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| Powering up the digits (Posted on 2009-04-12) |
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Each of x1, x2 and x3 corresponds to a nonzero digit, whether same or different, in positive integer base b.
Determine all possible value(s) of b ≤ 100 such that this equation has at least one valid solution:
x1x2x3 =
x1x1 +
x2x2 +
x3x3
Note: x1x2x3 corresponds to concatenation of the three digits.
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Submitted by K Sengupta
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Rating: 5.0000 (1 votes)
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Solution:
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(Hide)
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b = 4, 25 and 91 are the only possible values ≤ 100.
There are precisely two solutions in base 4,and:
There is precisely one solution in base 25, and:
There is precisely one solution in base 91.
For an explanation, refer to the solution submitted by Charlie in this location.
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Daniel has extended the provisions corresponding to the given problem in his location, exploring the cases where each of x1, x2 and x3 is between 1 and 99, and has found a solution each for base 904 and base 215.
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