Question

The last two digits of the number ${23^{14}}$ are ?

Answer 1

Hint : In order to solve the given question , we should know the important concepts related to the question that is Binomial Expansion . A Binomial is an algebraic expression containing 2 terms . We use the binomial theorem to help us expand binomials to any given power without direct multiplication . To simplify this question , we need to solve it step by step and we will be using these concepts to get our required answer.

Complete step-by-step answer :
The question given to us is to determine the last two digits of ${23^{14}}$ .
We are going to first break the number to two terms to make it a binomial expression in order to apply binomial expansion .
So we can write ${23^{14}}$ as –

$= {[{\left( {23} \right)^2}]^7}$

$= {[529]^7}$

$= {\left( {530 - 1} \right)^7}$

Now applying the concept of Binomial Expansion , ${(x + y)^n} = \sum\limits_{k = 0}^n {(\mathop {}\limits_k^n ){x^{n - k}}{y^k}}$ , we get-
${\left( {a{\text{ }} + {\text{ }}b} \right)^n}\; = {\text{ }}{a^n}\; + {\text{ }}\left( {^n{C_1}} \right){a^{n - 1}}b{\text{ }} + {\text{ }}\left( {^n{C_2}} \right){a^{n - 2}}{b^2}\; + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^n{C_{n - 1}}} \right)a{b^{n - 1}}\; + {\text{ }}{b^n}$

$= {}^7{C_0}{(530)^7}{( - 1)^0} + {}^7{C_1}{(530)^6}{( - 1)^1} + {}^7{C_2}{(530)^5}{( - 1)^2} + ...............{}^7{C_7}{(530)^0}{( - 1)^7}$

$= {}^7{C_0}{(530)^7} - {}^7{C_1}{(530)^6} + {}^7{C_2}{(530)^5}{( - 1)^2} + ............... + 1 \times 1 \times ( - 1)$

Since any exponent 0 raised to any number always gives 1 .

$= {}^7{C_0}{(530)^7} - {}^7{C_1}{(530)^6} + {}^7{C_2}{(530)^5}{( - 1)^2} + ............... + 1 \times 1 \times ( - 1)$

$= {}^7{C_0}{(530)^7} - {}^7{C_1}{(530)^6} + {}^7{C_2}{(530)^5}{( - 1)^2} + ............... + {}^7{C_6}{(530)^1}{( - 1)^6} - 1$

$= {}^7{C_0}{(530)^7} - {}^7{C_1}{(530)^6} + {}^7{C_2}{(530)^5}{( - 1)^2} + ............... + {}^7{C_5}{(530)^2}{( - 1)^5} + 7 \times 530 - 1$

$= {}^7{C_0}{(53)^7} \times {(10)^7} - {}^7{C_1}{(53)^6} \times {(10)^6} + ......... - {}^7{C_5}{(53)^2}{(10)^2} + 3710 - 1$

Now as we got our last expansion as multiple of ${\left( {10} \right)^2}$ that is we can say that the whole expansion is going to be ${\left( {10} \right)^2}k$ .

$= {}^7{C_0}{(53)^7} \times {(10)^7} - {}^7{C_1}{(53)^6} \times {(10)^6} + ......... - {}^7{C_5}{(53)^2}{(10)^2} + 3710 - 1$

$= {\left( {10} \right)^2}k + 3710 - 1$

$= {\left( {10} \right)^2}k + 3709$

$= 100 \times k + 3709$

$= k \times 100 + 3709$]

So, the expression comes out to be $= k \times 100 + 3709$. If we see the last two digits $k \times 00 + 09 = k \times 09$
Therefore , the last two digits of the number ${23^{14}}$ are $09$ .
So, the correct answer is “$09$”.

Note : Remember the fact that the binomial coefficients that are at an equal distance from the beginning and the end are considered equal .
Learn that in the expansion of ${(a + b)^n}$ , the total number of terms is equal to $(n + 1)$ .
Keep in mind that in ${(a + b)^n}$ the sum of the exponents of a and b is always considered as n.
Always try to understand the mathematical statement carefully and keep things distinct .
Remember the properties and apply appropriately .
Choose the options wisely , it's better to break the question and then solve part by part .
Cross check the answer and always keep the final answer simplified .