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 Deux Power Divisibility (Posted on 2015-04-12)
N is a positive integer and F(N) denotes the sum of the base ten digits of N.

Derive an algorithm to determine the values of N such that F(2N) is divisible by 10.

 No Solution Yet Submitted by K Sengupta No Rating

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 A sequence and some thoughts. | Comment 1 of 3
The digital roots of the powers of 2 follow the cycle 1,2,4,8,7,5.  (Of course multiples of 3 are not possible so 3,6,9 are not in this cycle.)
So F(2N) cannot be a multiple of 30.

There doesn't seem to be much of a pattern to those N's that otherwise comply except that as N increases they seem to get more common.

`N  F(2N) 6 1013 2026 4041 5046 7051 8057 8058 70`

OEIS does not contain this sequence, but I did use https://oeis.org/A001370 to compile it.

The problem asks for an algorithm.  I don't know of a simpler way than just directly computing F(2N).

However, if I were told that a number, X was in the sequence I could probably guess F(2X)if X isn't too big.

For example if I were told F(2100) was in the sequence (I don't know if it really is) I'd look at the cycle to see it's digital root is 7.  So F(2100)would have to be 70 or 160 or 250, etc..

2100 has 31 digits.

31*an average digit of 4.5=139.5 which is closest to 160.

However, from the OEIS mentioned above:

It is believed that a(n) ~ n*9*log_10(2)/2, but this is an open problem.

So even this idea may not work for large N.

 Posted by Jer on 2015-04-12 10:39:57

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