Solve 2+0:(2^{4}+3*5-3^2)-[2^2*5+(2^3*7*3*0)-2*3^4]

$$2 + 0 : ( 2 ^ { 4 } + 3 \cdot 5 - 3 ^ { 2 } ) - [ 2 ^ { 2 } \cdot 5 + ( 2 ^ { 3 } \cdot 7 \cdot 3 \cdot 0 ) - 2 \cdot 3 ^ { 4 } ]$$

Evaluate

$$144$$

Solution Steps

Calculate 2 to the power of 4 and get 16.

$$2+\frac{0}{16+3\times 5-3^{2}}-\left(2^{2}\times 5+2^{3}\times 7\times 3\times 0-2\times 3^{4}\right)$$

Multiply 3 and 5 to get 15.

$$2+\frac{0}{16+15-3^{2}}-\left(2^{2}\times 5+2^{3}\times 7\times 3\times 0-2\times 3^{4}\right)$$

Add 16 and 15 to get 31.

$$2+\frac{0}{31-3^{2}}-\left(2^{2}\times 5+2^{3}\times 7\times 3\times 0-2\times 3^{4}\right)$$

Calculate 3 to the power of 2 and get 9.

$$2+\frac{0}{31-9}-\left(2^{2}\times 5+2^{3}\times 7\times 3\times 0-2\times 3^{4}\right)$$

Subtract 9 from 31 to get 22.

$$2+\frac{0}{22}-\left(2^{2}\times 5+2^{3}\times 7\times 3\times 0-2\times 3^{4}\right)$$

Zero divided by any non-zero number gives zero.

$$2+0-\left(2^{2}\times 5+2^{3}\times 7\times 3\times 0-2\times 3^{4}\right)$$

Add 2 and 0 to get 2.

$$2-\left(2^{2}\times 5+2^{3}\times 7\times 3\times 0-2\times 3^{4}\right)$$

Calculate 2 to the power of 2 and get 4.

$$2-\left(4\times 5+2^{3}\times 7\times 3\times 0-2\times 3^{4}\right)$$

Multiply 4 and 5 to get 20.

$$2-\left(20+2^{3}\times 7\times 3\times 0-2\times 3^{4}\right)$$

Calculate 2 to the power of 3 and get 8.

$$2-\left(20+8\times 7\times 3\times 0-2\times 3^{4}\right)$$

Multiply 8 and 7 to get 56.

$$2-\left(20+56\times 3\times 0-2\times 3^{4}\right)$$

Multiply 56 and 3 to get 168.

$$2-\left(20+168\times 0-2\times 3^{4}\right)$$

Multiply 168 and 0 to get 0.

$$2-\left(20+0-2\times 3^{4}\right)$$

Add 20 and 0 to get 20.

$$2-\left(20-2\times 3^{4}\right)$$

Calculate 3 to the power of 4 and get 81.

$$2-\left(20-2\times 81\right)$$

Multiply 2 and 81 to get 162.

$$2-\left(20-162\right)$$

Subtract 162 from 20 to get -142.

$$2-\left(-142\right)$$

The opposite of -142 is 142.

$$2+142$$

Add 2 and 142 to get 144.

$$144$$

Factor

$$2^{4}\times 3^{2}$$