Question
Write last two digits of the number 3400.
Write last two digits of the number 3400.
$3^{400}=(9)^{200}$
$=(10-1)^{200}$
$={^\text{200}}\text{C}_{\text{0}}(10)^{200}+{^\text{200}}\text{C}_{\text{1}}(10)^{199}(-1)^{1}.....\\+{^\text{200}}\text{C}_{\text{108}}(10)^{2}(-1)^{198}+{^\text{200}}\text{C}_{\text{199}}(10)^{1}(-1)^{199}+{^\text{200}}\text{C}_{\text{200}}(-1)^{200}$
$=100\Big[(10)^{198}+{^\text{200}}\text{C}_{\text{1}}(10)^{197}(-1)^{1}+......+{^\text{200}}\text{C}_{\text{198}}(-1)^{198}\Big]+200(10)^{199}+(-1)^{200}$
$=100\Big[(10)^{198}-{^\text{200}}\text{C}_{\text{1}}(10)^{197}+......+{^\text{200}}\text{C}_{\text{198}}-2(10)\Big]+1$
$=100(\text{a natural number})+1$
Hence, last two digits of the number 3400 is 01.
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