since $\displaystyle B \leq 9,\;m = 0$

Then in $\displaystyle {\color{blue}[3]}\!:\;\;\boxed{Y \,=\, 0}$

And in $\displaystyle {\color{blue}[4]}\!:\;\;A \:=\:6X$
. . Since $\displaystyle A \leq 9\,\text{ and }\,X \neq 0,\,\text{ then: }\:\boxed{X \,=\, 1}\;\;\boxed{A \,=\, 6}$

Hence, $\displaystyle {\color{blue}[5]}$ becomes: .$\displaystyle B \:=\:Z+4$

And we have: . $\displaystyle \begin{array}{c||cccccc}\hline Z & 0&1&2&3&4&5 \\ \hline B & 4&5&6&7&8&9 \\ \hline \end{array}$

The solutions are:

. . $\displaystyle \begin{array}{|ccc|ccc|}\hline 64_{16} &=& 100_{10} & 64_{10} &=& 100_8 \\ 65_{16} &=& 101_{10} & 65_{10} &=& 101_8 \\ 66_{16} &=& 102_{10} & 66_{10} &=& 102_8 \\ 67_{16} &=& 103_{10} & 67_{10} &=& 103_8 \\ 68_{16} &=& 104_{10} & 68_{10} &=& 104_8 \\ 69_{16} &=& 105_{10} & 69_{10} &=& 105_8 \\ \hline \end{array}$

3. Thank You very much Soroban. I was a little confused at first when I read over your solution. A quick read over the information on base 10 again solved my problems though Thanks Again

Edit: I take back my previous statement. Why did you subtract the second equation from the first? Also more importantly, why did you substitute [4] back into equation 2 and not equation 1?

4. Hello again, Twinsfan85!

Why did you subtract the second equation from the first?
Thought it was obvious . . . good opportunity to eliminate B's and Z's.

Alsowhy did you substitute [4] back into equation 2 and not equation 1?
Um ... Equation 2 had smaller coefficients . . .