Ram gives a six-digit number 468312 to Shyam to check the divisibility. Shyam tells Ram that the number is divisible by 57. Shyam asks Ram, "If we rearrange the digits of this number in descending order, then by which number will it be always divisible?"
- A2
- B3
- C19
- D17
Solution & Step-by-step Explanation
The given six-digit number is 468312.
Let's find the sum of the digits of this number:
Sum of digits=4+6+8+3+1+2=24
Since the sum of the digits (24) is divisible by 3, the number itself is divisible by 3.
When we rearrange the digits of a number in any order (including descending order), the set of digits remains exactly the same. Thus, the sum of the digits will remain 24.
Since the sum of the digits remains divisible by 3, the rearranged number will always be divisible by 3. It may not remain divisible by 2 (if it ends in an odd digit), nor by 19 or 57 because divisibility by those numbers depends on the specific positional arrangement of the digits.
Let's find the sum of the digits of this number:
Sum of digits=4+6+8+3+1+2=24
Since the sum of the digits (24) is divisible by 3, the number itself is divisible by 3.
When we rearrange the digits of a number in any order (including descending order), the set of digits remains exactly the same. Thus, the sum of the digits will remain 24.
Since the sum of the digits remains divisible by 3, the rearranged number will always be divisible by 3. It may not remain divisible by 2 (if it ends in an odd digit), nor by 19 or 57 because divisibility by those numbers depends on the specific positional arrangement of the digits.