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.
(In reply to
Answer by K Sengupta)
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
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