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Divisibility Sequence (Posted on 2025-01-30)

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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SImple way

| Comment 1 of 4

Note that 1007 = 53*19

The differences between consecutive terms of the sequence are

9010, 90100, 901000, 9010000,...

901=53*17 so the sequence starts with a multiple of 53 and goes up by multiples of 53. Therefore, every number is divisible by 53.

Posted by Jer on 2025-01-30 11:30:06

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