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Forty- niners (Posted on 2011-02-01) Difficulty: 3 of 5
Find a pair of two consecutive integers, such that the s.o.d. of each of them is a multiple of 49. i.e. SOD(N)=K*49 & SOD(N+1)=L*49: (K,L integers)

Rem1: S.o.d. of a number is the sum of said number's digits.
Rem2: Only the smallest values requested.

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: solution? | Comment 3 of 5 |
(In reply to solution? by Justin)

Indeed it must end with exactly 11 9's (unless you want to go all the way to 60 9's, which takes you over the number already found), so that the next number will have the same sod mod 49. These 11 9's come to 1 mod 49, so you need to add 48 to the sod.

The smallest preface you can add to the 11 9's that ends in other than 9 is 499998, so indeed Justin's 49,999,899,999,999,999 is the smallest.

  Posted by Charlie on 2011-02-01 16:15:47
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