Question

Given a number $x = {2^{48}} - 1$ , then between $5$ and $10$ , $x$ has/have
A. No factor
B. Only one factor
C. Two factors
D. Three factors

Answer 1

Hint: This question is based on the concept of binomial theorem and expansion to find factors and remainders of any binomial expression. It utilises further factorisation of an expression with higher powers until there are singular terms remaining in the factors. These terms are then the factor of the binomial expression. As it specified, that the facts need to be between $5$ and $10$ , we will factorise until we get single digit numbers.

Formula used: The identity used for this is the factorisation of a particular binomial expression, the difference of square of terms. This factorisation gives addition of terms as one factor and difference as another, that is,
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
Where $a$ and $b$ are the two terms.

Complete step-by-step solution:
Given that the number is $x = {2^{48}} - 1$ we need to factorise it to get single digit factors. We can do this by writing $48 = (24)(2)$ and repeat this with all even powers until we get the lowest possible factor.
$x = {2^{48}} - 1$
$x = {({2^{24}})^2} - 1$
$ \Rightarrow ({2^{24}} - 1)({2^{24}} + 1)$
Since $24$ can be further divided by $2$ , therefore we continue factorisation of the terms.
$ \Rightarrow ({({2^{12}})^2} - 1)({2^{24}} + 1)$
$ \Rightarrow ({2^{12}} - 1)({2^{12}} + 1)({2^{24}} + 1)$
On further factorisation,
$ \Rightarrow ({({2^6})^2} - 1)({2^{12}} + 1)({2^{24}} + 1)$
\[ \Rightarrow ({2^6} - 1)({2^6} + 1)({2^{12}} + 1)({2^{24}} + 1)\]
On even further factorisation,
\[ \Rightarrow ({({2^3})^2} - 1)({2^6} + 1)({2^{12}} + 1)({2^{24}} + 1)\]
\[ \Rightarrow ({2^3} - 1)({2^3} + 1)({2^6} + 1)({2^{12}} + 1)({2^{24}} + 1)\]
Since $3$ can no longer be factored any further evenly therefore,
\[ \Rightarrow (7)(9)({2^6} + 1)({2^{12}} + 1)({2^{24}} + 1)\]
As we can see from the above expansion of the binomial expansion $7$ and $9$ are the only viable factors of $x$ between $5$ and $10$
Therefore there are two factors between $5$ and $10$.

Hence, the correct option is ( C )

Note: In such questions, it is important to only solve the powers once you’ve factored all the way to the lowest possible power. If there were more factors possible in this case, chances are they would’ve been less than $5$.