The last two digits of 2015! + 3^{2015} are: | vedclass

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(D) We need to find the last two digits of $2015! + 3^{2015}$.Since $2015!$ contains factors like $2, 5, 10, 20$,etc.,it is a multiple of $100$ for an

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$07$

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(D) We need to find the last two digits of $2015! + 3^{2015}$.Since $2015!$ contains factors like $2, 5, 10, 20$,etc.,it is a multiple of $100$ for any $n \ge 10$. Thus,the last two digits of $2015!$ are $00$.Now,we find the last two digits of $3^{2015}$ by calculating $3^{2015} \pmod{100}$.Using Euler's totient theorem,$\phi(100) = 100(1 - 1/2)(1 - 1/5) = 40$.Since $\gcd(3, 100) = 1$,we have $3^{40} \equiv 1 \pmod{100}$.$2015 = 40 	imes 50 + 15$,so $3^{2015} \equiv 3^{15} \pmod{100}$.$3^4 = 81 \equiv -19 \pmod{100}$.$3^8 = (81)^2 = 6561 \equiv 61 \pmod{100}$.$3^{12} = 3^8 	imes 3^4 = 61 	imes 81 = 4941 \equiv 41 \pmod{100}$.$3^{15} = 3^{12} 	imes 3^3 = 41 	imes 27 = 1107 \equiv 07 \pmod{100}$.Therefore,the last two digits are $07$.

The last two digits of $2015! + 3^{2015}$ are:

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