[tex]\it \overline{ab} +\overline{ba} = 10a+b+10b+a=11a+11b=11(a+b)
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\it \overline{bc} +\overline{cb} = 10b+c+10c+b=11b+11c=11(b+c)
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\it \overline{ca} +\overline{ac} = 10c+a+10a+c=11c+11a=11(c+a)[/tex]
n = 11(a+b) + 11(b+c) + 11(c+a) = 11(a+b+b+c+c+a) =
=11(2a+2b+2c) = 11·2(a+b+c) = 22(a+b+c)
9|n ⇔ 9|22(a+b+c)⇔9|(a+b+c)⇔9|abc (barat)