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Tenacious Divisibility Treat (Posted on 2016-04-06) Difficulty: 2 of 5
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.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution my numbers | Comment 4 of 10 |

Going backward, using f(n)=f(n-1)*8+5       ...   8 times:

start with f(0) =0  you get    f(8)    11,983,725


start with f(0) =1  you get 28,760,941


start with f(0) =2 you get a number over 40M


  Posted by Ady TZIDON on 2016-04-06 12:25:44
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