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To solve the equation ( abcd \times 9 = dcba ), we can represent ( abcd ) as ( 1000a + 100b + 10c + d ) and ( dcba ) as ( 1000d + 100c + 10b + a ). This leads to the equation ( 9(1000a + 100b + 10c + d) = 1000d + 100c + 10b + a ). Through trial and error, you can find that ( a = 1 ), ( b = 0 ), ( c = 8 ), and ( d = 9 ) satisfy the equation, giving ( abc = 108 ) and ( d = 9 ).

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