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| A digital root-perfect power problem (Posted on 2008-07-01) |
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Let S[x] be the digital root function (also known as the repeated digital sum function), where one adds the digits of positive integer x, then adds the digits of the sum until obtaining a single-digit number. (For example, S[975] = 3 because 9 + 7 + 5 = 21 and 2 + 1 = 3).
Given S[aa] = 2, what is the smallest positive integer that a can be
such that a is a perfect power?
Note: a is a perfect power if there exist natural numbers m > 1, and k > 1 such that mk = a.
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Submitted by Dej Mar
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Rating: 4.0000 (2 votes)
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Solution:
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(Hide)
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a = 78125.
The values of S[gh] for integers g > 0 and h > 1 are cyclic; and, thus, the values for x for S[x] = 2 can be expressed as the union of the
values (2 + 9m)(7 + 6n) and (5 + 9m)(5 + 6n), where m
and n are integers ≥ 0.
g h=2 h=3 h=4 h=5 h=6 h=7
1: 1 1 1 1 1 1
2: 4 8 7 5 1 2
3: 9 9 9 9 9 9
4: 7 1 4 7 1 4
5: 7 8 4 2 1 5
6: 9 9 9 9 9 9
7: 4 1 7 4 1 7
8: 1 8 1 8 1 8
9: 9 9 9 9 9 9
Where 2 + 9m = 7 + 6n, we find that where n is an integer, m is not.
The equation can be rewritten as m = (5 + 6n)/9; and, as (5 + 6n) modulo 9 results in the cyclic period (5, 2, 8), it has no integer solution. Thus, we can eliminate any values of (2 + 9m)(7 + 6n) as a possible
solution for a.
For 5 + 9m = 5 + 6n, 9 is comprised of the factors
3 and 3, and 6 is comprised of the factors 3 and 2. The lowest common multiple is 18, therefore a = 5 + 18n.
As S[5 + 18n] = 5, we can use our S[gh] cyclic table above to note
if a can be expressed as a perfect power, that a must either
be expressed as (2 + 9m)(5 + 6n) or (5 + 9m)(7 + 6n).
From our expression 5 + 18n, we can determine that our solution must be (5 + 18*(4320)) = 57 = 78125. |
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