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Consecutively Divisible (Posted on 2013-03-18) Difficulty: 4 of 5
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    
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re:1/6 remaining Comment 3 of 3 |

Answer to B3:

P  = 550000000000                = 11×50000000000
P-1= 549999999999 : sod(P-1)= 99 = 11×9
P+1= 550000000001 : sod(P+1)= 11 = 11×1

For the smallest |sod(P-1) - sod(P+1)| with a positive multiple of 11 is 88 and the smallest sod(P+1) with a positive multiple of 11 is 11, the sod(P-1) = 99. As can be found, the smallest number for an sod(P-1) of 99 is a twelve-digit number. The smallest twelve-digit number that conforms to the specifications is 550000000000. 

Edited on March 27, 2013, 11:45 pm
  Posted by Dej Mar on 2013-03-20 00:55:05

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