Question

What is the remainder when ${7^{38}}$ is divided by $48$ ?

Answer 1

Hint : The given question requires us to find the remainder when ${7^{38}}$ is divided by $48$. So, we will use the binomial theorem of the given binomial expression. We can find the required binomial expansion by using the Binomial theorem. Binomial theorem helps us to expand the powers of binomial expressions easily and can be used to solve the given problem. We must know the formulae of combinations and factorials for solving the given question using the binomial theorem.

Complete step-by-step answer :
So, we will first simplify the expression and create a binomial expression with one term as $48$ so as to find the remainder.
So, we have, ${7^{38}} = {\left( {{7^2}} \right)^{19}}$.
Now, we know that square of $7$ is equal to $49$.
So, we get, ${7^{38}} = {49^{19}}$
Now, we split the term $49$ as $\left( {48 + 1} \right)$. Hence, we get,
$\Rightarrow {\left( {48 + 1} \right)^{19}}$
We have to find the binomial expansion of ${\left( {48 + 1} \right)^{19}}$ . So, using the binomial theorem, the binomial expansion of ${\left( {x + y} \right)^n}$ is $\sum\nolimits_{r = 0}^n {\left( {^n{C_r}} \right){{\left( x \right)}^{n - r}}{{\left( y \right)}^r}}$ .
So, the binomial expansion of ${\left( {48 + 1} \right)^{19}}$ is $\sum\nolimits_{r = 0}^{19} {\left( {^{19}{C_r}} \right){{\left( {48} \right)}^{19 - r}}{{\left( 1 \right)}^r}}$ .
Now, we have to expand the expression $\sum\nolimits_{r = 0}^{19} {\left( {^{19}{C_r}} \right){{\left( {48} \right)}^{19 - r}}{{\left( 1 \right)}^r}}$ and we are done with the binomial expansion of ${\left( {48 + 1} \right)^{19}}$ .
$\sum\nolimits_{r = 0}^{19} {\left( {^{19}{C_r}} \right){{\left( {48} \right)}^{19 - r}}{{\left( 1 \right)}^r}} = \left( {^{19}{C_1}} \right){\left( {48} \right)^{19 - 1}}{\left( 1 \right)^1} + \left( {^{19}{C_2}} \right){\left( {48} \right)^{19 - 2}}{\left( 1 \right)^2} + ...\left( {^{19}{C_{19}}} \right){\left( {48} \right)^{19 - 19}}{\left( 1 \right)^{19}}$
Since any power of one is equal to one itself. So, equating all the brackets with powers of $1$ as $1$.
Hence, $\sum\nolimits_{r = 0}^{19} {\left( {^{19}{C_r}} \right){{\left( {48} \right)}^{19 - r}}{{\left( 1 \right)}^r}} = \left( {^{19}{C_1}} \right){\left( {48} \right)^{19 - 1}} + \left( {^{19}{C_2}} \right){\left( {48} \right)^{19 - 2}} + ...\left( {^{19}{C_{19}}} \right){\left( {48} \right)^{19 - 19}}$ .
Now, we can see that all the terms of binomial expansion are divisible by $48$ except the last one. So, we get,
$\Rightarrow \sum\nolimits_{r = 0}^{19} {\left( {^{19}{C_r}} \right){{\left( {48} \right)}^{19 - r}}{{\left( 1 \right)}^r}} = 48\lambda + \left( {^{19}{C_{19}}} \right){\left( {48} \right)^{19 - 19}}$, where $\lambda$ is any integer.
Now, we know the combination formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
$\Rightarrow \sum\nolimits_{r = 0}^{19} {\left( {^{19}{C_r}} \right){{\left( {48} \right)}^{19 - r}}{{\left( 1 \right)}^r}} = 48\lambda + \dfrac{{19!}}{{19! \times 0!}} \times {\left( {48} \right)^{19 - 19}}$
Cancelling common factors in numerator and denominator, we get,
$\Rightarrow \sum\nolimits_{r = 0}^{19} {\left( {^{19}{C_r}} \right){{\left( {48} \right)}^{19 - r}}{{\left( 1 \right)}^r}} = 48\lambda + 1 \times {\left( {48} \right)^0}$
Now, we know that vany number raised to power zero is equal to one. So, we get,
$\Rightarrow \sum\nolimits_{r = 0}^{19} {\left( {^{19}{C_r}} \right){{\left( {48} \right)}^{19 - r}}{{\left( 1 \right)}^r}} = 48\lambda + 1$
Since, ${7^{38}}$ is of the form $48\lambda + 1$ where $\lambda$ is any integer. So, we get the remainder when ${7^{38}}$ is divided by $48$ as $1$.
So, the correct answer is “1”.

Note : The easiest way to solve such a problem is to apply the concepts of Binomial theorem as it is very effective in finding the binomial expansion. We should always verify the calculations so as to be sure of the final answer. Formulas of permutations and combinations must be remembered to get to the required answer.