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 Tenacious Divisibility Treat (Posted on 2016-04-06)
A positive integer less than 30 million is such that if we subtract 5 from it – the resulting number is divisible by 8.

At the first step, from the number (considered originally) diminished by 5 - we subtract the eighth part. We then obtain a number that also becomes divisible by 8 after 5 is subtracted from it.

At the second step, we derive another in the same way, namely by subtracting a eighth part from the number at the end of the first step diminished by 5. The resulting number is also divisible by 8 after subtracting 5.

The operation concludes at 8th step given that at the end of 7th step we get a number that is divisible by 8 after after subtracting 5.

Determine the positive integer initially before the first step.

*** The resulting number at the end of 8th step is NOT necessarily divisible by 8 after subtracting 5.

 No Solution Yet Submitted by K Sengupta No Rating

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 Help from a negative (solution) | Comment 3 of 9 |
One full step is represented by the recursion x_n+1 = (x_n - 5)*(7/8), for some unknown initial term x_0.  Notice that x_0 = -35 maps to x_1 = -35.  Then all x_n = -35 if x_0 = -35.

To get the positive integer solution sought add 8^8 to x_0.  Then x_0 = 8^8-35 = 16777181 and by repeated iteration x_8 = 7^8-35 = 5764766.

 Posted by Brian Smith on 2016-04-06 10:30:14

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