How do I solve this math problem?

The digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 are each used once in some order to compose a single 9-digit number. Starting from the leftmost digit, 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. What is the number??

PLEASE SHOW ALL YOUR WORK!!!

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381,654,729

While it would be impossible for me to show all of the work here, there are some basic rules that can help.

The first digit can be anything since all whole numbers can be divided by 1. For 2, make sure it ends in an even number. For the third position, just add up all three of the digits, if the sum is divisible by 3 then you are ok. For 4, check the last two numbers. If they are divisible by 4, it works. The 5th position has to be a 5 in order for it to be divisible by 5. So, the number will be: odd-even-odd-even-5-even-odd-even-odd. For 6, make sure it is divisible by 3 (see above) and it ends in an even number. No good tricks for 7 I am afraid, but you only have 2 odd numbers left, so it won't be hard to figure it out. On eights, just follow the rule for 4. All combinations will be divisible by nine, so don't worry here.

The first digit can be divided by any number from 1-8. The last number for the first two digit must be an even number. The sum of the first three digits must be divisible by three. The last two digits of the first four digits must be divided by four. The first five digits must be divided by five. For the first six digits, the sum must be divisible by three. For the first seven digits, form an alternating sum oof blocks of three from left to right. For the first eight digits, examine the last three to see if divisible by eight. The sum of the first nine digiys to be divided by nine.

The final answer is **381654729**

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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