Fun way to learn some basic number theory !!!
A test of divisibility by 3, 7 and 19:
The product of the prime numbers 3, 7 and 19 is 399. If a number 
(where b is a two-digit number and a is any positive integer) is divisible by 399 or by any of its divisors, then 
is divisible by the same number.
On your own, can u prove this? (Hint: Use 
as a link). Can you formulate and prove its converse?
Devise a simple test of divisibility by 3, 7 and 19.
More recreational problems in number theory are coming soon !!
Nalin Pithwa
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[If a number 10a+b when divided by c leaves remainder 0, number x(10a+b) will also be divisible by c.] So 100a+b and 400a+4b will be equally divisible by 3, 7 or 19.
Now if 399a is subtracted from 400a+4b, the remainder still says the same as 399 is divisible by 3, 7 and 19. [26 leaves remainder 2 when divided by 3. When subtracted by a multiple of 3(15), no. becomes 11 which still leaves 2 as remainder.] So a+4b is as divisible as 10a+b by a divisor c (here 399 or any factors.)