abcab is a multiple of 7

abc is the multiple of 9

cba is the multiple of 4

what is the minimum of a x b x c ? (a,b,c >0)

The answer is 126.

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

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