Question

Find which of the weights 1,3,9,27,81,....... lbs must be used to weigh ten thousand lbs, not more than one of each kind being used but in either scale that is necessary.

Answer 1

Hint: Firstly, we have to express 10,000 in the scale of 3. This is done using a long division method. . We have to divide 10,000 by 3 and write its quotient below 10,000 and the remainder to the right of the quotient. Then, we have to divide this quotient by 3 and write its result below and the corresponding remainder on the right. We have to perform this step till we get the quotient as 1. Whenever we get a remainder greater than 1, say p, we have to add $\left( p-1 \right)$ to the quotient since only one weight of each kind can only be used. We have to represent 10000 in the form $N={{a}_{n}}{{r}^{n}}+{{a}_{n-1}}{{r}^{n-1}}+...+{{a}_{2}}{{r}^{2}}+{{a}_{1}}r+{{a}_{0}}$ , where ${{a}_{0}},{{a}_{1}},..,{{a}_{n}}$ are the digits by which N is to be expressed and r is the radix of the scale. The negative sign shows that corresponding weights are to be placed at the opposite scale to those indicated by the positive remainders.

Complete step by step answer:
We have expressed 10,000 on the scale of 3. We have to use a long division method. We have to divide 10,000 by 3 and write its quotient below 10,000 and the remainder to the right of the quotient. Then, we have to divide this quotient by 3 and write its result below and the corresponding remainder on the right. We have to perform this step till we get the quotient as 1.
$\begin{align}
  & 3\left| \!{\underline {\,
  10000 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3333 \,}} \right. ...1 \\
 & 3\left| \!{\underline {\,
  1111 \,}} \right. ...0 \\
 & 3\left| \!{\underline {\,
  370 \,}} \right. ...1 \\
 & 3\left| \!{\underline {\,
  123 \,}} \right. ...1 \\
 & 3\left| \!{\underline {\,
  41 \,}} \right. ...0 \\
 & 3\left| \!{\underline {\,
  14 \,}} \right. ...-1 \\
 & 3\left| \!{\underline {\,
  5 \,}} \right. ...-1 \\
 & 3\left| \!{\underline {\,
  2 \,}} \right. ...-1 \\
 & \text{ }\text{ }1...-1 \\
\end{align}$
On dividing 41 by 3, we will get the remainder as 2. Since only one weight of each kind can only be used we will write the quotient as 14 and -1 as the remainder.
We know that for a number $N={{a}_{n}}{{r}^{n}}+{{a}_{n-1}}{{r}^{n-1}}+...+{{a}_{2}}{{r}^{2}}+{{a}_{1}}r+{{a}_{0}}$ , where ${{a}_{0}},{{a}_{1}},..,{{a}_{n}}$ are the digits by which N is to be expressed and r is the radix of the scale. We have to write the remainder as ${{a}_{0}},{{a}_{1}},..,{{a}_{n}}$ .
We can write $\begin{align}
  & \Rightarrow 10000=1\times {{3}^{9}}+-1\times {{3}^{8}}+-1\times {{3}^{7}}+-1\times {{3}^{6}}+-1\times {{3}^{5}}+0\times {{3}^{4}}+1\times {{3}^{3}}+1\times {{3}^{2}}+0\times 3+1 \\
 & \Rightarrow 10000=1\times {{3}^{9}}+-1\times {{3}^{8}}+-1\times {{3}^{7}}+-1\times {{3}^{6}}+-1\times {{3}^{5}}+1\times {{3}^{3}}+1\times {{3}^{2}}+1 \\
\end{align}$
The negative sign shows that corresponding weights are to be placed at the opposite scale to those indicated by the positive remainders.
Thus, weights ${{3}^{9}},{{3}^{3}},{{3}^{2}},1$ have to placed in one scale and ${{3}^{8}},{{3}^{7}},{{3}^{6}},{{3}^{5}}$ in the other scale.

Note: Students must be thorough in dividing numbers by a weight using a long division method. They must never forget to add $p-1$ to the quotient, in case p is greater than 1, where p is the remainder (say). If we were not given that only one weight of each kind must be used, we can take the remainder as it is.