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        riddles >> medium >> Digit sum of three consecutive numbers
        
(Message started by: Christine on Mar 19^th, 2013, 10:38am)

Title: Digit sum of three consecutive numbers
Post by Christine on Mar 19^th, 2013, 10:38am

Is there a method to finding a positive integer x such that each of DigitSum(x), DigitSum(x-1) and DigitSum(x+1) is divisible by a certain prime number?

Say the prime number is 7 or 17, how to find the smallest integer x?

Title: Re: Digit sum of three consecutive numbers
Post by towr on Mar 19^th, 2013, 11:11am

Either ds(x-1) and ds(x), or ds(x) and ds(x+1) are consecutive numbers; so they can't be divisible by the same number (>1).

Title: Re: Digit sum of three consecutive numbers
Post by pex on Mar 19^th, 2013, 11:15am

on 03/19/13 at 10:38:59, Christine wrote:

If I understand the question correctly, this does not seem possible. At least one of the differences DigitSum(x)-DigitSum(x-1) and DigitSum(x+1)-DigitSum(x) is going to be 1, so those two digit sums will be coprime.

Edit: not fast enough...

Title: Re: Digit sum of three consecutive numbers
Post by Christine on Mar 19^th, 2013, 12:02pm

Sorry I got it wrong.

This is what I found so far:

It is possible to find 3 consecutive integers such two of them have their digitsum divisible by a certain prime p and other integer divisible by the same prime p.

For example,
156999, 157000 and 157001

DigitSum(156999) = 1+5+6+9+9+9 = 39 = 3*13
DigitSum(157000) = 1+5+7+0+0+0 = 13
157001 = 13 * 12077

I don't whether this is a known or perhaps interesting result. And how to explain this?

But I find it very cool.

Title: Re: Digit sum of three consecutive numbers
Post by pex on Mar 19^th, 2013, 12:25pm

Aha. Barring errors in my program that took all of two minutes to write, the smallest triple for p=7 is
DigitSum(609999) = 42 = 6*7
DigitSum(610000) = 7
610001 = 87143*7.
~~I don't really have a better method than brute-force searching over multiples of p that end in 1, though.~~ Of course it'd be much faster to generate the middle number directly and check x+1 for divisibility. If a solution for p=19 exists, its middle number is going to end in at least 17 zeros, because that's the smallest integer for which 9k-1 is divisible by 19.

For p=17:
DigitSum(17899) = 34 = 2*17
DigitSum(17900) = 17
17901 = 1053*17

Title: Re: Digit sum of three consecutive numbers
Post by towr on Mar 20^th, 2013, 12:22am

Then for 19, I'd say
19899999999999999999 => 9*19
19900000000000000000 => 19
19900000000000000001 = 1047368421052631579*19

However, I don't think it was Christine's intent that it's necessarily the first two where you take the digit sum. So you might get a smaller result by taking the digit sum of the first and last, or the last two.

Title: Re: Digit sum of three consecutive numbers
Post by pex on Mar 20^th, 2013, 12:30am

on 03/20/13 at 00:22:31, towr wrote:

However, I don't think it was Christine's intent that it's necessarily the first two where you take the digit sum. So you might get a smaller result by taking the digit sum of the first and last, or the last two.

I had convinced myself that that couldn't happen last night, but I don't recall why, and it's not making much sense now that it's morning. So... you're probably right.

Title: Re: Digit sum of three consecutive numbers
Post by towr on Mar 20^th, 2013, 10:20am

For example:
ds(419) = 2*7
420 = 10*7
ds(421) = 1*7

Title: Re: Digit sum of three consecutive numbers
Post by Christine on Mar 20^th, 2013, 5:18pm

on 03/20/13 at 10:20:44, towr wrote:

For example:
ds(419) = 2*7
420 = 10*7
ds(421) = 1*7

Nice! is the triplet (419, 420, 421) the smallest divisible by 7?

what are the smallest four consecutive integers with the same property?

Title: Re: Digit sum of three consecutive numbers
Post by pex on Mar 21^st, 2013, 12:26am

on 03/20/13 at 17:18:52, Christine wrote:

Nice! is the triplet (419, 420, 421) the smallest divisible by 7?

Yes. This time I simply checked all triplets starting at (1,2,3) to be sure. 8)

on 03/20/13 at 17:18:52, Christine wrote:

what are the smallest four consecutive integers with the same property?

Can there be any? Out of those four, only one can be divisible by 7, so three would need to have digit sums divisible by 7. But at least two of these three have the same digits except for the least significant place, which cannot differ by 7 either.

Title: Re: Digit sum of three consecutive numbers
Post by ravibhole_1 on Apr 7^th, 2013, 9:53am

It is not possible