a nine digit number is formed using each of the digit 1,2,3,...,9 exactly once. for n= 1,2,3,...,9,n divides the first n digits of the number. find the number.
To solve this problem, we will go step by step and build the number digit by digit.
Let's start with the first condition: for n = 1, the first digit must be divisible by 1. Since all digits are distinct and range from 1 to 9, the first digit can only be 1.
Now, let's move on to n = 2. The first two digits of the number must be divisible by 2. Since the number starts with 1, the second digit must be an even number. The only even digit available is 2.
Now, for n = 3, the first three digits of the number must be divisible by 3. The digits 1, 2, and 3 are already included, so the third digit can be any number between 4 and 9 that is divisible by 3. The only number available is 6.
For n = 4, the first four digits must be divisible by 4. The number formed so far is 126, and we need to find a fourth digit that makes the number divisible by 4. The two-digit number formed by the third and fourth digits (26) must be divisible by 4. The only number that satisfies this condition is 26.
Continuing this process, for n = 5, the number formed so far is 12626. To make the first five digits divisible by 5, the last digit must be 5 since any number ending in 5 is divisible by 5.
Now, for n = 6, the number formed so far is 126265. To make the first six digits divisible by 6, the two-digit number formed by the fifth and sixth digits (65) must be divisible by 6. The only number that satisfies this condition is 65.
For n = 7, the number formed so far is 12626565. To make the first seven digits divisible by 7, we need to find a three-digit number formed by the fifth, sixth, and seventh digits that is divisible by 7. The number that satisfies this condition is 656.
Next, for n = 8, the number formed so far is 126265656. To make the first eight digits divisible by 8, the three-digit number formed by the sixth, seventh, and eighth digits (565) must be divisible by 8. The only number that satisfies this condition is 565.
Lastly, for n = 9, we are left with digit 9. Thus, the final number that satisfies all the given conditions is 1262656569.
So, the nine-digit number formed using each digit 1, 2, 3, ..., 9 exactly once and satisfying the given divisibility conditions is 1262656569.