Question

If we are given $2\times 1=81;3\times 2=278;2\times 5=8125$ , then find the value of $1\times 3$ .

Answer 1

Hint: We have to find a common rule from the given results. We can see from the given data that $a\times b={{a}^{3}}{{b}^{3}}$ . Here, the RHS forms the digits and is not the result of multiplication of ${{a}^{3}}$ and ${{b}^{3}}$ . Using this rule, we can find the value of $1\times 3$ .

Complete step-by-step solution:
We are given that $2\times 1=81$ . We can see that the digits of the result, that is, 81, can be written as of 2 to the power 3 and 1 to the power 3. This means that ${{2}^{3}}=8$ and ${{1}^{3}}=1$ .
$\Rightarrow 2\times 1={{2}^{3}}{{1}^{3}}=81$
Let us consider $3\times 2=278$ . We know that ${{3}^{3}}=27$ and ${{2}^{3}}=8$ . Therefore we can write the given result as shown below.
\[\Rightarrow 3\times 2={{3}^{3}}{{2}^{3}}=278\]
We are also given that $2\times 5=8125$ . We know that ${{2}^{3}}=8$ and ${{5}^{3}}=125$ . Therefore we can write the given result as follows.
$\Rightarrow 2\times 5={{2}^{3}}{{5}^{3}}=8125$
In general, we can say that $a\times b={{a}^{3}}{{b}^{3}}$ .
Now, let us find the value of $1\times 3$ .
$\Rightarrow 1\times 3={{1}^{3}}{{3}^{3}}=127$
Therefore, the value of $1\times 3$ is 127.

Note: Students may be confused by the term in the RHS of the rule formed. They must note that we are not multiplying ${{a}^{3}}$ and ${{b}^{3}}$ in the rule $a\times b={{a}^{3}}{{b}^{3}}$ . Instead, we are forming digits of the given operations. The key to the questions of this type is that we have formed a common rule from the given results. The properties of multiplication fail in these data, that is, commutative property is not applicable. For example, if we were to find $3\times 1$ the result will not be the same.
$\Rightarrow 3\times 1={{3}^{3}}{{1}^{3}}=271$