Question 772586
The ages will be either
A) 3 consecutive odd numbers along with the 2 even numbers in between,
or
B) 3 consecutive even numbers and the two odd numbers in between.
 
The prime factorization of 6,720 is
{{{6720=2^6*3*5*7}}}
We cannot regroup the prime factors into 5 factor as described in A) above, because alonng with 3, 5, and 7, we would need 6, and that would require an extra 3 as a prime factor.
WE must use 5 and 7 as our ofdd numbers and that means 4,6,and 8, as our even numbers.
{{{6720=2^6*3*5*7=2^2*5*(2*3)*7*2^3=highlight(4*5*6*7*8)}}} is the product needed.
The children's ages are 4, 5, 6, 7, and 8.
 
We could have said
{{{x}}}= age of the middle child, so
{{{x-2}}}= age of the youngest child,
{{{x-1}}}= age of the second youngest child,
{{{x+1}}} and {{{x+2}}} being the ages of the two oldest ones.
Then the product would be
{{{x(x-1)(x+1)(x-2)(x+2)=x(x^2-1)(x^2-4)}}}
We could write {{{x(x^2-1)(x^2-4)=6720}}} as an equation, but that is not very helpful.
At most we could have said
{{{6720=x(x^2-1)(x^2-4)<x^5}}},
and knowing that {{{5^5=3125<6720}}} would have suggested {{{x>=6}}}.