Solve (27+1)*(2^2+1)*(2^4+1)*(2^8+1)

$$( 27 + 1 ) \times ( 2 ^ { 2 } + 1 ) \times ( 2 ^ { 4 } + 1 ) \times ( 2 ^ { 8 } + 1 )$$

Evaluate

$$611660$$

Solution Steps

Add 27 and 1 to get 28.

$$28\left(2^{2}+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)$$

Calculate 2 to the power of 2 and get 4.

$$28\left(4+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)$$

Add 4 and 1 to get 5.

$$28\times 5\left(2^{4}+1\right)\left(2^{8}+1\right)$$

Multiply 28 and 5 to get 140.

$$140\left(2^{4}+1\right)\left(2^{8}+1\right)$$

Calculate 2 to the power of 4 and get 16.

$$140\left(16+1\right)\left(2^{8}+1\right)$$

Add 16 and 1 to get 17.

$$140\times 17\left(2^{8}+1\right)$$

Multiply 140 and 17 to get 2380.

$$2380\left(2^{8}+1\right)$$

Calculate 2 to the power of 8 and get 256.

$$2380\left(256+1\right)$$

Add 256 and 1 to get 257.

$$2380\times 257$$

Multiply 2380 and 257 to get 611660.

$$611660$$

Factor

$$2^{2}\times 5\times 7\times 17\times 257$$