The last two digits of the number 3^{400} are | vedclass

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(D) We need to find the last two digits of $3^{400}$,which is equivalent to finding $3^{400} \pmod{100}$.$3^{400} = (3^4)^{100} = 81^{100}$.Using the

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$01$

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(D) We need to find the last two digits of $3^{400}$,which is equivalent to finding $3^{400} \pmod{100}$.$3^{400} = (3^4)^{100} = 81^{100}$.Using the Binomial Theorem,we can write $81^{100} = (1 + 80)^{100}$.Expanding this using the binomial expansion: $(1 + 80)^{100} = \binom{100}{0} + \binom{100}{1} 	imes 80 + \binom{100}{2} 	imes 80^2 + \dots + \binom{100}{100} 	imes 80^{100}$.$= 1 + 100 	imes 80 + \frac{100 	imes 99}{2} 	imes 6400 + \dots$$= 1 + 8000 + (	ext{terms containing } 80^2, 80^3, \dots)$.Since $80^2 = 6400$,all terms from the third term onwards are divisible by $100$.Thus,$3^{400} \equiv 1 + 8000 \pmod{100}$.$3^{400} \equiv 1 + 0 \pmod{100}$.Therefore,the last two digits are $01$.

The last two digits of the number $3^{400}$ are:

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