Question 973649
Find and justify a rule that produces all 3 digit multiples of 7 with an FDS of 7:
The FDS is the remainder of dividing a number by 9,
so a 3-digit number with an FDS of 7 can be written as
{{{9K+7}}} for some positive integer {{{K}}} .
If the number is also a multiple of 7,
then it can be written as {{{7N}}} for some positive integer {{{N}}} .
So,
{{{9K+7=7N}}}--->{{{9K=7N-7}}}--->{{{9K=7(N-1)}}}
That means that K must be a multiple of 7,
which could be written as {{{K=7P}}} for some positive integer {{{P}}} .
Now we have that the number is
{{{9K+7=9(7P)+7=63P+7}}} ,
which gives us a simple rule to find all the multiples of 7 with an FDS of 7.
We start with {{{P=2}}} ,
because we need {{{63P+7>=100}}} for a 3-digit number, and
{{{63P+7>=100}}}--->{{{63P>=100-7}}}--->{{{63P>=93}}}--->{{{P>=93/63=31/21=about 1.5}}} .
{{{P=2}}} yields {{{63P+7=63*2+7=126+7=133}}} ,
and that is our first 3-digit multiple of 7 with an FDS of 7.
Then we keep repeatedly adding {{{63}}} to get all the other 3-digit multiple of 7 with an FDS of 7,
until we get to {{{63*15+7=952}}} ,
which is the greatest 3-digit multiple of 7 with an FDS of 7,
because {{{63P+7<1000}}}--->{{{63P<993}}}--->{{{P<993/63=331/21=about15.5}}}
 
Find and justify a rule that produces all 3 digit multiples of 8 with an FDS of 8:
The same reasoning tells you that they will be numbers of the form
{{{9*8*P+8=72P+8}}} .
So, we start with {{{P=2}}} to get {{{72P+8=144+8=152}}} ,
and we repeatedly add {{{72}}} all the way up to
{{{72*13+8=944}}} .