The numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 have to be used once to create a 9 digit number such that the first digit is divisible by 1, the first two are divisible by 2, the first three digits are divisible by 3, and so on ending with 9-digit number being divisible by 9. There are many numbers that satisfy all the given conditions.
All numbers are divisible by 1. The left most digit can be either of the nine. A number is divisible by 2 if the last digit is either of 0, 2, 4, 6 or 8. The second digit can be either of these. A number is divisible by 3 if the sum of the last 2 digits is divisible by 3. A number is divisible by 9 if the sum of the last three digits is divisible by 9.
Choose any three digits that form a number divisible by 9. For instance, 126, 621 etc. These form the last digits of the number. Now if we take 126, of the remaining digits any can be the first digit of the number. Take 4. The first 2 digits form a number divisible by 2, this can be 34. To make the number formed by the first 3 digits one that is divisible by 3, the third digit can be 8. Use the other 3 digits in any order.
This gives one of the possible numbers as 348759126
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