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

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 –
$= {[{\left( {23} \right)^2}]^7}$
$= {^7}$
$= {\left( {530 - 1} \right)^7}$
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 .