Solve 9^2-sqrt{36}+lfloor40div8+(5*4)+24rfloor

$${ 9 }^{ 2 } - \sqrt{ 36 } + \lfloor 40 \div 8+(5 \times 4)+24 \rfloor$$

Evaluate

$$124$$

Solution Steps

Calculate 9 to the power of 2 and get 81.

$$81-\sqrt{36}+\lfloor \frac{40}{8}+5\times 4+24\rfloor$$

Calculate the square root of 36 and get 6.

$$81-6+\lfloor \frac{40}{8}+5\times 4+24\rfloor$$

Subtract 6 from 81 to get 75.

$$75+\lfloor \frac{40}{8}+5\times 4+24\rfloor$$

Divide 40 by 8 to get 5.

$$75+\lfloor 5+5\times 4+24\rfloor$$

Multiply 5 and 4 to get 20.

$$75+\lfloor 5+20+24\rfloor$$

Add 5 and 20 to get 25.

$$75+\lfloor 25+24\rfloor$$

Add 25 and 24 to get 49.

$$75+\lfloor 49\rfloor$$

The floor of a real number a is the largest integer number less than or equal to a. The floor of 49 is 49.

$$75+49$$

Add 75 and 49 to get 124.

$$124$$

Factor

$$2^{2}\times 31$$