Find a number containing every digit 0-9 exactly once, such that for every 1≤N≤10, the leftmost N digits comprise a number, divisible by N.
SOLUTION
Let the number be ABCDEFGHIJ.
The digit J must be 0, so that the number is divisible by 10.
The digit E must be 5, so that the number comprised of the first 5 digits is divisible by 5.
The digits B, D, F, H, J must be even, and therefore the digits A, C, E, G, I must be odd.
The numbers CD and GH must be divisible by 4 and the number FGH must be divisible by 8. Since C and G are odd, D and H must be 2 and 6, or vice-versa.
Since ABC, ABCDEF, and ABCDEFGHI are all divisible by 3, we see that A+B+C, D+E+F, and G+H+I are divisible by 3 as well. Therefore, D+F+5 is divisible by 3 and since D and F are even, we have either D=2, H=6, F=8, B=4 or D=6, H=2, F=4, B=8.
D=2, H=6, F=8, B=4. Since FGH is divisible by 8 and G is odd, we have G=1 or G=9. Also, G+I+6 is divisible by 3 and since I is odd, we have G=1, I=3. The only possibilities for ABCDEFG are 7492581630 and 9472581630, but they are not divisible by 7.
D=6, H=2, F=4, B=8. Since FGH is divisible by 8 and G is odd, we have G=3 or G=7. Also, G+I+2 is divisible by 3 and since I is odd, we have G=3, I=1 or G=3, I=7 or G=7, I=3 or G=7, I=9. The only possibilities for ABCDEFG are 9876543, 7896543, 1896543, 9816543, 1896547, 9816547, 1836547, 3816547. Out of these, only 3816547 is divisible by 7.
