SOLUTION: find greatest number of four digits which when divided by 3,5,7,9 leaves remainder 1,3,5,7 respectively

Let's first look at the properties of ALL such 
integers, not necessarily the greatest one with 
four digits, not necessarily even one with four 
digits, and not necessarily even one that is 
positive.

Theorem:
If two integers p and q are such that p mod r = q mod r,
then they differ by a multiple of r.

Proof:

p mod r = q mod r

(p mod r) - (q mod r) = (0 mod r)

(p - q) mod r = 0 mod r

Thus they must differ by a multiple of r.

Therefore if p and q both satisfy the given
conditions, then they must differ by some 
multiple of 3, which is also a multiple of 5, 
which is also a multiple of 7, and which is 
also a multiple of 9.

Since a multiple of 9 is also a multiple of 3, 
we only need to multiply 5x7x9 = 315 to find the 
smallest number two such numbers can differ by.
This suggests an arithmetic sequence of values
satisfying the given conditions. 

Now let's find ANY integer meeting the 
given requirements.

If p is such an integer then there exist 
integers a,b,c,d such that 

p = 3a + 1 = 5b + 3 = 7c + 5 = 9d + 7

Let's investigate to find out if there exists 
a simple solution where a = b = c = d. If so then

p = 3a + 1 = 5a + 3 = 7a + 5 = 9a + 7

Now we are in luck here because setting any of
the 4 expressions on the right equal to each
other gives a = -1 and thus p = -2

So the arithmetic sequence with first term

a1 = p = -2 and common difference d = 315 will be 
an arithmetic sequence of integers meeting the 
given requirements.

The nth term of an arithmetic sequence is given by

an = a1 + (n-1)d

an = -2 + (n-1)(315)

an = -2 + 315n - 315

an = -317 + 315n

Now since the term of this sequence we are seeking
is the largest one with 4 digits, we require that it
be less than 10000.

So 

an = -317 + 315n < 10000
            315n < 10317
               n < 32

So the largest value of n we can use
is n = 32.

So

a32 = -317 + 315(32)

a32 = 9763 
 
That's it!

Edwin