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

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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.

**:**

__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 –

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

\[ = {[529]^7} \]

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

\[{\left( {a{\text{ }} + {\text{ }}b} \right)^n}\; = {\text{ }}{a^n}\; + {\text{ }}\left( {^n{C_1}} \right){a^{n - 1}}b{\text{ }} + {\text{ }}\left( {^n{C_2}} \right){a^{n - 2}}{b^2}\; + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^n{C_{n - 1}}} \right)a{b^{n - 1}}\; + {\text{ }}{b^n}\]

$= {}^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 .

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