All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Digital Sum 11 (Posted on 2008-12-08)
Two consecutive numbers both have a digital sum which is a multiple of 11. What is the smallest pair of numbers?

Note: the digital sum is the number equalling the sum of all the digits of the number. For example the digital sum of 123456 is 21.

 See The Solution Submitted by Brian Smith Rating: 2.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Puzzle Solution Comment 5 of 5 |

Let the consecutive numbers be C and C+1, where C contains n digits and:
C= A1A2....An. let the digital sum of any given positive integer be given by d(x).

Now, if An < 9, it follows that d(C+1) = d(C) + 1, and accordingly if precisely one of C and C+1 is divisible by 11, the other will remain indivisible by 11. This leads to a contradiction.

Thus An=9, so that the last digit of C+1 is 0. Therefore the last digit of C must be 9. Let the total number of consecutive 9's from the right of C be t.

Then, C = A1A2....A(n-t)99...9, and:
C+1 = A1A2....A(n-t)+1 00....0
-> d(C) - d(C+1) = 9t-1

Since each of d(c) and d(C+1) is divisible by 11, it follows that 9t-1 is divisible by 11. The minimum value of t for which this is possible occurs at t=5, giving:

d(C) - d(C+1) = 9*5-1 = 44

Now, d(C) = Sum(A(1), A(n-5)) + 9*5
= 45 + Sum(A(1), A(n)), and:

d(C+1) = Sum(A(1), A(n-5)) + 1

The  minimum value of Sum(A(1), A(n-5)) for which both d(C) and d(C+1) is divisible by 11 is thus 10.

Suppose in this situation A(n-5) = 9. If so, then the consecutive number of 9's from the right is 6, instead of 5, which leads to a contradiction.
Obviously the first digit in C will be minimum, when the value of A(n-5) is allowed to be the maximum possible but in conformity with the restriction A(n-5)< 9. Hence, A(n-5) = 8.

It is now trivial to obserce that C is minimized whenever n=7, so that: A1 = 10-8= 2.

Accordingly, C = 2899999, so that:
C+1 = 2899999 = 1 = 2900000

Edited on January 28, 2009, 11:48 am
 Posted by K Sengupta on 2009-01-28 11:41:28

 Search: Search body:
Forums (0)