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 Consecutively Divisible (Posted on 2013-03-18)
PART A:

P is a positive integer such that each of sod(P-1), sod(P) and P+1 is divisible by 13. Determine the minimum value of P.

What if each of P-1, sod(P) and sod(P+1) is divisible by 13? How about each of sod(P-1), P and sod(P+1) being divisible by 13?

PART B:

What will be the respective answers to Part A, when the words "divisible by 13" are replaced with "divisible by 11"?

*** sod(x) refers to the sum of the digits in the base ten expansion of x.

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: 5/6 of computer solution | Comment 2 of 3 |
(In reply to 5/6 of computer solution by Charlie)

Looking for minimums and patterns for A3 and B3.

In order for both sod(P-1) and sod(P+1) to be divisible by the given integer, |sod(P-1) - sod(P+1)| must also be divisible by the integer.
1/13 = 0.[076923]...
The first occurance for
|sod(P-1) - sod(P+1)| =
7 @ P =      3×13 =      39
16 @ P =     23×13 =     299
25 @ P =    923×13 =   11999
34 @ P =   6923×13 =   89999
52 @ P =  76923×13 =  999999
43 @ P = 100000×13 = 1300000
...
52 is the first multiple of 13 that appears, thus P > 999999.
As Charlie has found, P = 39000000.

1/11 = 0.[09]...
The first occurance for
|sod(P-1) - sod(P+1)| =
16 @ P =         9×11 =         99
7 @ P =        10×11 =        110
34 @ P =       909×11 =       9999
25 @ P =      1000×11 =      11000
52 @ P =     90909×11 =     999999
43 @ P =    100000×11 =    1100000
70 @ P =   9090909×11 =   99999999
61 @ P =  10000000×11 =  110000000
88 @ P = 909090909×11 = 9999999999
...
88 is the first multiple of 11 that appears, thus P > 9999999999.
It appears Charlie's program looked only up to 11×99999999, thus it was unable to reveal the solution for B3.

Edited on March 19, 2013, 9:11 am
 Posted by Dej Mar on 2013-03-19 08:18:32

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