The sum of 3-digit numbers abc, bca and cab is always divisible by:
- A35
- B41
- C37
- D31
Solution & Step-by-step Explanation
A 3-digit number abc can be written as 100a+10b+c.
Similarly, bca=100b+10c+a and cab=100c+10a+b.
Sum of these numbers:
Sum=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)
Sum=111a+111b+111c=111(a+b+c)
We know that 111=3×37.
Therefore, the sum is always divisible by 37.
Similarly, bca=100b+10c+a and cab=100c+10a+b.
Sum of these numbers:
Sum=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)
Sum=111a+111b+111c=111(a+b+c)
We know that 111=3×37.
Therefore, the sum is always divisible by 37.