Solve 6(7+1)(7^2+1)(7^4+1)(1^8+1)+1

$$6 ( 7 + 1 ) ( 7 ^ { 2 } + 1 ) ( 7 ^ { 4 } + 1 ) ( 1 ^ { 8 } + 1 ) + 1$$

Evaluate

$$11529601$$

Solution Steps

Add 7 and 1 to get 8.

$$6\times 8\left(7^{2}+1\right)\left(7^{4}+1\right)\left(1^{8}+1\right)+1$$

Multiply 6 and 8 to get 48.

$$48\left(7^{2}+1\right)\left(7^{4}+1\right)\left(1^{8}+1\right)+1$$

Calculate 7 to the power of 2 and get 49.

$$48\left(49+1\right)\left(7^{4}+1\right)\left(1^{8}+1\right)+1$$

Add 49 and 1 to get 50.

$$48\times 50\left(7^{4}+1\right)\left(1^{8}+1\right)+1$$

Multiply 48 and 50 to get 2400.

$$2400\left(7^{4}+1\right)\left(1^{8}+1\right)+1$$

Calculate 7 to the power of 4 and get 2401.

$$2400\left(2401+1\right)\left(1^{8}+1\right)+1$$

Add 2401 and 1 to get 2402.

$$2400\times 2402\left(1^{8}+1\right)+1$$

Multiply 2400 and 2402 to get 5764800.

$$5764800\left(1^{8}+1\right)+1$$

Calculate 1 to the power of 8 and get 1.

$$5764800\left(1+1\right)+1$$

Add 1 and 1 to get 2.

$$5764800\times 2+1$$

Multiply 5764800 and 2 to get 11529600.

$$11529600+1$$

Add 11529600 and 1 to get 11529601.

$$11529601$$

Factor

$$23\times 501287$$