Question 1066754
THE IDEA:
To be divisible by the relatively prime numbers {{{8}}} and {{{9}}} ,
a number has to be divisible by {{{8*9=72}}}
 
THE HARD WAY:
Dividing {{{"111,111,000"}}} by 72,
we get {{{"1,543,208"}}} as the quotient,
with {{{24}}} as a remainder,
so {{{"1,543,208"*72}}} is less than {{{"111,111,000"}}} ,
but the next multiple of {{{72}}} is
{{{("1,543,208"+1)*72="1,543,209"*72=highlight("111,111,048")}}} .
 
EASIER:
Dividing {{{"111,111,000"}}} by 72,
we get {{{"1,543,208"}}} as the quotient,
with {{{24}}} as a remainder,
so adding {{{72-24=48}}} to {{{"111,111,000"}}} 
we get the next multiple of {{{72}}} :
{{{"111,111,000"+48=highlight("111,111,048")}}} .
 
WITHOUT DIVIDING:
Like all numbers ending in {{{"000"}}} ,
{{{"111,111,000"}}} is divisible by {{{8}}} ,
because {{{1000}}} is divisible by {{{8}}} .
When dividing by {{{9}}} ,
the remainder can be found by adding the digits,
and repeating the digit adding with the result,
as many times as needed until you get a single digit result.
The final result is the remainder, unless it is {{{9}}} .
If the final result is {{{9}}} ,
then the number is divisible by {{{9}}} ,
and the remainder is {{{0}}} .
The sum of the digits of {{{"111,111,000"}}} is {{{6}}} ,
so when dividing {{{"111,111,000"}}} by {{{9}}} ,
we get a quotient {{{N}}} , and the remainder is {{{6}}} .
So, {{{"111,111,000"=9*N+6=3*3*N+3*2=3(3n+2)}}} is a multiple of {{{8}}} ,
and it is even a multiple of {{{3}}},
but it is not a multiple of {{{9}}} .
If I add a multiple of {{{8}}} that is also a multiple of {{{3}}} ,
such as {{{3*8=24}}} or {{{2*3*8=48}}} ,
the sum will also be a multiple of {{{8}}} and a multiple of {{{3}}} .
One of those sums must be a multiple of {{{9}}} too.
{{{"111,111,000"+24="111,111,024"}}} is not a multiple of {{{9}}} ,
because adding digits we get
{{{(1+1+1+1+1+1)+2+4=6+2+4=12}}} and {{{1+2=3}}} ,
but {{{"111,111,000"+48=highlight("111,111,048")}}} is a multiple of {{{9}}} ,
because {{{(1+1+1+1+1+1)+4+8=6+4+8=18}}} and {{{1+8=9}}} .