Resolver (2^{2}+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1

$$( 2 ^ { 2 } + 1 ) ( 2 ^ { 4 } + 1 ) ( 2 ^ { 8 } + 1 ) ( 2 ^ { 16 } + 1 ) ( 2 ^ { 32 } + 1 ) + 1$$

Calcular

$$6148914691236517206$$

Pasos da solución

Calcula 2 á potencia de 2 e obtén 4.

$$\left(4+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Suma 4 e 1 para obter 5.

$$5\left(2^{4}+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Calcula 2 á potencia de 4 e obtén 16.

$$5\left(16+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Suma 16 e 1 para obter 17.

$$5\times 17\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Multiplica 5 e 17 para obter 85.

$$85\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Calcula 2 á potencia de 8 e obtén 256.

$$85\left(256+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Suma 256 e 1 para obter 257.

$$85\times 257\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Multiplica 85 e 257 para obter 21845.

$$21845\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Calcula 2 á potencia de 16 e obtén 65536.

$$21845\left(65536+1\right)\left(2^{32}+1\right)+1$$

Suma 65536 e 1 para obter 65537.

$$21845\times 65537\left(2^{32}+1\right)+1$$

Multiplica 21845 e 65537 para obter 1431655765.

$$1431655765\left(2^{32}+1\right)+1$$

Calcula 2 á potencia de 32 e obtén 4294967296.

$$1431655765\left(4294967296+1\right)+1$$

Suma 4294967296 e 1 para obter 4294967297.

$$1431655765\times 4294967297+1$$

Multiplica 1431655765 e 4294967297 para obter 6148914691236517205.

$$6148914691236517205+1$$

Suma 6148914691236517205 e 1 para obter 6148914691236517206.

$$6148914691236517206$$

Factorizar

$$2\times 3^{2}\times 19\times 43\times 5419\times 77158673929$$