Question

a) {2^4 + 7× [31 + 10 × (3^7: 3^5 + 3^2: 3^1 ) - 49]} : 73; b) {[(2^15×3^13×5^7): (2^13 × 3^12×5^5 ) – 2^4×3: 8] : 49-2}:2019^0 c) (3 · 10°3 + 2×10^2 + 4): 3^2× 2 + [20 + (202 : 2 - 51) - 2019^0].​

Answer 1

a)

$({2^4 + 7× [31 + 10 × (3^7: 3^5 + 3^2: 3^1 ) - 49]} ): 73 =$

${(2^4 + 7× [31 + 10 × (3^2 + 3^1 ) - 49])} : 73 =$

$[{2^4 + 7× (31 + 10 × 12 - 49)]} : 73 =$

$[{2^4 + 7× (31 + 120 - 49)]} : 73 =$

$[{2^4 + 7× (151 - 49)]} : 73 =$

$({2^4 + 7× 102)} : 73 =$

$({16 + 714)} : 73 = 730 \div 73 = 10$

b)

$({[( {2}^{15} × {3}^{13} ×5^7): ( {2}^{13} × {3}^{12} ×5^5 ) – 2^4×3: 8] : 49-2}):2019^0 =$

${[ ({2}^{15 - 13} × {3}^{13 - 12} × {5}^{7 - 5} – 16×3: 8): 49-2] }:2019^0 =$

${[ ( {2}^{2} × {3}^{1} × {5}^{2} – 48: 8): 49-2] }:1 =$

${[ ( 4 ×3 × 25 – 6): 49-2] }:1 =$

${[ ( 300 – 6): 49-2] }:1 =$

$(294 \div 49 - 2) \div 1 = (6 - 2) \div 1 = 4$

c)

$(3 · {10}^{3} + 2×10^2 + 4): 3^2× 2 + [20 + (202 : 2 - 51) - 2019^0]=$

$(3000+ 200 + 4): 9× 2 + [20 + (101 - 51) - 1]=$

$3204: 9× 2 + (20 + 50 - 1)=$

$356× 2 + (70 - 1)= 712 + 69 = 781$