1,甲的卡片如果个位和百位互换,和乙的卡片就相同。 2,丙的卡片是甲的卡片的三倍,乙的四倍。 let the three numbers be A=abc, B = def and C = ghi condition 1 gives cba=def condition 2 gives ghi=3(abc) = 4(def) from 2 we can infer that a <=3 and d<=2; because of g<=9 combining 1 & 2 we have 3(abc) =4(cba). Re-write it in 10-base number notation 3(a100+b10+c) = 4(c100+b10+a)www.ddhw.com collecting terms we have 296a-10b-397c=0 this is a Diophantine equation that we need to solve for possitive integers a,b,c with the condition of a<=3, b<=9 and c<=9 re-rarrange the Diophantine equation b= (296/10)a-(397/10)c or b= 29a-39c+(6/10)a-(7/10)c
1) case a=1 b= 29-39c+(6-7c)/10 soluiton for c is c=8, so that b = 29-39x8 -5 = -288 a=2, b=-288 and c=8 do not satisfy.
2) case a= 2 b= 59-39c+(2-7c)/10 solution for c is c=6, so that b=59-39x6-4 = -179 a=2, b=-179 and c=6 do not satisfy.www.ddhw.com
3) case a=3 b= 88-39c+(8-7c)/10 solution for c is c=4, so that b=88-39x4-2 = -70 a=3, b=-70 and c=4 do not satisfy. we have exhausted all the possible cases for a and found no solution. Hence there is no solution for the problem given in 10base number system. There might be a solution in other number system, which I haven't tried yet.
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