Numbers in different bases

**Twinsfan85** · March 12th 2010, 12:06 PM

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).

**Soroban** · March 12th 2010, 04:29 PM

Hello, Twinsfan85!

Find all three-digit numbers $XYZ$ and two-digit numbers $AB$
. . such that: . $AB_{16} \:=\:XYZ _{10}\;\text{ and } \;AB_{10} \:=\: XYZ_8$

Note that: . $\begin{array}{ccccc} A,B & \in & \{0,1,2,3,\hdots,9\} & A \neq 0 \\ X,Y,Z & \in & \{0,1,2,\hdots,7\} & X \neq 0\end{array}$

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

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

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

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

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

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

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

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

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

The solutions are:

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

**Twinsfan85** · March 12th 2010, 06:20 PM

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?

**Soroban** · March 12th 2010, 07:31 PM

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 . . .

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