Question 983976
"I am the number which will make a rectangle 3 times wide" may mean that
if the number is the length of a rectangle,
it will be twice the whole number measurement of the width of that rectangle.
In other words, I take that to mean that the number is a multiple of {{{3}}} .
That means {{{3}}} is a factor of the mystery number.


"I am the number which has one of my factor as 4" means that {{{4=2^2}}} is a factor of the mystery number.


"I am the number which is not a multiple of 5 or 7" tells us that 5 and 7 are not prime factors of that mystery number, but it is not a particularly useful clue.
Other prime numbers could be factors, like 11, 13, 17, 19, 23, etc .


"I am the number which has exactly 8 factors" is an important clue.
A number {{{N}}} that has {{{4=2^2}}} and {{{3}}} , but not {{{5}}} or {{{7}}} as factors
has a prime factorization of the form
{{{N=2^a*3^b*11^c*13^d*17^e*"..."}}} , with {{{a>=2}}} and {{{b>=1}}}.
The factors of {{{N}}} , with their prime factorizations, range from
{{{1=2^0*3^0*11^0*13^0*17^0*"..."}}} to {{{N=2^a*3^b*11^c*13^d*17^e*"..."}}} .
There are
{{{a+1>=2+1=3}}} different possible exponents for {{{2}}} ,
{{{b+1.=1+1=2}}} different possible exponents for {{{3}}} ,
{{{c+1}}} different possible exponents for {{{11}}} ,
{{{d+1}}} different possible exponents for {{{13}}} ,
and so on.
With all the combined possible choices, we can make
{{{(a+1)*(b+1)*(c+1)*(d+1)*(e+1)*"..."}}} factors.
If {{{N}}} has exactly {{{8=4*2}}} factors,
knowing that {{{a+1>=3}}} and {{{b+1>=2}}} ,
it is obvious that it must be
{{{a+1=4}}}<--->{{{a=4-1=3}}} ,
{{{b+1=2}}}<--->{{{b=2-1=1}}} ,
{{{c+1=1}}}<--->{{{c=0}}} ,
{{{d+1=1}}}<--->{{{d=0}}} ,
{{{e+1=1}}}<--->{{{e=0}}} , and so on.
So, {{{N=2^3*3^1*11^0*13^0*17^0*"..."=2^3*3=8*3=highlight(24)}}} .