A 5 Digit number

$abcab = a \times 10^{4} + b \times 10^{3} + c \times 10^{2} + a \times 10 + b$

Which gives modulo 7, and using first property:

$4a + 6b + 2c + 3a + b = 2c = 0 ( mod 7)$

Since $0 < c < 10$ we deduce $c = 7$

From second property we deduce $a + b + 7 = 0 ( mod 9)$

Since $0 < a,b < 10$ , we have either $a + b = 2$ or $a + b = 11$ . The first case give the triplet $(1,1,7)$ which does not satisfy third property.

The second case gives the following triplets:

$(2,9,7)$

$(4,7,7)$

$(6,5,7)$

$(8,3,7)$

Of which only first and third complies with third property, with first having the minimal product.

Hence the value: 126

divisibility rule for 4 is last two digits divisible by 4 and a,b,c>0 so this 18 possibilities.. and divisibility rule for 9 is sum of the digits is divisible by 9... so make a such that 18 possibilities are divisible by 9 and check them for 7