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Question Number 134037 by benjo_mathlover last updated on 26/Feb/21

Answered by john_santu last updated on 27/Feb/21

49^(303) .3993^(202) .39^(606) = (7^2 )^(303) . 3^(606) .13^(606) .3^(202) .11^(606) = (7×13×11)^(606) . 3^(808) = 1001^(606) .(3^(200) )^4 .3^8 [ note by 3^(400) ≡ 1 mod 1000 ] ≡ 1×1×3^8 (mod 1000) ≡ 6561 (mod 1000) ≡ 561 (mod 1000). Hence it′s the last three digits are 561.

49303.3993202.39606=(72)303.3606.13606.3202.11606=(7×13×11)606.3808=1001606.(3200)4.38[noteby3400≡1mod1000]≡1×1×38(mod1000)≡6561(mod1000)≡561(mod1000).Henceit′sthelastthreedigitsare561.

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