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Question
Write last two digits of the number 3400.
Solution
\[3^{400} = \left( 9 \right)^{200} \]
\[ = \left( 10 - 1 \right)^{200} \]
\[ =^{200} C_0 \left( 10 \right)^{200} +^{200} C_1 \left( 10 \right)^{199} \left( - 1 \right)^1 + . . . . . +^{200} C_{198} \left( 10 \right)^2 \left( - 1 \right)^{198} +^{200} C_{199} \left( 10 \right)^1 \left( - 1 \right)^{199} +^{200} C_{200} \left( - 1 \right)^{200} \]
\[ = 100\left[ \left( 10 \right)^{198} +^{200} C_1 \left( 10 \right)^{197} \left( - 1 \right)^1 + . . . . . +^{200} C_{198} \left( - 1 \right)^{198} \right] + 200 \left( 10 \right)^1 \left( - 1 \right)^{199} + \left( - 1 \right)^{200} \]
\[ = 100\left[ \left( 10 \right)^{198} -^{200} C_1 \left( 10 \right)^{197} + . . . . . +^{200} C_{198} - 2\left( 10 \right) \right] + 1\]
\[ = 100(\text{ a natural number } ) + 1\]
Hence, last two digits of the number 3400 is 01.
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