The Missing Divisor

$784={ 2 }^{ 4 }\times { 7 }^{ 2 }\\ We\quad deduce\quad that\quad N={ 2 }^{ a }\times { 7 }^{ b }\quad where\quad (a+1)(b+1)=6\quad because\quad it\quad has\quad 6\quad factors.\\ Let\quad the\quad product\quad of\quad the\quad six\quad factors\quad be\quad P\\ P={ 2 }^{ 0 }({ 7 }^{ 0 })\times { 2 }^{ 0 }({ 7 }^{ 1 })\times ...\times { 2 }^{ 0 }({ 7 }^{ b })\times { 2 }^{ 1 }({ 7 }^{ 0 })\times ...\times { 2 }^{ 1 }({ 7 }^{ b })\times ...\times { 2 }^{ a }({ 7 }^{ b })\\ \quad ={ 2 }^{ (b+1)\frac { a }{ 2 } (a+1) }\times { 7 }^{ (a+1)\frac { b }{ 2 } (b+1) }\\ \quad ={ 2 }^{ 3a }\times { 7 }^{ 3b }\\ \Rightarrow P\quad is\quad a\quad perfect\quad cube\Rightarrow P={ 2 }^{ 6 }\times { 7 }^{ 3 }={ 2 }^{ 3(2) }\times { 7 }^{ 3(1) }\quad since\quad (2+1)(1+1)=6\\ Hence\quad the\quad missing\quad divisor\quad is\quad P\div 784=\boxed { 28 }$

Let the six divisors of N be a, b, c, d, e, and f. Then, N = ab = cd = ef and N^3 = abcdef. Since we know that the product of five of them is 784, abcde = 784 = (N^2)(e). Factoring 784, we have 784 = (2^4)(7^2). N^3 = abcdef = 784f. The value of f that would make 784f a perfect cube is (2^2)(7) = 28.

First write out 784 as a product of 1 and its prime factors

$784=1\times2\times2\times2\times2\times7\times7$

One way to group these as a product of five different factors is

$784=1\times2\times4\times14\times7$

It is easy to see that this is the only way to combine these prime factors into a product of five different numbers.

So now we know that five of the divisors of N are 1,2,4,7 and 14.

The remaining factor is N itself, which must be the smallest non-trivial multiple of each of these divisors.

And so $N=\boxed{28}$

784 = 2x2x4x7x7, so prime factors are 2 and 7, now N has 6 divisors which can be expressed as (1+2)(1+1) so one prime number has power 2 and other has 1 but 784 = 2x2x4x7x7 so N = 2^2 × 7. Its divisors are 1, 2, 4, 7, 2×7, 4×7. You can observe that missing number is 4×7 = 28.