Solve 35divlceil(54div6+48div4-19)div2+2*(24-2*11)+2rceil+(75-15*3)div(24div4

$$35 \div \lceil (54 \div 6+48 \div 4-19) \div 2+2 \times (24-2 \times 11)+2 \rceil +(75-15 \times 3) \div (24 \div 4$$

Evaluate

$$10$$

Solution Steps

Divide 54 by 6 to get 9.

$$\frac{35}{\lceil \frac{9+\frac{48}{4}-19}{2}+2\left(24-2\times 11\right)+2\rceil }+\frac{75-15\times 3}{\frac{24}{4}}$$

Divide 48 by 4 to get 12.

$$\frac{35}{\lceil \frac{9+12-19}{2}+2\left(24-2\times 11\right)+2\rceil }+\frac{75-15\times 3}{\frac{24}{4}}$$

Add 9 and 12 to get 21.

$$\frac{35}{\lceil \frac{21-19}{2}+2\left(24-2\times 11\right)+2\rceil }+\frac{75-15\times 3}{\frac{24}{4}}$$

Subtract 19 from 21 to get 2.

$$\frac{35}{\lceil \frac{2}{2}+2\left(24-2\times 11\right)+2\rceil }+\frac{75-15\times 3}{\frac{24}{4}}$$

Divide 2 by 2 to get 1.

$$\frac{35}{\lceil 1+2\left(24-2\times 11\right)+2\rceil }+\frac{75-15\times 3}{\frac{24}{4}}$$

Multiply 2 and 11 to get 22.

$$\frac{35}{\lceil 1+2\left(24-22\right)+2\rceil }+\frac{75-15\times 3}{\frac{24}{4}}$$

Subtract 22 from 24 to get 2.

$$\frac{35}{\lceil 1+2\times 2+2\rceil }+\frac{75-15\times 3}{\frac{24}{4}}$$

Multiply 2 and 2 to get 4.

$$\frac{35}{\lceil 1+4+2\rceil }+\frac{75-15\times 3}{\frac{24}{4}}$$

Add 1 and 4 to get 5.

$$\frac{35}{\lceil 5+2\rceil }+\frac{75-15\times 3}{\frac{24}{4}}$$

Add 5 and 2 to get 7.

$$\frac{35}{\lceil 7\rceil }+\frac{75-15\times 3}{\frac{24}{4}}$$

The ceiling of a real number a is the smallest integer number greater than or equal to a. The ceiling of 7 is 7.

$$\frac{35}{7}+\frac{75-15\times 3}{\frac{24}{4}}$$

Divide 35 by 7 to get 5.

$$5+\frac{75-15\times 3}{\frac{24}{4}}$$

Multiply 15 and 3 to get 45.

$$5+\frac{75-45}{\frac{24}{4}}$$

Subtract 45 from 75 to get 30.

$$5+\frac{30}{\frac{24}{4}}$$

Divide 24 by 4 to get 6.

$$5+\frac{30}{6}$$

Divide 30 by 6 to get 5.

$$5+5$$

Add 5 and 5 to get 10.

$$10$$

Factor

$$2\times 5$$