Smallest integer with 21 factors??

1. Factorize $d$ as a product of his prime divisors: ${ { d = p }_{ 1 } }^{ { a }_{ 1 } } { { \times p }_{ 2 } }^{ { a }_{ 2 } } { { \times p }_{ 3 } }^{ { a }_{ 3 } } { { \times }... }$

2. convert this factorization in another arithmetically equivalent factorization, composed of non-powered monotonically decreasing and not necesarilly prime factors... ${ { d = p }_{ 1 } }^{ { a }_{ 1 } } { { \times p }_{ 2 } }^{ { a }_{ 2 } } { { \times p }_{ 3 } }^{ { a }_{ 3 } } { { \times }... } = { { d = b }_{ 1 } \times }{ b }_{ 2 } \times { b }_{ 3 } \times ...$ such that ${ { b }_{ 1 } \ge }{ b }_{ 2 } \ge { b }_{ 3 }\ge ...$ You must realize that for every given d, there are several arithmetically equivalent factorisations that can be done: by example: In this case $d = 21$ then there are 2 equivalent factorisations: $d=7\times 3=21$

3. $N$ is the minimal number resulting of computing ${ 2 }^{ { b }_{ 1 }-1 } \times { 3 }^{ { b }_{ 2 }-1 } \times 5^{ { b }_{ 3 }-1 } \times ...$ for all the equivalent factorizations of $d$ . Working the same example: $N(21) =$ the minimal of these $\{ { 2 }^{ 6 } \times { 3 }^{ 2 } , { 2 }^{ 20 }\} ={ 2 }^{ 6 } \times { 3 }^{ 2 } =576$

that's 2^6 x 3^2 = 576