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Divisibility Sequence (Posted on 2025-01-30) Difficulty: 2 of 5
Show that every number in the sequence

1007, 10017, 100117, 1001117, 10011117, . . .

is divisible by 53.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Another solution | Comment 2 of 4 |
Define the sequence recursively, a(n) = 10*a(n-1) - 53 with initial a(1) = 1007.  1007 = 53*19.  The first term is a multiple of 53.  

Now create an induction hypothesis: if a(n-1) is a multiple of 53 then a(n) is a multiple of 53.  Assume a(n-1) = 53*k, so a(n) = 10*53*k - 53 = 53*(10*k-1).  Therefore a(n) is a multiple of 53.

Since the first term is a multiple of 53 our induction hypothesis applies and we can subsequently say every term in the sequence is a multiple of 53. QED

Fixed typo noticed by Steven.

Edited on January 31, 2025, 11:07 am
  Posted by Brian Smith on 2025-01-30 11:36:57

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