Solve (100-99)^102div(199-200)^211+(199-399)^2

$$( 100 - 99 ) ^ { 102 } \div ( 199 - 200 ) ^ { 211 } + ( 199 - 399 ) ^ { 2 }$$

Evaluate

$$39999$$

Solution Steps

Subtract 99 from 100 to get 1.

$$\frac{1^{102}}{\left(199-200\right)^{211}}+\left(199-399\right)^{2}$$

Calculate 1 to the power of 102 and get 1.

$$\frac{1}{\left(199-200\right)^{211}}+\left(199-399\right)^{2}$$

Subtract 200 from 199 to get -1.

$$\frac{1}{\left(-1\right)^{211}}+\left(199-399\right)^{2}$$

Calculate -1 to the power of 211 and get -1.

$$\frac{1}{-1}+\left(199-399\right)^{2}$$

Divide 1 by -1 to get -1.

$$-1+\left(199-399\right)^{2}$$

Subtract 399 from 199 to get -200.

$$-1+\left(-200\right)^{2}$$

Calculate -200 to the power of 2 and get 40000.

$$-1+40000$$

Add -1 and 40000 to get 39999.

$$39999$$

Factor

$$3\times 67\times 199$$