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

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

Calcular

$$18446744073709551616$$

Pasos da solución

Suma 2 e 1 para obter 3.

$$3\left(2^{2}+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Calcula 2 á potencia de 2 e obtén 4.

$$3\left(4+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Suma 4 e 1 para obter 5.

$$3\times 5\left(2^{4}+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Multiplica 3 e 5 para obter 15.

$$15\left(2^{4}+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Calcula 2 á potencia de 4 e obtén 16.

$$15\left(16+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Suma 16 e 1 para obter 17.

$$15\times 17\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Multiplica 15 e 17 para obter 255.

$$255\left(2^{8}+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Calcula 2 á potencia de 8 e obtén 256.

$$255\left(256+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Suma 256 e 1 para obter 257.

$$255\times 257\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Multiplica 255 e 257 para obter 65535.

$$65535\left(2^{16}+1\right)\left(2^{32}+1\right)+1$$

Calcula 2 á potencia de 16 e obtén 65536.

$$65535\left(65536+1\right)\left(2^{32}+1\right)+1$$

Suma 65536 e 1 para obter 65537.

$$65535\times 65537\left(2^{32}+1\right)+1$$

Multiplica 65535 e 65537 para obter 4294967295.

$$4294967295\left(2^{32}+1\right)+1$$

Calcula 2 á potencia de 32 e obtén 4294967296.

$$4294967295\left(4294967296+1\right)+1$$

Suma 4294967296 e 1 para obter 4294967297.

$$4294967295\times 4294967297+1$$

Multiplica 4294967295 e 4294967297 para obter 18446744073709551615.

$$18446744073709551615+1$$

Suma 18446744073709551615 e 1 para obter 18446744073709551616.

$$18446744073709551616$$

Factorizar

$$2^{64}$$