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2015-01-01T04:24:51+05:30

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A b c = a^1 + b^2 + c^3

99 a + b (10 - b) = c ( C+1)(c-1) 

Applying some limits of  0 <= b <= 9 and 0 <= C<= 9  0 <= a <= 9

       for   a = 0,     0 6 3 = 0^1 + 6^2 + 3^3

       for   a = 1 ,    1 7 5 = 1^1 + 7^2 + 5^3

       for a = 5       5 9 8 = 5^1 + 9^2 + 8^3
 
In three digits the above are the only other numbers other than given numbers in the question.

So we find in 4 digit numbers.

a b c d = a^1 + b^2 + c^3 + d^4

999 a + b (100 - b) = d (d-1)(d^2+d+1) + c (c^2 - 10)

It is not easy to solve this easily.


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