Start with N, a six digit number. Subtract three from it, and then the new number formed should be divisible by 7. Take the six sevenths of this number. Call this new number i.e,  "\frac{6}{7}(N - 3)")
as N1, and it should be divisible by 7. Continue this process till you form N6. Find N and N6. How long you can go like this?
4 comments:
is there any condition that all N1 N2... N6 to be six digited neumbers??
is N1 divisible by 7 or 6/7(N1-3) divisible by 7??
It is not stated conditionally, but all are six digit numbers.
N1=6/7(N-3)
N2=6/7(N1-3)
As all the numbers involved are whole numbers, the numbers must be divisible by seven at each stage.
823525 (7^7 - 18)
705876
605034
518598
444510
381006
326574
279918 (6^7 - 18)
Started with a hunch that powers of 7 must be a good place to start and then used excel :)
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