Solve 6^3+lfloor8^2+(3^4+7^2)+1+{2}^4rfloor-6

$${ 6 }^{ 3 } + \lfloor { 8 }^{ 2 } +( { 3 }^{ 4 } + { 7 }^{ 2 } )+1+ { 2 }^{ 4 } \rfloor -6$$

Evaluate

$$421$$

Solution Steps

Calculate 6 to the power of 3 and get 216.

$$216+\lfloor 8^{2}+3^{4}+7^{2}+1+2^{4}\rfloor -6$$

Calculate 8 to the power of 2 and get 64.

$$216+\lfloor 64+3^{4}+7^{2}+1+2^{4}\rfloor -6$$

Calculate 3 to the power of 4 and get 81.

$$216+\lfloor 64+81+7^{2}+1+2^{4}\rfloor -6$$

Add 64 and 81 to get 145.

$$216+\lfloor 145+7^{2}+1+2^{4}\rfloor -6$$

Calculate 7 to the power of 2 and get 49.

$$216+\lfloor 145+49+1+2^{4}\rfloor -6$$

Add 145 and 49 to get 194.

$$216+\lfloor 194+1+2^{4}\rfloor -6$$

Add 194 and 1 to get 195.

$$216+\lfloor 195+2^{4}\rfloor -6$$

Calculate 2 to the power of 4 and get 16.

$$216+\lfloor 195+16\rfloor -6$$

Add 195 and 16 to get 211.

$$216+\lfloor 211\rfloor -6$$

The floor of a real number a is the largest integer number less than or equal to a. The floor of 211 is 211.

$$216+211-6$$

Add 216 and 211 to get 427.

$$427-6$$

Subtract 6 from 427 to get 421.

$$421$$

Factor

$$421$$