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        riddles >> easy >> multiples of 2-digit numbers
        
(Message started by: Christine on Jul 1st, 2014, 6:50pm)
Title: multiples of 2-digit numbers
Post by Christine on Jul 1st, 2014, 6:50pm
Is there a multiple of 11 that ends in 11 and whose sum of digits is 11

e.g.

912 is the smallest multiple of 12 ending in 12 whose sum of digits is 12

Are there any 2-digit numbers that don't satisfy these conditions?
Title: Re: multiples of 2-digit numbers
Post by dudiobugtron on Jul 1st, 2014, 7:08pm

on 07/01/14 at 18:50:40, Christine wrote:
Are there any 2-digit numbers that don't satisfy these conditions?

I'm having trouble interpreting this question.  It seems like for all 2-digit numbers which end in 11, their digits sum to 2, which is not 11.  So all 2 digit numbers don't satisfy the conditions.
Title: Re: multiples of 2-digit numbers
Post by Christine on Jul 1st, 2014, 7:30pm
I meant, from the list of all the digit numbers which do not satisfy the conditions.

e.g. 12, 13, 14, 15, 16, 17, 18, 19 do satisfy the conditions of my puzzle.
Title: Re: multiples of 2-digit numbers
Post by dudiobugtron on Jul 1st, 2014, 7:54pm

on 07/01/14 at 19:30:12, Christine wrote:
I meant, from the list of all the digit numbers which do not satisfy the conditions.

e.g. 12, 13, 14, 15, 16, 17, 18, 19 do satisfy the conditions of my puzzle.

Thanks for the clarification, I'm still a bit stuck though.  What are the conditions of your puzzle?  Apologies for not understanding it!
Title: Re: multiples of 2-digit numbers
Post by Christine on Jul 1st, 2014, 8:50pm

on 07/01/14 at 19:54:37, dudiobugtron wrote:
 What are the conditions of your puzzle?  


Find, when possible,  multiples of N (all 2-digit numbers) that end in N and sum of the digits of the multiple is N.
Title: Re: multiples of 2-digit numbers
Post by towr on Jul 1st, 2014, 10:36pm

on 07/01/14 at 18:50:40, Christine wrote:
Is there a multiple of 11 that ends in 11 and whose sum of digits is 11?
Not even if it didn't end in 11.
For multiples of eleven, if you add the digits in the odd position (SO) and subtract the digits in the even position (SE) , the result has to be a multiple of 11.
Because you want the sum to be 11, the only valid multiple is 0, but 0 is even, and 11 is not. You can't have SO+SE=11 and SO-SE=0.
Title: Re: multiples of 2-digit numbers
Post by dudiobugtron on Jul 1st, 2014, 11:59pm
towr - I think your proof isn't robust enough, since 11 and -11 are also valid multiples.  These could be achieved with SO = 11 and SE = 0, or SE = 11 and SO = 0.


on 07/01/14 at 20:50:24, Christine wrote:
Find, when possible,  multiples of N (all 2-digit numbers) that end in N and sum of the digits of the multiple is N.


OOOhh, right now I get it.  You wanted to find 2 digit numbers that worked like 12 does.  I thought you meant 2 digit numbers which were multiples of 11.  Apologies for the confusion.
Title: Re: multiples of 2-digit numbers
Post by Grimbal on Jul 2nd, 2014, 6:10am

on 07/01/14 at 22:36:17, towr wrote:
Not even if it didn't end in 11.
For multiples of eleven, if you add the digits in the odd position (SO) and subtract the digits in the even position (SE) , the result has to be a multiple of 11.
Because you want the sum to be 11, the only valid multiple is 0, but 0 is even, and 11 is not. You can't have SO+SE=11 and SO-SE=0.

What is wrong with SO=11 and SE=0?
101010101010101010101 is a multiple of 11.  It just doesn't end in 11.

PS: oops, I didn't see dudiobugtron already made that point.
Title: Re: multiples of 2-digit numbers
Post by rloginunix on Jul 2nd, 2014, 9:19am
Dangit. How did I miss this one? Checked the forum last night and saw nothing ...

You can use an iterative approach:

1). A 3-digit number "abc" (in base 10) is divisible by 11 iff: a - b + c = 11*k, but b == c, hence a = 11*k where "a" is a 0-9 digit. Can't do it unless a == k == 0.

2). A 4-digit number "abcd": a - b + c - d = 11*k, c == d, hence a - b = 11*k, where "a" and "b" are 0-9 digits. Again, can't do it unless a == b == k == 0. (Not even considering the sum of digits yet).

3). A 5-digit number "abcde": a - b + c - d + e = 11*k, d == e, hence a - b + c == 11*k.

Aha! Now you see a pattern emerging: to construct a smallest number (divisible by 11) satisfying the "last two digits are 11" requirement you have to make "a" as small as possible, zero out the "b" and add 9 in the odd position:

20911

I think that should be the smallest one. However, the sum of digits is not 11. You can eliminate the "smallest" requirement and use the above method to construct an arbitrary number (like Grimbal did with a long string of zeros and ones) but all three requirements, 1) divisible by 11, 2) last two digits are 11 (or simply equal), 3) digits sum to 11, can't be satisfied.
Title: Re: multiples of 2-digit numbers
Post by towr on Jul 2nd, 2014, 10:44am

on 07/01/14 at 23:59:02, dudiobugtron wrote:
towr - I think your proof isn't robust enough, since 11 and -11 are also valid multiples.  These could be achieved with SO = 11 and SE = 0, or SE = 11 and SO = 0.
Hmm, yes. I guess I can't drop the "ending in 11" condition, which makes SE and SO >= 1 And therefore SE and SO <= 10


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