Solve 13+[6+2*(2^34+3*2)-72]+9div3

$$13 + [ 6 + 2 \cdot ( 2 ^ { 3 } 4 + 3 \cdot 2 ) - 72 ] + 9 \div 3$$

Evaluate

$$26$$

Solution Steps

Add 13 and 6 to get 19.

$$19+2\left(2^{3}\times 4+3\times 2\right)-72+\frac{9}{3}$$

Calculate 2 to the power of 3 and get 8.

$$19+2\left(8\times 4+3\times 2\right)-72+\frac{9}{3}$$

Multiply 8 and 4 to get 32.

$$19+2\left(32+3\times 2\right)-72+\frac{9}{3}$$

Multiply 3 and 2 to get 6.

$$19+2\left(32+6\right)-72+\frac{9}{3}$$

Add 32 and 6 to get 38.

$$19+2\times 38-72+\frac{9}{3}$$

Multiply 2 and 38 to get 76.

$$19+76-72+\frac{9}{3}$$

Add 19 and 76 to get 95.

$$95-72+\frac{9}{3}$$

Subtract 72 from 95 to get 23.

$$23+\frac{9}{3}$$

Divide 9 by 3 to get 3.

$$23+3$$

Add 23 and 3 to get 26.

$$26$$

Factor

$$2\times 13$$