Question

N=3la201+5la609+6la309;Ultima cifra a Nr=?

Answer 1

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Answer 2

$N=3^{201} +5 ^{609}+6^{309} \\ \\ U( 3^{201}) =u(3 ^{50*4+1})=3 \\ \\ U( 5^{609}) =6 \\ \\ U( 6^{309}) =5 \\ \\ N=( 3^{201} +5 ^{609}+6^{309})=3+6+5=14$

ultima cifra fiind 4