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A digital root-perfect power problem (Posted on 2008-07-01) Difficulty: 4 of 5
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.

See The Solution Submitted by Dej Mar    
Rating: 4.0000 (2 votes)

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Hints/Tips Missed it by that much (spoiler) | Comment 1 of 7
Well, the S[x] is just the remainder Mod 9.

Therefore, S[xy] = S[x]*S[y].

And what possible values mod 9 could "a" have?

Let's first consider the powers (mod 9) of any number which equals 2 mod 9.:

2*2 mod 9 = 4
2^3 mod 9 = 8
2^4 mod 9 = 2*8 mod 9 = 7
2^5 mod 9 = 2*7 mod 9 = 5
2^6 mod 9 = 2*5 mod 9 = 1
2^7 mod 9 = 2*1 mod 9 = 2

so the powers of 2 mod 9 are 2,4,8,7,5,1,and repeating thereafter with period 6

For all numbers, the powers (mod 9) are:

0:  0, 0, 0 , etc.
1:  1, 1, 1, etc.
2: 2, 4, 8, 7, 5, 1, repeating thereafter with period 6
3: 3, 0, 0, 0, etc.
4: 4, 7, 1, repeating thereafter with period 3
5: 5, 7, 8, 4, 2, 1, repeating thereafter with period 6
6: 6, 0, 0, 0, etc.
7: 7, 4, 1 repeating thereafter with period 3
8: 8, 1, 8, 1, repeating thereafter with period 3

Note that 2 appears in only 2 of these lists.  So, the only values of a which could possibly = 2 mod 9 when raised to the a-th power are = (mod 9) to 2 or 5.

If a = 2 mod 9, then the power it is raised to (which also a) must be 1, or 6, or 13, or 1 mod 6 (because of the repeating period).
But, in order to equal 2 mod 9, a (mod 18) must equal 2 or 11.
And, in order to equal 1 mod 6, a (mod 18) must equal 1 or 7 or 13.
Therefore a is not equal to 2 mod 9.

Therefore, a must equal 5 mod 9.
And the power it is raised to (which also a) must equal 5 mod 6.
And these two things can only happen simultaneously if a (mod 18) = 5.

Therefore, any and all positive numbers which equal 5 mod 18 will satisfy S[a^a] = 2

And the solution to the puzzle is the smallest perfect power which equals 5 mod 18, (unless I've made a mistake).

And that's as far as I've gotten.

Although it wouldn't be hard to work out (using the method above) to work out the smallest such number.  m and k must both be less than 18, and m is not divisible by a factor of 18 (2 or 3),
so m must be 5 or 7 or 11 or 13 or 17.  I'll finish this after supper.

  Posted by Steve Herman on 2008-07-01 19:48:23
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