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

seo-qna
Last updated date: 23rd Jul 2024
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Answer
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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 –
\[   = {[{\left( {23} \right)^2}]^7} \]
\[   = {[529]^7} \]
\[   = {\left( {530 - 1} \right)^7}  \]
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 .