The product of the last two digits of (1919)^ | vedclass

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(D) We need to find the last two digits of $(1919)^{1919}$.This is equivalent to finding $(1919)^{1919} \pmod{100}$.$(1919)^{1919} \equiv (19)^{1919}

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(D) We need to find the last two digits of $(1919)^{1919}$.This is equivalent to finding $(1919)^{1919} \pmod{100}$.$(1919)^{1919} \equiv (19)^{1919} \pmod{100}$.Since $19^2 = 361 \equiv 61 \pmod{100}$,$19^4 = (61)^2 = 3721 \equiv 21 \pmod{100}$.$19^8 = (21)^2 = 441 \equiv 41 \pmod{100}$.$19^{10} = 19^8 	imes 19^2 = 41 	imes 61 = 2501 \equiv 01 \pmod{100}$.Thus,$(19)^{10} \equiv 1 \pmod{100}$.Now,$19^{1919} = (19^{10})^{191} 	imes 19^9 \equiv 1^{191} 	imes 19^9 \pmod{100}$.$19^9 = 19^8 	imes 19 = 41 	imes 19 = 779 \equiv 79 \pmod{100}$.The last two digits are $79$.The product of the last two digits is $7 	imes 9 = 63$.

The product of the last two digits of $(1919)^{1919}$ is . . . . . . .

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