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Solar Flare
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snowtiger:
Well what I need to know is if a 9 digit number consists of each digit from 1 to 9 and the 9 digit number is divisible by 9. If you remove the rightmost digit, the remaining 8 digit number will be divisible by eight. If you remove the next rightmost digit, the remaining 7 digit number will be divisible by 7. If you remove the next rightmost digit, the remaining 6 digit number will be divisible by 6. This process continues to one. What is the original nine-digit number?
let the first digit be a1, second a2 . . .
1) You know a2, a4, a6 and a8 are even
2) So you know a1, a3, a5, a7 and a9 are odd.
3) You know a5 = 5.
Now because of 2), you know 4 divides 10 * a3 +a5. So a4 is 2 or 6. Similarly, a8 is 2 or 6. So a2 is 4 or 8 and a6 is 4 or 8.
3 divides 100* a7 + 10 * a8 + a9 (why?) and from this you can deduce a7 is 3, 7 or 9 (not 1).
3 divides 100* a4 + 10 * a5 + a6 (why) so those numbers are
258 or 654
Playing around you get the first three numbers (a1-a3) are one of
147
381
387
741
783
789
981
987
the second three are one of
258
654
The final three are
321
327
723
729
963
Now since these numbers have to be unique, (one of each) you can eliminate some of them. For example, the first three cannot be 387 (why) or 783 (why)
So picking combination where we do not have repeats, the possible numbers are
147258963 (fails division by 7)
381654729
741258963 (fails division by 7)
789654321 (fails division by 7)
981654327 (fails division by 7)
987654321 (fails division by 7)
So the answer is 381654729
HTH
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