A(=abcde) is a 5-digit natural number with no leading zeros and e>a. B(=eca), 3-digit natural number is a multiple of 9 while C(=db), 2-digit natural number is a multiple of 3 but not 9.
Show that x⁵-y⁵=A has no integer solutions if sum of digits of C is not 6.

Note: A(=abcde) means Ten Thousand's digit of A:a, Thousand's digit of A:b and so on unit's digit of A:e.

Working the problem backwords, find the combinations of integers [x,y], (including [-x,y],[x,-y],[-x-y]) that yield a 5-digit A:

[1,7] = {16806,16808}
[1,8] = {32767,32769}
[1,9] = {59048,59050}
[1,10] = {99999}
[2,7] = {16775,16839}
[2,8] = {32736,32800}
[2,9] = {59017,59081}
[2,10] = {99968}
[3,7] = {16564,17050}
[3,8] = {32525,33011}
[3,9] = {58806,59292}
[3,10] = {99757}
[4,7] = {15783,17831}
[4,8] = {31744,33792}
[4,9] = {58025,60073}
[4,10] = {98976}
[5,6] = {10901}
[5,7] = {13682,19932}
[5,8] = {29643,35893}
[5,9] = {55924,62174}
[5,10] = {96875}
[6,7] = {24583}
[6,8] = {24992,40544}
[6,9] = {51273,66825}
[6,10] = {92224}
[7,8] = {15961,49575}
[7,9] = {42242,75856}
[7,10] = {83192}
[8,9] = {26281,91817}
[8,10] = {67232}
[9,10] = {40951}
[10,11] = {61051}

From the first clue ( e > a ), many of these can be eliminated, leaving the following list:

A = {13682, 15783, 16564, 16775, 16806, 16808, 16839, 19932, 24583, 29643, 31744, 32525, 32736, 32767, 32769, 49575, 58806, 59017, 59048, 61051}

This list yields the following values for B:

B = {261, 371, 451, 571, 681, 881, 981, 291, 352, 362, 473, 553, 673, 773, 973, 554, 685, 705, 805, 106}

Of these values, only two are evenly divisible by 9.

A = {13682, 16839};  B = {261, 981};  C = {13,16}

Of these C-values, neither are divisible by three.  Thus, the remaining clues are not needed.

Posted by hoodat on 2007-09-18 13:59:46