N is a multiple of 72. Also, it is known that all the digits of N are ...
We know that the number is a multiple of 72. Therefore, the number must be a multiple of both 8 and 9.
If the number contains all the digits from 0 to 9, the sum of the digits will be 9*10/2 = 45.
Therefore, a number containing all the ten digits will be divisible by 9.
For a number to be divisible by 8, the last 3 digits should be divisible by 8.
Now, we have to find the largest possible number. Therefore, the left-most digit must be 9, the second digit from the left should be 8 and so on.
We'll get 9876543210 as the number. However, the last 3 digits are not divisible by 8. We must try to make the last 3 digits divisible by 8 without altering the position of other numbers. '120' is divisible by 8.
Therefore, the largest number with different digits divisible by 72 is 9876543120.
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N is a multiple of 72. Also, it is known that all the digits of N are ...
Question: What is the largest possible value of N if N is a multiple of 72 and all the digits of N are different?
Approach:
To find the largest possible value of N, we need to determine the arrangement of the digits such that N is divisible by 72.
Divisibility Rule of 72:
A number is divisible by 72 if it is divisible by both 8 and 9.
Divisibility Rule of 8:
A number is divisible by 8 if the last three digits of the number form a multiple of 8.
Divisibility Rule of 9:
A number is divisible by 9 if the sum of its digits is divisible by 9.
Finding the Largest Possible Value of N:
1. We start by considering the divisibility rule of 8. Since the last three digits of N must form a multiple of 8, we need to find a combination of digits that satisfies this condition.
2. We can start with the largest digit, i.e., 9, and check if the remaining digits can form a multiple of 8.
3. To form a multiple of 8, the last three digits of N can be 912, 928, 936, 952, 968, or 984.
4. Next, we consider the divisibility rule of 9. Since all the digits of N must be different, the sum of the digits must be a multiple of 9.
5. The sum of the digits from 1 to 9 is 45, which is divisible by 9. However, if we include 0 in the sum, the sum becomes 45 + 0 = 45, which is not divisible by 9. Therefore, we cannot include 0 in N.
6. We can arrange the remaining digits (1 to 9) in descending order to form the largest possible value of N.
Final Answer:
The largest possible value of N is 9876543120.