If there is a special nine - digit - number where are distinct digits.
If the number formed by first
one digit ('A') is divisible by 1
two digits ('AB') is divisible by 2
three digits ('ABC') is divisible by 3
first four digits ('ABCD') is divisible by 4
and so on till 9 ...
Then find the nine - digit - number
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We have the following constraints:
(i) The digit B must be even
(ii) A + B + C must be divisible by 3
(iii) C D must be divisible by 4
(iv) E must be 5
(v) F must be even and D + E + F must be divisible by 3
(vi) A + G − D + 2 ( E − B ) + 3 ( F − C ) must be divisible by 7
(vii) F G H must be divisible by 8
Using all these, we get the number A B C D E F G H I ≡ 3 8 1 6 5 4 7 2 9 .