Find a digit number made from all digits such that the number made from the first digits is divisible by , from the first is divisible by and so on.
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Firstly, I eliminated possibilities for each digit of a b c d e f g h i j :
j = 0 ⇒ e = 5
( b , d , f , h ) ∈ ( 2 , 4 , 6 , 8 ) ⇒ ( a , c , g , i ) ∈ ( 1 , 3 , 7 , 9 )
4 ∣ a b c d ⇒ 4 ∣ c d , so c ∈ ( 1 , 3 , 7 , 9 ) ⇒ d ∈ ( 2 , 6 )
8 ∣ a b c d e f g h ⇒ 8 ∣ f g h , so f ∈ ( 2 , 4 , 6 , 8 ) , g ∈ ( 1 , 3 , 7 , 9 ) ⇒ h ∈ ( 2 , 6 )
Finally, ( d , h ) ∈ ( 2 , 6 ) ⇒ ( b , f ) ∈ ( 4 , 8 ) .
Then, I made lists of each n -digit number which was valid:
a b ∈ ( 1 4 , 1 8 , 3 4 , 3 8 , 7 4 , 7 8 , 9 4 , 9 8 )
a b c ∈ ( 1 4 7 , 1 8 3 , 1 8 9 , 3 8 1 , 3 8 7 , 7 4 1 , 7 8 3 , 7 8 9 , 9 8 1 , 9 8 7
I'll skip a few lines, because d just tacks ( 2 , 6 ) onto each and e sticks on a 5 , so each set has 2 0 elements apiece.
a b c d e f ∈ ( 1 4 7 2 5 8 , 1 8 3 6 5 4 , 1 8 9 6 5 4 , 3 8 1 6 5 4 , 3 8 7 6 5 4 , 7 4 1 2 5 8 , 7 8 3 6 5 4 , 7 8 9 6 5 4 , 9 8 1 6 5 4 , 9 8 7 6 5 4 )
a b c d e f g ∈ ( 1 4 7 2 5 8 3 , 3 8 1 6 5 4 7 , 7 8 3 6 5 4 9 )
a b c d e f g h = 3 8 1 6 5 4 7 2 ⇒ a b c d e f g h i j = 3 8 1 6 5 4 7 2 9 0