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

01

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03

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09

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none of these

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Solution

The correct option is A $09$
$(23)^{14}$
$= (23^{2})^{7}$
$= (529)^{7}$
$= (530 - 1)^{7}$
$= - (1 - 530)^{7}$
$= - [1 -^{7} C_{1} (530)^{1} +^{7} C_{2} (530)^{2} . . . -^{7} C_{7} (530)^{7}]$
$= - 1 +^{7} C_{1} (530)^{1} -^{7} C_{2} (530)^{2} . . . +^{7} C_{7} (530)^{7}$
$= - 1 + (3710) -^{7} C_{2} (530)^{2} . . . +^{7} C_{7} (530)^{7}$
Hence ending digit will be given by
$3710 - 1$
$= 3709$
Therefore the last two digits of $23^{14}$ is $09.$

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The last two digits of the number 2314 are

The correct option is A 09(23)14=(232)7=(529)7=(530−1)7=−(1−530)7=−[1−7C1(530)1+7C2(530)2...−7C7(530)7]=−1+7C1(530)1−7C2(530)2...+7C7(530)7=−1+(3710)−7C2(530)2...+7C7(530)7Hence ending digit will be given by3710−1=3709Therefore the last two digits of 2314 is 09.

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