Question 1016707
a) The ordinal number of the first ball with all 3 designs is the least common multiple of {{{red(18)}}} , {{{blue(15)}}} , and {{{green(30)}}} .
That number is {{{highlight(90)=3*green(30)=6*blue(15)=5*red(18)}}} .
The {{{highlight(90th)}}} ball has all 3 designs.
How do we find it? Let me count the ways.
1) We could start checking increasing multiples of {{{green(30)}}} until we find one that is also a multiple of {{{red(18)}}} , and {{{blue(15)}}} .
{{{1*green(30)=2*blue(15)}}} is already a multiple of {{{blue(15)}}} ,
so all multiples of {{{green(30)}}} will be multiple of {{{blue(15)}}} ,
but {{{1*green(30)}}} is not a multiple of {{{red(18)}}} ,
and neither is {{{2*green(30)=60}}} ,
but {{{3*green(30)=5*red(18)=highlight(90)}}} is the first common multiple of {{{red(18)}}} , {{{blue(15)}}} , and {{{green(30)}}} that we find.
2) We could start from the prime factorizations:
{{{red(18)=2*9=2*3^2}}} ,
{{{blue(15)=3*5}}} , and
{{{green(30)=3*10=2*5*3}}} ,
so the least common multiple must contain the prime factors {{{2}}} , {{{3}}} , and {{{5}}} ,
and we need the {{{3}}} to be squared, as it is in {{{red(18)=2*3^2}}} ,
so the least common multiple is {{{2*3^2*5=2*9*5=10*9=highlight(90)}}} .
 
b) We already saw in 1) above that all multiples of {{{green(30)}}} are multiples of {{{blue(15)}}} ,
so the balls number {{{green(30)=2*blue(15)}}} and {{{60=2*green(30)=4*blue(15)}}} have two designs:
polka dots, and stars, but not stripes.
So by the time the first ball with all three designs, the 90th ball, is printed,
at least {{{highlight(2)}}} balls, the 30th and the 60th one, will have been printed with only two designs:
polka dots, and stars, but not stripes.
Are there any other balls before the 90th ball that have two designs?
Maybe stripes and polka dots, or stripes and stars?
NO, there is none.
Any multiple of {{{red(18)=2*9=2*3^2}}} , and {{{blue(15)=3*5}}} ,
or {{{red(18)=2*9=2*3^2}}} , and {{{green(30)=3*10=2*5*3}}} ,
must be a multiple of all three numbers.
It must have {{{2}}} , {{{3^2}}} , and {{{5}}} in its factorization,
so it will be a multiple of {{{2*3^2*5=90}}} .
If a ball has stripes and one another design printed, it has all three designs.