Question

${3}^{5} \times {27}^{2} \div {3}^{10} + ({5}^{3} ) ^{7} \div {25}^{10} - 8 = \\ {5}^{1000} + {5}^{1001} + {5}^{1002} divizibil \: cu \: 31$

Answer 1

3^5 × 27^2 : 3^10 + (5^3)^7 : 25^10 - 8 =

3^5 × (3^3)^2 : 3^10 + 5^21 : (5^2)^10 - 8 =

3^5 × 3^6 : 3^10 + 5^21 : 5^20 - 8

3^11 : 3^10 + 5^1 - 8 =

3 + 5 - 8 =

0
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5^1000 + 5^1001 + 5^1002 =

5^1000 × (1 + 5^1 + 5^2) =

5^1000 × 31 divizibil cu 31
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